Write an exponential function for a graph that passes through the points (2,16) and (3,32).

Answer:

125

Step-by-step explanation:

y = 1k/x^3  I am trying to find the constant k.  I will use the information that I have been given to find k

1 = k/5^3

1 = k/125  Multiply both sides by 125

125 = k

Now, we will find y when x = 1

y = 125/1^3

y = 125

The value of y is 125 when x is 1.

Write the function y as a function of x, i.e., y = f(x), and then solve for x as a function of y, to determine the inverse of a function.

y = 1k/x^3  I am trying to find the constant k.  I will use the information that I have been given to find k

1 = k/5^3

1 = k/125

Multiply both sides by 125

125 = k

Now, we will find y when x = 1

y = 125/1^3

y = 125

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Answer:

V≈3053.63

A≈1017.88

Step-by-step explanation:

V=[tex]\frac{4}{3}[/tex]π [tex]r^{3}[/tex]=[tex]\frac{4}{3}[/tex]·π·[tex]9^{3}[/tex] ≈3053.62806

A=4π[tex]r^{2}[/tex]=4·π·[tex]9^{2}[/tex] ≈1017.87602

Answer:

I'm pretty sure it is √10, 3.12, 3.24, 31

Step-by-step explanation:

I don't know if you messed up but I got a question similar that is the length 31 in your quesion is actually 3 1/4 in my question. but i did the math and 31 is still the greatest number.

Shipping points are independent organizational entities within which processing and monitoring of the deliveries, as well as goods issue, is carried out. A delivery is processed by one shipping point only.

"FOB shipping point" or "FOB origin" means the buyer is at risk once the seller ships the product. The purchaser pays the shipping cost from the factory and is responsible if the goods are damaged while in transit. "FOB destination" means the seller retains the risk of loss until the goods reach the buyer.

Please check the attached file for a complete answer.

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Answer: C

Step-by-step explanation:

[tex]\frac{5\pi}{4}[/tex] is in the third quadrant, so both x and y are negative.

Therefore, the only possible answer is C.

(i) 57students liked at least one game.

(ii) 7 students who liked both the games.

(iii) 27 students liked football.

(iv) 37 students liked cricket.

(v) Below we represented the result in a Venn diagram.​

What is Venn diagram?

John Venn popularized the Venn diagram in the 1880s, which is a type of diagram that displays the logical relationship between sets. The illustrations are used in probability, logic, statistics, linguistics, and computer science to demonstrate basic set relationships and to teach rudimentary set theory.

Part (i)- We have to get number of people who like at least one game, there are total 75 students and only 18 did not like any of two games, so students who like at least one game are-

75-18 = 57

Part (ii)- There are 57 students who like at least one game, from the question statement we know that 20 students like football only and 30 liked cricket only, so student who like both games are-

57-20-30 = 7

Part (iii)- There are 7 students who liked both games and 20 who liked football only, so student who liked football are-

20+7 = 27

Part (iv)- There are 7 students who liked both games and 30 who liked cricket only, so student who liked cricket are-

30+7=37

Part (v)- Venn Diagram will look like-

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Complete Question- In a group of 75 students, 20 liked football only, 30 liked cricket only and 18 did not like any of two games? (i) How many of them liked at least one game? (ii) Find the number of students who liked both the games. (iii) How many of them liked football? (iv) How many of them liked cricket? (v) Represent the result in a Venn diagram.​

The necessary solution that would be used to fill in the blanks would be

The way that Quintin would be able to do this would be the center of the circumferential triangle is equidistant from the vertices.

The center is going to be the intersection of the bisectors of the three sides of the given triangle.

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perpendicular bisectors, is the center of, circumscribed

Answer:

In the image

Hope this helps!

Step-by-step explanation:

1. perpendicular

2. is the center of

3. circumscribed

Answer:

144=12², so 144^3/2=(12²)^(3/2)=12³=1728.

this answer is not from the internet

go check yourselves

mark me brainliest

El precio resultante con descuento de 12 % es igual a 342.32 pesos.

En este problema tenemos que determinar el precio resultante, el cual es igual al precio inicial menos el descuento. A continuación, se presenta la siguiente expresión:

x = 389 · (1 - 12/100)

x = 389 - 46.68

x = 342.32

El precio resultante con descuento de 12 % es igual a 342.32 pesos.

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The volume of the cuboid is 132 cubic inches.

Given length of cuboid 7 1/3 inches, breadth being 6 2/3 inches , height being 2 7/10 inches.

Volume is amount of liquid a container can hold in its capacity. Cuboid is a 3 dimensional figure having length , breadth and height.Volume of cuboid is the product of length, breadth and height of cuboid.

We have to calculate volume of cuboid.

We have to make a simple fraction showing length , breadth and height.

Length=(3*7+1)/3=22/3

Breadth=(3*6+2)/3=20/3

Height=(10*2+7)/10=27/10

Volume=22/3*20/3*27/10

=11880/90

=132 cubic inches.

Hence the volume of cuboid having length of 22/3 inches, breadth of 20/3 inches and height of 27/10 is 132 cubic inches.

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Question is incomplete as it should include figure.

The synthetic division's representation of the dividend is 2x3 + 10x2 + x + 5.

Given that

An L shape is created when two lines intersect vertically and horizontally.

The shape has entries in two rows.

Entries in row 1 are 2, 10, 1, and 5.

Blank, -10, and 0 are the entries in row 2.

A simplified method of dividing a polynomial with another polynomial equation of degree one is known as synthetic division.

On the exterior, to the left of the form, is entry number 5.

The entry stands for the divisor's zero.

If the variable is x, then this entry to the variable is;

2x³+10x²+x+5

The dividend is thus represented by synthetic division as

2x³+10x²+x+5

The Question is incomplete And complete question is given below!!

What dividend is represented by the synthetic division below? A vertical line and horizontal line combine to make a L shape. There are two rows of entries within the shape. Row 1 has entries 2, 10, 1, 5. Row 2 has entries blank, negative 10, 0, negative 5. Entry negative 5 is on the outside to the left of the shape, and a third row of entries is outside and below the shape. Row 3 has entries 2, 0, 1, 0. Negative 10 x squared minus 5 2 x cubed 10 x squared x 5 2 x squared 1 2 x cubed x.

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Answer:

The answer on Edge is A= 1

Step-by-step explanation:

Answer:

The solution is 25, 36, 49

Step-by-step explanation:

a = 1

b = 3

c = 5-3

= 2

________________________________

Un = [tex]\sf a + (n-1) \times b + \frac{(n-1)(n-2)}{2} \times c[/tex]

Un = [tex]\sf 1 + (n-1) \times 3 + \frac{(n-1)(n-2)}{2} \times 2[/tex]

Un = [tex]\sf 1 + 3n-3 + (n-1)(n-2)[/tex]

Un = [tex]\sf 1 + 3n-3 + n^{2} - 2n - n + 2[/tex]

Un = [tex]\sf n^{2} + 3n - 2n - n + 2 + 1 - 3[/tex]

Un = [tex]\sf n^{2} + 0n + 0[/tex]

Un = [tex]\sf n^2[/tex]

________________________________

Un = n²

U5 = 5²

= 25

U6 = 6²

= 36

U7 = 7²

= 49

________________________________

So, the next three members are 25, 36, 49

If x - 4 is a factor then x = 4 is a root:

f(4) = 4^3 - 2 * 4^2 + 5 * 4 + 1

= 64 - 32 + 20 + 1

= 53

since f(4) is not zero then 4 is not a root of the f(x)

Also, x - 4 is not a factor

In algebra, the remainder theorem, or Bezout's small theorem, applies the principle of polynomial division. It states that the remainder of dividing the polynomial f (x) by the linear polynomial {\ display style x-r} is equal to {\ display style f (r). }

In algebra, a factor set is a set of factors and roots of. Polynomial. This is a special case of the remainder theorem. The factor theorem shows that the polynomial f (x) has a factor only if f = 0.

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The quadrants of the points that needs to be graphed  are:

Since the points that was given are:

A(2, 4)        B(0, -3)  

C(1, -1)       E(-4,1)    

G(-3,-2)      I(0, 2)

D(3, 3)        F(2,0)  

H(-2, 3)     J(-1,-4)

Then, we plot all the coordinate above on a graph (see image attached)

Hence, from the image attached graph, the categories that can be seen are:

Quadrant 1 is  Points A and D

Quadrant 2 is  Points E and H

Quadrant 3 is Points G and J

Quadrant 4 is Points C

x-axis-  Point F

y-axis- Point B and I

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Answer: Add 5 to the product of 3 and 6, then subtract 4

Step-by-step explanation:

     We will use the order of operations. Looking at 5 + (3 x 6) − 4, we would:

[1] Multiply 3 x 6

[2] Add 5 to that product

[3] Subtract 4 from that value

     This means the answer to your question is:

Add 5 to the product of 3 and 6, then subtract 4

Answer: B

i DiD tHe TeSt

The function that is the solution of the differential y" + y = sin(x) is: y(x) = -1/2(x cos x)

A function is an expression, rule, or law in mathematics that describes a connection between one factor (the independent variable) and another variable (the dependent variable).

Take a look at the the following differential equation:

y" + y = sin (x)

The auxiiliary equation is

m² + 1 = 0

m² + 1-1 = -1

m² + 0 = -1

m² = -1

m = ±√-1

m = ±i

So, the complimentary function is yₐ (x) = c₁ cos x + c₂ sin x

Let the particular integral be:

yₙ (x) = A cos x + B sin x

yₙ '(x) = - A sin x + B cos x

yₙ ''(x) = - A cos x + B cos x

yₙ ''(x) = - (A cos x + B cos x)

After we have substituted yₙ (x); and yₙ''(x) in the given differential equation

y'' + y = sin (x)

= - (A cos x + B cos x) + (A cos x + B cos x)  = sin(x)

0 = sin (x)

If we take the particular integral to be:
yₙ (x) =  x(A cos x + B cos x)

yₙ '(x) = x(-A sin x + B cos x) + A cos x + B cos x

yₙ ''(x) = x(-A cos x - B sin x) - A sin x + B cos x + -A sin x + B cos x

Substitute yₙ (x), yₙ''(x) into the stated differential equation

y'' + y = sin (x)

x (-Acosx - Bsin x) - Asinx + Bcosx + (-Asinx + Bcosx) - x (Acosx + Bsin x) = sin (x)

-Axcosx - Bxsin x - Asinx + Bcosx -Asinx + Bcosx - Axcosx + Bxsin x = sin (x)

-2Asinx + 2Bcosx = sin(x)

Compare the coefficients of like terms on both sides of the equation

-2A = 1, B = 0

A = -1/2, B = 0

Substitute A = -1/2, B =0 into the assumed solution.

yₙ(x) = x((-1/2)cosx + (0) sinx)

= -1/2xcosx +0

= -(1/2)xcosx

Now, the general solution for the given differential equation is:

y(x) = yₓ(x) +yₙ (x)

y (x) - c₁cosx + c₂sin x -1/2x cosx

Hence, the solution is:

y(x) = -1/2xcosx

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Full Question:

Which of the following functions are solutions of the differential equation y'' + y = sin x? (Select all that apply.)

A) y = 1 2 x sin x

B) y = cos x

C) y = x sin x − 5x cos x

D) y = − 1 /2 x cos x

E) y = sin x

Surface area of the rectangular solid = 416 in.².

Volume = 480 in.³.

Surface area = 2(wl+hl+hw)

Volume = (length)(width)(height).

Given the following:

Length (l) = 12 in.

Width (w) = 10 in.

Height (h) = 4 in.

Surface area = 2(wl+hl+hw) = =2·(10·12+4·12+4·10) = 416 in.².

Volume = (12)(10)(4) = 480 in.³.

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The given radii of 4 and 6 and the sector angle of 112°, gives;

[tex] C.\: \frac{56 \cdot \pi}{9} [/tex]

The possible figure that relates to the question can be described as follows;

Two concentric circles having radii of AE = 4, and AB = 6, respectively.

Sector of the circles formed by lines AC and AB, with angle of sector = 112°

The shaded portion is described by arcs FE and CB

Therefore;

Area of the sectors are found as follows;

[tex]AFE = \frac{112 ^\circ}{360 ^\circ} \times \pi \times {4}^{2} = \frac{224 \cdot \pi}{45} [/tex]

[tex]ACB = \frac{112 ^\circ}{360 ^\circ} \times \pi \times {6}^{2} = \frac{56 \cdot \pi}{5} [/tex]

Area of the shaded portion, A is therefore;

[tex] A= \frac{56 \cdot \pi}{5} - \frac{224 \cdot \pi}{45} =\frac{56 \cdot \pi}{9} [/tex]

[tex] Shaded \: area, \: A= \frac{56 \cdot \pi}{9} [/tex]

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So lets try to prove it,

So let's consider the function f(x) = x^2.

Since f(x) is a polynomial, then it is continuous on the interval (- infinity, + infinity).

Using the Intermediate Value Theorem,

it would be enough to show that at some point a f(x) is less than 2 and at some point b f(x) is greater than 2. For example, let a = 0 and b = 3.

Therefore, f(0) = 0, which is less than 2, and f(3) = 9, which is greater than 2. Applying IVT to f(x) = x^2 on the interval [0,3}.

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Answer:

f(1) = 81

Step-by-step explanation:

f(n + 1) = 1/3 f(n)

⇔ f(n) = 3 × f(n + 1)

……………………………

if f(3) = 9  ⇒  f(2) = 3 × f(3) = 3 × 9 = 27

Then

f(1) = 3 × f(2) = 3 × 27 = 81

The value of y from the given expression is 5

Given:

BD = 16

BC = 2y - 2

CD = y + 3

Since AC is a perpendicular bisector of BD:

BC  =  CD

2y - 2 = y + 3

2y - y = 3 + 2

y  =  5

Hence, the value of y from the given expression is 5

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The solution to the system of equations is given as follows:

x = 1 and x = 3.

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

In this problem, the two equations are:

[tex]\frac{1}{x - 2} = x - 2[/tex]

Applying cross multiplication:

(x - 2)(x - 2) = 1

x² - 4x + 4 = 1

x² - 4x + 3 = 0

(x - 1)(x - 3) = 0

Hence the solutions are:

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Check the picture below, so the parabola looks more or less like so, with the vertex half-way from the focus point and the directrix.

Now, the distance from the directrix to the focus point is only 1 unit, so half that, or namely the "p" distance is 1/2 unit, and since the parabola is opening downwards, "p" is negative, so

[tex]\textit{vertical parabola vertex form with focus point distance} \\\\ 4p(y- k)=(x- h)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h,k+p)}\qquad \stackrel{directrix}{y=k-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\cap}\qquad \stackrel{"p"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\begin{cases} h=0\\ k=-\frac{1}{2}\\[1em] p=-\frac{1}{2} \end{cases}\implies 4\left( -\frac{1}{2} \right)\left( ~~y-\left( -\frac{1}{2} \right) ~~ \right)~~ = ~~(x-0)^2 \\\\\\ -2\left( y+\frac{1}{2} \right)=x^2\implies y+\cfrac{1}{2} =\cfrac{x^2}{-2}\implies y = -\cfrac{1}{2}x^2-\cfrac{1}{2}[/tex]

now, we could also write that as either

[tex]y = -\cfrac{1}{2}(x^2+1)\qquad or\qquad y = \cfrac{1}{2}(-x^2-1)[/tex]

The given triangle is a right triangle

The coordinates are given as:

D (0,n), E(m,n), F(m,0)

The length is calculated using:

[tex]l=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2[/tex]

So, we have:

[tex]DE=\sqrt{(0-m)^2 + (n-n)^2} = m[/tex]

[tex]DF=\sqrt{(0-m)^2 + (n-0)^2} = \sqrt{m^2 + n^2[/tex]

[tex]EF=\sqrt{(m-m)^2 + (n-0)^2} = n[/tex]

The slope is calculated using:

[tex]m = \frac{y_2 -y_1}{x_2 - x_1}[/tex]

So, we have:

[tex]DE = \frac{n -n}{0- m} = 0[/tex]

[tex]DF = \frac{n -0}{0- m} = -\frac nm[/tex]

[tex]EF = \frac{n -0}{m- m} = \mathbf{unde fined}[/tex]

This is calculated using

[tex]m = 0.5 * (x_1 + x_2, y_1 + y_2)[/tex]

So, we have:

[tex]DE = 0.5 * (0 + m, n+ n)=(0.5m, n)[/tex]

[tex]DF = 0.5 * (0 + m, n+ 0)=(0.5m, 0.5n)[/tex]

[tex]EF = 0.5 * (m + m, n+ 0)=(m, 0.5n)[/tex]

In (a), we have:

Lengths

[tex]DE= m[/tex]

[tex]DF = \sqrt{m^2 + n^2[/tex]

[tex]EF= n[/tex]

Slope

[tex]DE = 0[/tex]

[tex]DF = -\frac nm[/tex]

[tex]EF = \mathbf{unde fined}[/tex]

The sides are not equal.

However, the 0 and the undefined slope implies that the triangle is a right triangle because the sides are perpendicular

It should be noted that the triangle cannot be graphed because the coordinates are not numeric

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Answer: No

Step-by-step explanation:

There is only one pair of congruent angles that can be determined, but to prove triangles similar, there needs to be two pairs of congruent angles.

Answer:

42.1

Step-by-step explanation:

The formula to calculate distance is √[(x₂ - x₁)² + (y₂ - y₁)²]. Using (22,27) as our x1 and y1, and (2,-10) as our x2 and y2.

Therefore our formula is √[(2 - 22)² + (-10 - 27)²].

(2 - 22)² = 400

(-10 - 27)² = 1369

(Make sure for both of these you put the negative number in parenthesis and the exponent outside them, if you are using a calculator of some sort)

Then we add and square root

√[400 + 1369] = 42.0594816896

The distance between the points (22,27) and (2,-10) is 42.1 units

The coordinates of the points are (22,27) and (2,-10)

The distance is calculated as:

[tex]d = \sqrt{(x_2 -x_1)^2 + (y_2 -y_1)^2[/tex]

So, we have:

[tex]d = \sqrt{(22 -2)^2 + (27 +10)^2[/tex]

Evaluate

[tex]d = \sqrt{1769[/tex]

Take the square root

d = 42.1

Hence, the distance between the points (22,27) and (2,-10) is 42.1 units

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  [a] x = 175 - 7.5p

For a, we are isolating the x variable to one side of the equation.

        0.4x + 3p = 70

        0.4x = - 3p + 70

        x =  - 7.5p + 175

  [b] x = 115

For b, we are solving for x when p = 8.

        x =  - 7.5p + 175

        x =  - 7.5(8) + 175

        x = 115

The relationship that can be modeled by the function shown in the table is y = 4·√x

f(x) is defined as a product of 4 into √x

4 × √1 = 4

4 × √4 = 8

4 × √9 = 12

4 × √16 = 16

4 × √25= 20

Therefore, the relationship is defined as y is 4 times the square root of x that is, y = f(x) = 4√x

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By applying the Theorem of Intersecting Secant to circle Q, an equation which is correct about the measure of ∠MNP is: A. m∠MNP = One-half(x – y).

The Theorem of Intersecting Secant states that when two (2) lines intersect outside a circle, the measure of the angle formed by these lines is equal to one-half (½) of the difference of the two (2) arcs it intercepts.

For this exercise, the following points should be noted:

By applying the Theorem of Intersecting Secant to circle Q shown in the image attached below, we can infer and logically deduce that angle MNP will be given by this formula:

m∠MNP = One-half(x – y).

m∠MNP = ½(x – y)

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