A triangle has angles that measure 60°, 70°, and z. What is z?

The expressions are factorized to;

1.(x + 1)(x + 2)

2.(x + 2) (x + 8)

3. (x -4) (x -6)

Using the factorization method of solving quadratic equations, we have;

1.  x² + 3x + 2

Find the pair factors of 2 that adds up to 3 and  substitute the values, we get

x² + 2x + x + 2

Factorize in pairs

x(x + 2) + 1( x+ 2)

(x + 1)(x + 2)

2. x² +10x+16

Find the pair factors

x² + 8x + 2x + 16

Factorize in pairs

x(x + 8) + 2( x+ 8)

(x + 2) (x + 8)

3. X²-10x+24

Find the pair factors

x² - 6x - 4x + 24

Factor in pairs

x(x -6) - 4(x -6)

(x -4) (x -6)

Hence, use the factorization  method in the expressions

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In a 3×4 matrix with reduced row echelon form given by U=[1,0,5,1;0,1,1,1;0,0,0,0], where a1=[−1,3,−2], and a2=[2,-3,-1]  is given and the value of a3 = [-5, -1, 1, 0], a4 = [-1, -1, 0, 1] .

Since the reduced row echelon form of A has only two pivot columns, the matrix A can have at most two linearly independent columns. We already know that a1 and a2 are linearly independent (neither is a scalar multiple of the other), so we can use them as a basis for the column space of A.

To find a3 and a4, we need to find two vectors that are linearly independent of a1 and a2 and that span the null space of A (i.e., are solutions to the homogeneous equation Ax=0).

To find a basis for the null space of A, we can set the non-pivot columns of A equal to free variables, and then solve for the corresponding pivot variables in terms of the free variables. In this case, the non-pivot columns are columns 3 and 4, so we set x3=t and x4=s for some scalars t and s, and solve for x1 and x2:

x1 = -5t - s

x2 = -t - s

Thus, the general solution to Ax=0 can be written as a linear combination of the vectors [-5, -1, 1, 0] and [-1, -1, 0, 1]. We can check that these two vectors are linearly independent, so they form a basis for the null space of A. We can use them as a3 and a4:

a3 = [-5, -1, 1, 0]

a4 = [-1, -1, 0, 1]

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____The given question is incorrect, the correct question is given below:

Let A be a 3×4 matrix with reduced row echelon form given by U=[1,0,5,1;0,1,1,1;0,0,0,0]. If a1=[−1,3,−2], and a2=[2,-3,-1] Find a3 and a4

Therefore, the probability that Bob will be hit by a car while crossing the point A is 1 [tex]-e^{-4t}[/tex] , where t is the time it takes him to cross the point A in seconds.

If the cars follow a Poisson process with rate 4 cars per minute, then the time between successive cars follows an exponential distribution with parameter λ = 4. Therefore, the probability density function of the time between cars is given by:

f(t) = λe^(-λt)

Since Bob takes t seconds to cross the point A, the probability that he will be hit by a car during this time is the probability that a car arrives within t seconds of his crossing, which is given by the cumulative distribution function:

F(t) = P(X ≤ t) = ∫[0, t] f(x)dx = ∫[0, t] 4e⁽⁻⁴ˣ⁾dx

Evaluating this integral, we get:

F(t) = [tex]-e^{-4t}[/tex]

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Referring to the given statements,

a) p → q: "If it is Friday, then I don't go to class."

b)  -q → -p:  "If I don't go to class, then it is not Friday."

c)  p ∧ q :  "It is both Friday and I don't go to class."

d)  p V -q: "Either it is Friday, or I don't go to class (or both)."

In this problem, the propositions are:

p : It is Friday

q :  I don't go to class

Hence,

a)  p → q is an implication and read as: "If it is Friday, then I don't go to class."

b). -q → -p

-q is negation of q and read as: I go to class

-p is negation of p and read as: It is not Friday

Therefore, -q → -p means "If I don't go to class, then it is not Friday."

c)  p ∧ q is a conjunction. p ∧ q can be read as: "It is both Friday and I don't go to class."

d)  p V -q  is a disjunction. p V -q can be read as: "Either it is Friday, or I don't go to class (or both)."

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The SAS congruence criteria, option (c), is the appropriate response. The acronym SAS stands for "side-angle-side," and it denotes that two sides and the included angle of one triangle must be congruent to two sides and the incorporated angle of another triangle for the triangles to be considered congruent.

"Right-angle-hypotenuse-side" (or option a) refers to a particular instance of the SAS criteria that only applies to right triangles.

The acronym ASA (option b) stands for "angle-side-angle," and it denotes that two angles and the included side of one triangle must be congruent with two angles and the included side of some other triangle for the triangles to be considered congruent.

Side-side-side, or SSS (option d), refers to the situation in which all three sides of one triangle are congruent with all three sides of another triangle.

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The question is -

Two triangles are congruent if two angles and the side included between them in one triangle are equal to the two angles and the side included between them in the other triangle. This is known as:

a. RHS congruence criterion

b. ASA congruence criterion

c. SAS congruence criterion

d. SSS congruence criterion

Answer: 1.75 yards

Step-by-step explanation:

You need to divide 7 by 4 to get 4 equal sections.

7/4 = 1.75

Therefore, 1.75 is your answer

Area of figure in the graph is 5 square unit.

Correct option is C.

Area is defined as the total space taken up by a flat surface or shape of an object. The space enclosed by the boundary of a plane figure is called its area. The area of a figure is the number of unit squares that cover the surface of a closed figure. Area is measured in square units like cm² and m².

Given, By the graph

Figure consists of one triangle and one rectangule

height of the triangle = 2 unit

length of base of triangle = 3 unit

Area of triangle

= (1/2) × height × base

= (1/2) × 2 × 3

= 3 unit²

Length of rectangle = 2 unit

Width of rectangle = 1 unit

Area of the rectangle

= length × width

= 2 × 1

= 2 unit²

Total area of the figure

= area of triangle + area of rectangle

= 3 + 2

= 5 unit²

Hence, 5 square unit is area of the figure in the graph.

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Answer: please look at the image for the answer

Step-by-step explanation:

The Mean number of major hurricanes gulf can expect in a season is 3.

What is probability?

A probability is a number that reflects the chance or likelihood that a particular event will occur. Probabilities can be expressed as proportions that range from 0 to 1, and they can also be expressed as percentages ranging from 0% to 100%.

What is mean?

The mean (aka the arithmetic mean, different from the geometric mean) of a dataset is the sum of all values divided by the total number of values. It's the most commonly used measure of central tendency and is often referred to as the “average.”

Given:

Gulf of Mexico averages 3 major Hurricanes

Since, It is stated that the Gulf experiences 3 major hurricanes on an average in a season. Since the average is the expected no. of observations the mean no. of major hurricanes Gulf expect is 3.

Hence, the Mean number of major hurricanes gulf can expect in a season is 3.

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Based on the graph, the slope of this line is equal to 2.

In Mathematics, the slope of any straight line can be determined by using this mathematical equation;

Slope (m) = (Change in y-axis, Δy)/(Change in x-axis, Δx)

Slope (m) = (y₂ - y₁)/(x₂ - x₁)

Based on the information provided in the graph, we can logically deduce the following data points on the line:

Substituting the given data points into the slope formula, we have the following;

Slope, m = (y₂ - y₁)/(x₂ - x₁)

Slope, m = (6 + 8)/(5 + 2)

Slope, m = 14/7.

Slope, m = 2.

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The well ordering property states that every non-empty set of positive integers contains a least element. This property can be used to show the existence and uniqueness of the greatest common divisor (GCD) of two positive integers.

a) To show that S is non-empty, we note that the element 0 is a member of S, since 0 = 0 + 0*b.

b) The well ordering property guarantees that S has a least element, which we denote as c.

c) Suppose d is a common divisor of a and b. We know that a = dq and b = dr for some integers q and r. Then c = a + bt = dq + drt = d(q + rt). Thus, d must be a divisor of c since it divides the terms on the right side of the equation.

d) Assume that c does not divide a. Then a = qc + r where 0 < r < c. This means r is also a member of S, contradicting the choice of c as the least element of S.

e) From (c) and (d) we can conclude that the GCD of a and b exists. To show that it is unique, we note that if a and b have two GCDs, d and e, then d and e must both divide c. Since c is the least possible value, this implies that d and e must actually be the same number, thus showing the GCD of a and b is unique.

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The capacitance is [tex]1.58*10^{-12} F[/tex], the charge is 12000 e, the charge density is [tex]4.80*10^{7} C/m^{2}[/tex], and the electric field is [tex]3.92*10^{6} N/C[/tex].

Here's how you can solve the above problems:

(a) To find the capacitance of a parallel-plate capacitor, we use the formula:

C = ε0A/d

where ε0 is the vacuum permittivity constant [tex](8.85*10^{-12}F/m )[/tex], A is the area of the plates, and d is the distance between the plates. Substituting the given values, we get:

C =[tex](8.85 * 10^{-12} F/m) (2.50* 10^{-4}m^{2} )/(1.40*10^{-7}m )[/tex]

= [tex]1.58*10^{-12} F[/tex]

(b) To find the amount of charge on the positive plate of the capacitor, we use the formula:

Q = CV

where Q is the charge, C is the capacitance, and V is the voltage across the capacitor. Substituting the values from (a) and the given voltage, we get:

Q = [tex](1.58*10^{-12}F )(3.00 V)[/tex]

= [tex]4.74*10^{-11} C[/tex]

= 12000 e

(c) To find the charge density on the positive plate, assuming the density is uniform, we divide the charge on the plate by the area of the plate:

ρ = Q/A = [tex](12000e)/(2.50*10^{-4}m^{2} )[/tex]

= [tex]4.80*10^{7} C/m^{2}[/tex]

(d) To find the magnitude of the electric field between the plates, we use the formula:

E = V/d

where E is the electric field, V is the voltage across the capacitor, and d is the distance between the plates. Substituting the given values, we get:

E = [tex](3.00 V)/(1.40*10^{-7}m )[/tex]

= [tex]2.14*10^{-7} V/m[/tex]

= [tex]3.92*10^{6} N/C[/tex]

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Answer:

2 pieces of sc and 1 piece of rc

Step-by-step explanation:

Answer:

Below

Step-by-step explanation:

You COULD calculate the lengths of the three sides and then use Heron's Formula to calculate the area.....

 Here is perhaps an easier way (see image)

  calculate the entire area of the rectangle then subtract the 3 smaller RIGHT triangles from the total

Area of rectangle =   6 x 7 = 42 units^2

  triangles    1/2 ( 3 x 7    +    3 x 3    +    4 x 6 ) = 27 units^2

42 - 27 = 15 units^2   = area of given triangle

Answer:

15 square units

Step-by-step explanation:

i am using Heron's formula for this question

Firstly, lets calculate the lengths of the triangle. We can do this by using the formula:

[tex]length=\sqrt{(y_{2}\\-y_{1}\\)^{2}+(x_{2}\\-x_{1}\\)^{2} }[/tex]

Calculating the length between D (3,-9) and E (9,-5):

[tex]length(A)=\sqrt{(-9-(-5))^{2}+(3-9)^{2} } \\length(A)=2\sqrt{13}[/tex]

Calculating the length between F (6,-2) and E (9,-5):

[tex]length(B)=\sqrt{(-2-(-5))^{2}+(6-9)^{2} }\\length(B)=3\sqrt{2}[/tex]

Calculating the length between D (3,-9) and F (6,-2):

[tex]length(C)=\sqrt{(-9-(-2))^{2}+(3-6)^{2} }\\length(C)=\sqrt{58}[/tex]

Now, we can find the semi-perimeter with the formula:

[tex]s=\frac{A+B+C}{2}[/tex]

where A, B and C are the lengths

The semi-perimeter is:

[tex]s=\frac{2\sqrt{13}+3\sqrt{2}+\sqrt{58} }{2}\\s=9.534758... (store \, the \, full \, value \, in \, your \, calculator)[/tex]

The formula for the area of the triangle:

[tex]A=\sqrt{s(s-A)(s-B)(s-C)}\\A=\sqrt{9.534758...(9.534758...-2\sqrt{3})(9.534758...-3\sqrt{2})(9.534758...-\sqrt{58} ) }\\A=15[/tex]

To determine if a linear transformation is onto, we need to check if every vector in the range has at least one pre-image in the domain.

Let T be a linear transformation from R^3 to R^2 given by:

T (x, y, z) = (2x - y, x + 3y - 4z)

(a) To check if T is one-to-one, we need to check if for any two distinct vectors (x1, y1, z1) and (x2, y2, z2) in R^3, we have T (x1, y1, z1) ≠ T (x2, y2, z2). Equivalently, we can check if T (x1, y1, z1) - T (x2, y2, z2) ≠ 0 for any two distinct vectors (x1, y1, z1) and (x2, y2, z2) in R^3.

So, let (x1, y1, z1) and (x2, y2, z2) be two distinct vectors in R^3. Then:

T (x1, y1, z1) - T (x2, y2, z2) = (2x1 - y1, x1 + 3y1 - 4z1) - (2x2 - y2, x2 + 3y2 - 4z2)

= (2x1 - y1 - 2x2 + y2, x1 + 3y1 - 4z1 - x2 - 3y2 + 4z2)

= (2(x1 - x2) - (y1 - y2), x1 - x2 + 3(y1 - y2) - 4(z1 - z2))

For T to be one-to-one, the only solution to T (x1, y1, z1) - T (x2, y2, z2) = 0 must be (x1, y1, z1) = (x2, y2, z2). So, we need to solve the equation above and check if the only solution is (x1, y1, z1) = (x2, y2, z2).

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The volume of a paraboloid with height h and base radius r can be found by treating it as a solid of revolution, rotating the parabolic cross-section about the y-axis.

The equation for the parabola is y = x^2 / (4h), so the equation for the cross-section is x = 2sqrt(hy). The volume can be found by integrating the area of the cross-section from y = 0 to y = h, and then multiplying by 2pi:

V_parab = 2pi * ∫[0,h] (2sqrt(hy))^2 dy, = 2pi * ∫[0,h] 4hy dy= 2pi * 2hy^2]_[0,h] = 4/3 * pi * h^3

The volume of a cylinder with the same height and base radius r is V_cyl = pi * r^2 * h. The ratio of the volume of the paraboloid to the volume of the cylinder is: V_parab / V_cyl = (4/3 * pi * h^3) / (pi * r^2 * h)

= (4/3 * h^2 * r)

For the second part, the volume of a sphere with radius r can be found by treating it as a solid of revolution, rotating a semicircle about its diameter.

The equation for a semicircle with radius r is y = sqrt(r^2 - x^2), so the equation for the cross-section is x = sqrt(r^2 - y^2). The volume can be found by integrating the area of the cross-section from y = -r to y = r, and then multiplying by 2pi: V_sphere = 2pi * ∫[-r,r] (sqrt(r^2 - y^2))^2 dy

= 2pi * ∫[-r,r] (r^2 - y^2) dy = 4/3 * pi * r^3

The volume of a cylinder with the same radius and height (which is equal to the diameter of the sphere) is V_cyl = pi * r^2 * 2r = 2pi * r^3. The ratio of the volume of the sphere to the volume of the cylinder is: V_sphere / V_cyl = (4/3 * pi * r^3) / (2pi * r^3) = 2/3

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ab is diagonalizable and can be written as a product of invertible matrices.

Since a and b are both diagonalizable, there exist invertible matrices u and v such that [tex]uau^{-1} and vbv^{-1}[/tex] are both diagonal matrices. Thus, we can rewrite ab as [tex]uau^{-1}vbv^{-1}[/tex]. Since matrix multiplication is associative, the order of multiplication does not matter and thus [tex]ab = u(au^{-1}vb)v^{-1}[/tex]. Since the [tex]au^{-1}vb[/tex] is also diagonalizable, it can be written as c, where c is a diagonal matrix. Thus, [tex]ab = ucv^{-1}[/tex], which is also a diagonal matrix. Therefore, ab is diagonalizable and can be written as a product of invertible matrices.

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Complete questiom:

Suppose that A and B are diagonalizable matrices of the same size, i.e.\ there exist invertible matrices U and V such that [tex]UAU^{-1} and VBV^{-1}[/tex]are diagonal. Show that AB = UDEV, where UDEV is also a diagonal matrix.

7 minutes will it be until there are 120 grams of the element remaining.

To determine the quantity or percentage of something in terms of 100, use the percentage formula. Per cent simply means one in a hundred. With the percentage formula, a number between 0 and 1 can be expressed. A number that is expressed as a fraction of 100 is what it is. It is mostly used to compare and determine ratios and is represented by the symbol %.

We have,

An element with a mass of 820 grams decays by 24% per minute. To the nearest minute.

Now, 100 - 24 = 76% of the element remains after each minute.

So, 820 x (0.76[tex])^t[/tex] = 120 where t is number of minutes.

(0.76[tex])^t[/tex]  = 0.146

Taking log both side we get

t log 0.76 = log 0.146

t = 7.00724

t = 7 minutes

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The mean of the Experimental Group is -3.14 and the standard deviation is 132.

We can use this data to calculate the range of pain for most people in this group.

Using the formula for standard deviation, we can find the lower and upper bounds of the range. The lower bound is the mean minus one standard deviation, or[tex]-3.14 - 132 = -135.14[/tex]. The upper bound is the mean plus one standard deviation, or [tex]-3.14 + 132 = 128.86[/tex].

When converted to the 1 - 10 pain scale, the lower bound is 1.82 and the upper bound is 4.46. Therefore, we can conclude that the range of pain for most people in the Experimental Group is 1.82 - 4.46.

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Answer: (0,0)

Step-by-step explanation: 3+5=8-4=4-4=0

Answer:

∠C = 35°

∠D = 76°

∠DEC = 69°

Step-by-step explanation:

    Exterior angle equals the sum of opposite interior angles (remote angles).

               ∠C + ∠D = ∠DEF

      3y + 5 + 7y + 6 =  111°

      3y + 7y + 5 + 6 = 111

Combine like terms,

              10y +  11 = 111

Subtract 11 from both sides,

                      10y = 111 - 11

                     10y  = 100

Divide both sides by 10

                       y = 100 ÷ 10

                       [tex]\boxed{y = 10}[/tex]

∠C = 3y + 5

      = 3*10 + 5

      = 30 + 5

[tex]\boxed{\bf \angle C = 35^\circ}[/tex]

∠D = 7y + 5

      = 7*10 + 6

     = 70 + 6

 [tex]\boxed{\bf \angle D = 76^\circ}[/tex]

∠C + ∠D + ∠DEC = 180°      {Angle sum property of triangle}

  35 + 76 + ∠DEC = 180

            111 + ∠DEC = 180

                    ∠DEC = 180 - 111

                   [tex]\boxed{\bf \angle DEC = 69^\circ}[/tex]

Answer:

t ≤ 8

Step-by-step explanation:

Equation Total

1 hour 32 + 9.4 41.4

2 hours 32 + 2(9.4) 50.8

3 hours 32 + 3(9.4.) 60.2

4 hours 32 + 4(9.4.) 69.6

5 hours 32 + 5(9.4.) 79.0

6 hours 32 + 6(9.4.) 88.4

7 hours 32 + 7(9.4.) 97.8

8 hours 32 + 8(9.4.) 107.2

So we know the maximum amount of time is 8 hours. So now let's express this as an inequality:

32 + 9.4t ≤ 107.2 Now let's solve for t

9.4t ≤ 75.2

t ≤ 8

Answer: 2 hrs

Step-by-step explanation:

4/5 + 4/5 = 1 3/5 + 2/5 (half of 4/5) = 2

         ↑

(4/5 of a room)

Let X be the number of different faces that appear (so X = 1, 2, or 3). The probability mass function of X is px(1) = 1/36, px(2) = 15/36, px(3) = 60/216 = 5/18.

To find the probability mass function px(k), where k is the number of different faces that appear, we can use the combinatorial approach:

When k = 1, there is only one possible combination of outcomes: all three tosses must have the same face, and there are 6 ways to choose which face it is. Therefore, the probability of k = 1 is:

px(1) = (6/6) * (1/6) * (1/6) = 1/36

When k = 2, there are three possible combinations of outcomes: the first two tosses can have the same face and the third a different face, the first and third tosses can have the same face and the second a different face, or the second and third tosses can have the same face and the first a different face.

There are 6 ways to choose which face appears twice and 5 ways to choose which face appears once. Therefore, the probability of k = 2 is:

px(2) = 3 * [(6/6) * (1/6) * (5/6)] = 15/36

When k = 3, there are 3! = 6 possible combinations of outcomes: each of the three tosses can have a different face. There are 6 ways to choose the face for the first toss, 5 ways to choose the face for the second toss, and 4 ways to choose the face for the third toss.

Therefore, the probability of k = 3 is:

px(3) = 6 * (6/6) * (5/6) * (4/6) = 60/216

Note that the probabilities add up to 1:

px(1) + px(2) + px(3) = 1/36 + 15/36 + 60/216 = 1

Therefore, the probability mass function of X is:

px(1) = 1/36

px(2) = 15/36

px(3) = 60/216 = 5/18

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The formula for a curve grade is based on a normal distribution with the average score being 0.

The formula for each grade is as follows:

A-grade: S + 20

B-grade: S + 10S< + 20

C-grade: S - 10S< + 20

D-grade: S - 20

F-grade: S < 0

The formula  is based on a normal distribution with the average score being 0.

For example, if the average score is 50 and your score is 55, then you would receive an A-grade because your score is 5 points above the average. This can be calculated by adding 20 points to your score of 55 resulting in a grade of[tex]75 (55 + 20 = 75)[/tex]. Similarly, if your score is 40, then you would receive a C-grade because your score is 10 points below the average. This can be calculated by subtracting 10 points from your score of 40 resulting in a grade of 30 (40 - 10 = 30).

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All possible fourth corners if  the following three points are corners of a parallelogram:(1,2,3),(3,4,7),(2,1,2) is (2, 3, 5) and  (3, 5/2, 9/2)

Let the corners be A(1,2,3);B(3,4,7);C(2,1,2)

Let the fourth vertex D =(x,y,z)

We know that the diagonals of a parallelogram bisect each other. So, the midpoint of AC is same as the midpoint of BD.

Mid point of two points (x 1,y 1,z1) and (x 2,y 2,z2)​  is calculated by the formula ( x 1+x 2/2 , y 1+y 2/2, z1+z2/2 ).

So, midpoint of AC= Mid point of BD

⇒( 1+3/2, 2+4/2, 3+7/2) =(2, 3, 5)

Hence, D=(2, 3, 5).

Same if,

Mid point of two points (x 1,y 1,z1) and (x 2,y 2,z2)​  is calculated by the formula ( x 1+x 2/2 , y 1+y 2/2, z1+z2/2 ).

So, midpoint of AC= Mid point of BD

⇒( 3+2/2, 4+1/2, 7+2/2) = (3, 5/2, 9/2)

Hence, D=  (3, 5/2, 9/2)

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The rate at which the water level is rising when the water level is 3 cm,

⇒ 0.65 cm / sec

Mathematical expression is defined as the collection of the numbers variables and functions by using operations like addition, subtraction, multiplication, and division.

Let us assume it is a regular pyramid. The (altitude) height "h" of apex from base square, slanting height and the side "a" of the base square rise uniformly and proportionally when we move from the apex point to the base.

Now, If height (altitude) h becomes x times, then side "a" of the base square becomes x times, the slanting height also becomes x times.  Thus, the area b of the base square rises x² times.

Δh / h = Δa / a  

Δb / b  = 2 Δa / a = 2 Δh / h    

Or,     h₁/h₂ = a₁/a₂

Since, We know that;

⇒ V = 1/3 b h = 1/3 a² h

⇒ d V / dt = 1/3 a² dh/dt  + 1/3 h a da/dt

                 = 1/3 a² dh /dt + 2/3 h a  (a/h) (dh/dt)

                 = 1/3 a² dh / dt + 2/3 a² dh/dt

Hence, We get;

dV/dt  = a² dh/dt        

dh/dt = (1/a²) dV/dt

We have;

h₀ = 5 cm and a₀ = 3 cm    

At the base;

When h = 5cm,  

Hence,  h / h₀ = a/a₀ ;

⇒ a = a₀ h / h₀ = 5×5/3 = 8.3 cm

And, dh/dt = 1/8.3² × 45 = 45/68.89 cm/sec = 0.65 cm / sec

Thus, The rate at which the water level is rising when the water level is 3 cm,

⇒ 0.65 cm / sec

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All the sample tests are stated below :

1) A two-independent-sample t test.

2) A one-sample t test.

3) A related-samples t test.

4)  A related-samples t test.

It is a subgroup of people with traits from a wider population. When population sizes are too big for the test to include all potential participants or observations, samples are utilized in statistical testing. A sample must be representative of the entire population and must not be biased toward any one trait.

A professor runs an experiment to see if front-row students perform better on an exam than back-row pupils.

This is a two-independent-sample t test, as there are two distinct groups (students in the front row and students in the back row) and the professor is interested in comparing the mean exam scores between them.

To determine whether students spend more than 30 minutes on an activity on average, a graduate student chooses a sample of 25 participants.

This is a one-sample t test, as there is only one group of participants and the graduate student is interested in comparing the mean time students attend to a task to a specific value (30 minutes).

To determine whether right-handed siblings exhibit higher emotional intelligence than left-handed siblings, a researcher pairs up right- and left-handed siblings.

This is a related-samples t test using a matched-pairs design, as the researcher is interested in comparing emotional intelligence scores of right-handed and left-handed siblings who are matched based on certain criteria (in this case, being siblings).

A local school's principal is curious to discover how much an honors class benefits the children there. He administers tests to honors English students before the start of the school year and again at the conclusion of the year to gauge their progress.

This is a related-samples t test using a repeated-measures design, as the principal is interested in comparing the test scores of the same group of students before and after taking the honors English class to measure how much they learned during the year.

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The required measures of the angles ∠PQS and ∠SQR are given as 51 and 49.

Complementary angles are two angles whose measures add up to 90 degrees. In other words, if angle A and angle B are complementary, then A + B = 90 degrees.

Here,
∠PQR is a right angle,
So ∠PQS and ∠SQR are supplementary angles, from the figure.

∠PQR = ∠PQS + ∠SQR
90 = 8x + 131 + 10x + 139
90 = 18x + 270
18x = -180
x = -10

Now,
∠PQS = 8x + 131 = -80 + 131 = 51°
Similarly,
∠SQR = 49°

Thus, the required measure of the angles ∠PQS and ∠SQR are given as 51 and 49.

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