What is the surface area of this square pyramid? Please help

Answer:

137

Step-by-step explanation:

VYXW is a parallelogram.

The opposite angles of a parallelogram are congruent, meaning their angle measurements are equal.

Following this information we can write an equation to find the value of x:

10x - 27 = 2x + 29 transfer like terms to the same side of the equation

10x - 2x = 29 + 27 add/subtract

8x = 56 divide both sides by 8

x = 7 replace x with 7 to find the given angle measures

now, the consecutive angles in a parallelogram are supplementary meaning their sum is equal to 180

m<V + m<W = 180

m<V + 2*7 + 29 = 180

m<V + 43 = 180 subtract 43 from both sides

m<V = 137

Answer:

x=40

Step-by-step explanation:

first you simplify both sides of the equation, then you isolate the variable

40/4= 10

10 times 3/2 = 15

Answer: -0.6

Step-by-step explanation:

The number of prints that the probability of producing at least one quality print is 22.

Given that the probability of producing a high-quality color print is 0.10.

The binomial distribution summarizes the number of trials or observations when each trial has an equal chance of reaching a certain value. The binomial distribution determines the probability of observing a particular number of positive results in a defined number of tests.

Probability of high-quality color print (P) = 0.1

Probability of producing not high-quality color print (1-P) = 0.9

Assume the no. of prints to produce 'n' and these are produced independently say 'x' be the no. of producing quality prints, which follows a binomial distribution.

The probability function of binomial distribution is

[tex]P(X=x)\left(\begin{array}{l}n\\ x\end{array}\right)P^x(1-P)^{n-x},x=0,1,2,.....,n[/tex]

Given that

P(X≥1)≥0.90

1-P(X=0)≥0.90

1-(1-P)ⁿ≥0.90

1-(0.90)ⁿ≥0.90

(0.90)ⁿ≥0.10

Now, taking log of both sides, we get

nlog(0.90)≥log(0.10)

n(0.1053)≥2.3025

n≥21.86

n≈22

Hence, the probability of producing a high-quality color print is 0.10 and the number of prints to produce so that the probability of producing at least one quality print is larger than 0.90 is 22.

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Answer:

[tex]5\frac{6}{11}[/tex]

Step-by-step explanation:

Number of Values = 11.

Mean = Sum of values / Number of Values

= [tex]\frac{3+2+4+6+5+6+1+7+9+8+10}{11} \\=\frac{61}{11} \\= 5\frac{6}{11}[/tex]

Graph is in the attached image.

Based on the percentage of people who actually voted and the margin of error, the data implies that people lied when responding to the survey.

The evidence that people lied about voting would bet be shown by the confidence interval. If the confidence interval of the survey does not have the actual percentage of people who voted in it, then people lied.

The confidence interval is:

= Average ± Margin of error

Solving:

= 70% + 3%

= 73%

= 70% - 3%

= 67%

Confidence interval is:

67% - 73%

The actual percentage of 61% who voted is not in this interval which means that some people lied about voting in the survey.

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The slope of the parallel line is 0 and the slope of the perpendicular line is undefined

The equation is given as:

y = 1

A linear equation is represented as:

y = mx + c

Where:

m represents the slope

By comparison, we have:

m = 0

The slope of the parallel line is:

Slope = m

This gives

Slope = 0

The slope of the perpendicular line is:

Slope =-1/m

This gives

Slope = -1/0

Slope = undefined

Hence, the slope of the parallel line is 0 and the slope of the perpendicular line is undefined

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The last three digits of the final number are 109.

A number is an arithmetic value that is used to calculate and represent a quantity.

The digit zero will no longer be utilized in any counting numbers, according to the Leader of Zip, since it stands for nothing. There can only be counting numbers with 1 digit.

From 1 to 11,

11 is the tenth counting number.

So for the case of first 1000 numbers,

Number of permutations =1000/10 =100

Since 101 to 109 contains a zero.

So, 100+9=109

So, 1000+109=1109

Thus, the last number is 1109, and its last three digits are 1,0,9.

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Answer:

The ¨mandatory dues¨ asked by the unions are not much (practically none), since it´s a voluntary system implemented, normally all the members pay a ¨agency fee¨ but not more than that, this changes depending on the personal involvement with the union.

a natural person can refuse the membership at any moment, without issues, and this way avoid any kind of activity with the union, including contributing with money (dues) or duties.

However, when this happens, the person it´s restricted to certain activities like participate in union elections, general or important meetings, and the contribution in the bargain for the collective elections is banned as well, these as main aspects to highlight.

If one doesn't pay they union dues, they will not be able to participate in collective bargaining with owners.

Only the real representatives of the union could participate in collective bargaining with owner (in which they could negotiate the new term of the salary with the owner). Without paying their dues, they're still haven't been acknowledged as the member of union.



Hope this helps have a great day!

For the given tree diagram of the experiment consisting of two trials, P(D) = 0.74

A method that can be infinitely repeated and has a clearly defined range of potential outcomes, or sample space, is referred to as an experiment or trial in probability theory. If there are multiple possible outcomes from an experiment, it is considered to be random; if there is just one, it is said to be deterministic.

Calculation of P(D) for the given experiment:

From the first branch of the tree, P(D) = 0.7 × 0.6

P(D) = 0.42

From the second branch of the tree, P(D) = 0.8 × 0.4

P(D) = 0.32

Hence, the probability of getting D in the experiment is,

P(D) = P(D) from first branch of the tree + P(D) from the second branch of the tree

P(D) = 0.42 + 0.32

P(D) = 0.74

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Question 6

1) [tex]\overline{AB} \cong \overline{BD}[/tex], [tex]\overline{CD} \perp \overline{BD}[/tex], O is the midpoint of [tex]\overline{BD}[/tex], [tex]\overline{AB} \cong \overline{CD}[/tex] (given)

2) [tex]\angle ABO, \angle ODC[/tex] are right angles (perpendicular lines form right angles)

3) [tex]\triangle ABO, \triangle CDO[/tex] are right triangles (a triangle with a right angle is a right triangle)

4) [tex]\overline{BO} \cong \overline{OD}[/tex] (a midpoint splits a segment into two congruent parts)

5) [tex]\triangle ABO \cong \triangle CDO[/tex] (LL)

Question 7

1) [tex]\angle ADC, \angle BDC[/tex] are right angles), [tex]\overline{AD} \cong \overline{BD}[/tex]

2) [tex]\overline{CD} \cong \overline{CD}[/tex] (reflexive property)

3) [tex]\triangle CDA, \triangle CDB[/tex] are right triangles (a triangle with a right angle is a right triangle)

4) [tex]\triangle ADC \cong \triangle BDC[/tex] (LL)

5) [tex]\overline{AC} \cong \overline{BC}[/tex] (CPCTC)

Question 8

1) [tex]\overline{CD} \perp \overline{AB}[/tex], point D bisects [tex]\overline{AB}[/tex] (given)

2) [tex]\angle CDA, \angle CDB[/tex] are right angles (perpendicular lines form right angles)

3) [tex]\triangle CDA, \triangle CDB[/tex] are right triangles (a triangle with a right angle is a right triangle)

4) [tex]\overline{AD} \cong \overline{DB}[/tex] (definition of a bisector)

5) [tex]\overline{CD} \cong \overline{CD}[/tex] (reflexive property)

6)  [tex]\triangle ADC \cong \triangle BDC[/tex] (LL)

7) [tex]\angle ACD \cong \angle BCD[/tex] (CPCTC)

Answer:

A) Domain: x∈R or in interval notation (- ∞, ∞)

Range: x∈R or in interval notation (- ∞, ∞)

Thus, they both the domain & the range of the first function are the same.

B) Domain: x∈R or in interval notation (- ∞, ∞)

Range: [-5,∞)

Using it's concepts, the domain and the range of the functions are given as follows:

a) Both all real numbers.

b) Domain all real numbers, range [tex]y \geq 5[/tex].

In this problem, neither function has any impediment, such as an even root or denominator, hence the domain is all real numbers. As for the range, a linear function assumes all real values, while a concave up quadratic function assumes values of the vertex, in this case y = 5, and greater, hence the range of the second function is [tex]y \geq 5[/tex].

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The slope of the parallel line is 0 and the slope of the perpendicular line is undefined

The equation is given as:

y = 1

A linear equation is represented as:

y = mx + c

Where:

m represents the slope

By comparison, we have:

m = 0

The slope of the parallel line is:

Slope = m

This gives

Slope = 0

The slope of the perpendicular line is:

Slope =-1/m

This gives

Slope = -1/0

Slope = undefined

Hence, the slope of the parallel line is 0 and the slope of the perpendicular line is undefined

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Answer:

A and B

Step-by-step explanation:

x < 100 means x is less than 100. Therefore, pick any x-values that are less than 100 which are x = 54 and x = 89.

Hence, A and B are the answer.

Hugh bough 4 books.

What is linear equation ?

The cost of magazines =

3.95(3) + 8.95b = 47.65

11.85 + 8.95b = 47.65

Subtract 11.85 from both sides.

8.95b = 35.8

Divide 8.95 from both sides.

b = 4

Therefore, Hugh bough 4 books.

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The complete question is -

Hugh bought some magazines that cost $3.95 each and some books that cost $8.95 each. He spent a total of $47.65. If Hugh bought 3 magazines, how many books did he buy? The equation that models the problem is 3.95m + 8.95b = 47.65, where m is the number of magazines and b is the number of books.

The area of the triangle abc in which bd is altitude and having coordinates a(-6,0),b(0,0),c(0,6),d(3,-3) is 18 square units.

Given the coordinates of abcd are a (-6,0) b (0,0) c (o,6) and d (3,-3).

We are required to find the area of the triangle when bd is the altitude of the triangle.

When we draw the triangle we will find that that the base is ac and altitude is given bd.

ac=[tex]\sqrt{(6-0)^{2} +(0+6)^{2} }[/tex]

=[tex]\sqrt{36+36}[/tex]

=[tex]\sqrt{72}[/tex]

=6[tex]\sqrt{2}[/tex]

bd=[tex]\sqrt{(-3-0)^{2} +(3-0)^{2} }[/tex]

=[tex]\sqrt{9+9}[/tex]

=[tex]\sqrt{18}[/tex]

=3[tex]\sqrt{2}[/tex]

Area=(Base*Height)/2

=(6[tex]\sqrt{2}[/tex]*3[tex]\sqrt{2}[/tex])/2

=(18*2)/2

=18 square units.

Hence the area of the triangle abc in which bd is altitude and having coordinates a(-6,0),b(0,0),c(0,6),d(3,-3) is 18 square units.

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Answer:

Step-by-step explanation:

hello :

the cannonball hit the ground when h=0

-5t² + 40t=0 means : -5t(t- 8)=0

t=8

The measure of each angle formed by the place where the median meets the vertex is 30 degrees

The median of an equilateral triangle divides the side and the angle into equal parts.

If the angle at the vertex is, the new angle would be

Angle = x/2

The angle at the vertex of an equilateral triangle is 60.

So, we have:

Angle = 60/2

Evaluate

Angle = 30

Hence, the measure of each angle formed by the place where the median meets the vertex is 30 degrees

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Answer:

b. x^2+4x

Step-by-step explanation:

So since you don't know the width, you can just represent it as a variable, and in this case that variable will be "x". Now using this variable we can use it to represent the length since there is a relationship between the width and length which is given. Since the length is 4 feet greater than the width, we can represent the length as (x+4). Now to find the area, we simply multiply the length * width, which is x * (x+4) = x^2 + 4x

When the balloon, rising at 10 ft/sec, rises to 100 feet, the rate of change of the angle of elevation is 0.05 rad/sec

The rate at which the balloon rises = 10 ft/sec

Horizontal distance of the balloon from the observer, d = 100 feet

Therefore;

[tex] \frac{dy}{dt} = 10[/tex]

The angle of elevation can be presented as follows;

[tex] tan(\theta) = \frac{y}{100} [/tex]

[tex] y = 100 \times tan(\theta) [/tex]

Therefore;

[tex] \frac{dy}{dt} = \mathbf{ 100\cdot sec^2 \frac{d \theta}{dt}} [/tex]

Which gives;

[tex] \frac{d \theta}{dt}= \frac{ \frac{d y}{dt}}{100} \times cos^2 \theta [/tex]

[tex] \mathbf{ \frac{d \theta}{dt}} = \frac{ 10}{100} \times cos^2 \theta [/tex]

When the balloon is 100 ft. from the ground, we have;

Therefore;

[tex] \frac{d \theta}{dt}= \frac{ 10}{100} \times \left(\frac{1}{\sqrt{2}} \right)^2 [/tex]

[tex] \frac{ 10}{100} \times \left(\frac{1}{\sqrt{2}} \right)^2 = \frac{ 1}{20} [/tex]

Which gives;

[tex] \frac{d \theta}{dt}=\frac{ 1}{20} = 0.05 [/tex]

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Step-by-step explanation:

[tex]12 {x}^{2} - 6( {a}^{2} + {b}^{2} )x + 3 {a}^{2} {b}^{2} = 0[/tex]

Using quadratic formula, we get

[tex]x = 6( {a}^{2} + {b}^{2} )± \frac{ \sqrt{( - 6) {}^{2}( {a}^{2} + {b}^{2}) {}^{2} - 4(12)(3 {a}^{2} {b}^{2} )} }{24} [/tex]

[tex]x = 6( {a}^{2} + {b}^{2} )± \frac{ \sqrt{36( {a}^{2} + {b}^{2} ) {}^{2} - 144 {a}^{2} {b}^{2} } }{24} [/tex]

[tex]x = 6( {a}^{2} + {b}^{2} )± \frac{ \sqrt{36( ({a}^{2} + {b}^{2}) {}^{2} - 4 {a}^{2} {b}^{2} }) }{24} [/tex]

[tex]x = 6( {a}^{2} + {b}^{2} )± \frac{6 \sqrt{ ({a}^{2} + {b}^{2} ) {}^{2} - 4 {a}^{2} {b}^{2} } }{24} [/tex]

[tex]x = 6( {a}^{2} + {b}^{2} )± \frac{ \sqrt{( {a}^{2} + {b}^{2} ) {}^{2} - 4 {a}^{2} {b}^{2} } }{4} [/tex]

Answer: x₁=6b²,  x₂=6a².

Step-by-step explanation:

[tex]12x^2-6*(a^2+b^2)+3a^2b^2=0\\D=(6*(a^2+b^2))^2-4*12*3a^2b^2\\D=6^2*(a^2+b^2)^2-144a^2b^2\\D=36*(a^4+2a^2b^2+b^4)-144a^2b^2\\D=36a^4+72a^2b^2+b^4-144a^2b^2\\D=36a^4-72a^2b^2+36b^4\\D=(6a^2)^2-2*6a^2*6b^2+(6b^2)^2\\D=(6a^2-6b^2)^2\\\sqrt{D}=\sqrt{(6a^2-6b^2)^2}=|6a^2-6b^2|=б(6a^2-6b^2).\\\displaystyle x_{1,2}=\frac{-(-6*(a^2+b^2))б(6a^2-6b^2)}{2} .\\x_1=\frac{6a^2+6b^2-6a^2+6b^2}{2} \\x_1=\frac{12b^2}{2} \\x_1=6b^2.\\x_2=\frac{6a^2+6b^2+6a^2-6b^2}{2} \\x_2=\frac{12a^2}{2} \\x_2=6a^2.[/tex]

The total bill plus interest on the credit card is $3,250

What is credit card?

Credit card gives its holder the opportunity to make purchases on credit such that amount spent and the interest that accrues thereon would be paid by the credit card holder at a future time.

In this case, the function shows the relationship between the amount owed and the interest so as to give the outstanding balance.

f(p) = p(1.30)

p=amount owed=$2,500

f(p)=$2,500*(1.30)

f(p)=$3,250

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Answer:

X=10

Step-by-step explanation:

You can subtract 180 by 110 to get 70. Then you can do 8x+30+70=180
Add 30+70 to get 100
Subtract 100 to 180 to get 80.
8x=80
You divide 8 from both sides which leaves you with x=10.

Answer:

x=10

Step-by-step explanation:

See attached image

The permutations is computed below.

1. Evaluate 9P4

9P4 = 9!/(9 - 4)

= 9!/5!

= (9 × 8 × 7 × 6)

= 3024

2. How many ways can 5 students sit on a row of 15 chairs for a photograph?

= 15!(15 - 5)

= 15!/10!

= (15 × 14 × 13 × 12 × 11)

= 360360

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This exercise involves the construction of a blueprint. To do so, we need to solve for angles using the Cos Rule. See the step by steps procedure below.

A blueprint is a masterplan that highlights how a structure will be look like when it is completed.

From the above, see attached the first simple quadrilateral QRST (Draft I) that is indicates the above information.

Next Step

We have to derive ∠T, ∠ S, and ∠R.
In order to achieve the above, we must split the quadrilateral into two triangles.

As noted in the image, the quadrilateral is bisected from R to T.

∠R and ∠T are also divided into R1 R2 and T1, T2 respectively.

Next step - Derive ∠T1 and ∠R1 .

To do the above,  we must deploy the Pythagoras Theorem. (PT)

PT requires:

TR² = QT² + QR²

TR² = (210)² + (515)²

TR² = 44,100 + 265,225

TR² =  309,325

TR = √309,325

TR = 556.17

Next step, Derive T1

We can derive T1 using the cos rule:

Cos T1 = QR/QT

= 515/556.17

Cos T1 = 0.926

Therefore

T1 = Cos⁻¹(0.926)

T1  = 22.18°

Next we derive R1

R1 = 180° - (90° + 22.18°)
R1 = 67.82°

In ΔTRS, by using Cosine Rule,

Cos T2 = [TS² + QR² - RS²]/2(TS) * (TR)

Cos T2 = [(232)² + (556.16)² - (542)²]/ 2 * (232) x (556.16)

Cost T2 = 69,373.945/258,058

Cost T2 ≈ 0.269

Therefore,

T2 = Cos⁻¹(0.269)

T2 ≈ 74.4°

Applying the same process, we can derive R2

Cos R2 = TR² + RS² - TS²

Cos R2 = [(556.16)² + (542)² - (232)²] / 2 (556.16) * (542)

Cost R2 = 0.911

R2 = Cos ⁻1 (0.911)

R2 = 24.34°

From the above, we can execute the following iterations:

∠S = 180° - (R2 + T2)

= 180 ° - (24.34 + 74.4)
∠C = 81.26°

Likewise:

∠R = R1 + R2
= 67.82 + 24.34
∠B = 92.16°

Lastly:

∠T = T1 + T2

= 22.18 + 74.40

∠T = 96.48°

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Full Question:

A client has just inherited a property. The deed is quite old, but she knows where the front two corners of the property are and that the distance between them is 210 feet. The shape is an irregular quadrilateral and the other three sides have a distance of 515 feet, 232 feet, and 542 feet in a clockwise direction from the front left corner. More importantly, it is known that the front left corner of the property is a right angle.

-Construct a blueprint of the property labeling the measures of all sides and angle of the property.

The surface area of the composite figure is 444m².

Given a composite figure which is shown in the question.

The  area of ​​a shape (in square units) is the number of unit squares required to cover the entire area without gaps or overlaps. If a hologram has planes, those faces are called faces.

Firstly, we will find the area of front and end triangles by using the formula

Area=(1/2)×b×h and we will multiply this area by 2 because we are finding the area of two same triangles.

Here, h=6 and b=16 and substitute these values in the formula, we get

A₁=2×(1/2)×16×6

A₁=2×8×6

A₁=96m²

Now, we will find the area of the left and right side rectangles which joined both the triangles.

We will find the area by using the formula Area=l×b and we will multiply this area by 2 because we are finding the area of two same rectangles.

here, l=10 and b=5 and substitute these values in the formula, we get

A₂=2×10×5

A₂=100m²

Further, we will find the area of the front and end side rectangles that joined both by the base of the triangles.

We will find the area by using the formula Area=l×b and we will multiply this area by 2 because we are finding the area of two same rectangles.

here, l=16 and b=4 and substitute these values in the formula, we get

A₃=2×16×4

A₃=128m²

Furthermore,  we will find the area of the left and right side rectangles which joined by the front and end rectangles.

We will find the area by using the formula Area=l×b and we will multiply this area by 2 because we are finding the area of two same rectangles.

here, l=4 and b=5 and substitute these values in the formula, we get

A₄=2×4×5

A₄=40m²

Now, we will find the area of the base of the composite figure which is rectangle.

We will find the area by using the formula Area=l×b.

here, l=16 and b=5 and substitute these values in the formula, we get

A₅=16×5

A₅=80m²

So, the surface area of the given composite figure will be

Surface area=A₁+A₂+A₃+A₄+A₅

Surface area=96m²+100m²+128m²+40m²+80m²

Surface area=444m²

Hence, the surface area of the given composite figure is 444m².

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The country is 3250 miles in distance from east to west, and 3400 miles from north to south.

Given Information

It is given that,

The distance of the country on map from east to west, l = 65 inches

The distance of the country on map from north to south, b = 68 inches

As per the given scaling of map, 1 inch = 50 miles

How large is the actual country?

Thus, the actual distance from east to west, L = l × 50 miles

L = 65 × 50 miles

L = 3250 miles

The actual distance from north to south, B = b × 50 miles

B = 68 × 50 miles

B = 3400 miles

Therefore, the actual size of the country is 3250 miles in distance from east to west, and 3400 miles in distance from north to south.

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