NSPECTIONS Mrs. Sanchez is inspecting a shed to determine if it is safe to use for storing football equipment. Mrs. Sanchez mapped the top view of the ceiling walls of the shed on a coordinate plane. If one of the walls lies from (–6, 11.5) to (2, 9.5) and the second wall lies from (–1, –2.5) to (2, 9.5), are the walls perpendicular? Explain your reasoning.

From the properties of a right triangle is found that the missing length x= [tex]5\sqrt{3}[/tex].

A triangle is classified as a right triangle when it presents one of your angles equal to 90º.  The greatest side of a right triangle is called the hypotenuse. And, the other two sides are called legs.

The math tools applied for finding angles or sides in a right triangle are the trigonometric ratios or the Pythagorean Theorem.

The Pythagorean Theorem says: (hypotenuse)²= (leg1)²+(leg2)² . And the main trigonometric ratios are: sin (x) , cos  (x) and tan  (x) .

The sum of the internal angles of a right triangle also is equal to 180°.

The question gives a figure which is compound for two right triangles.

For each triangle is informed two angles, then:

       45+90+ x = 180

       135 + x =180

        x=180-135

        x=45°

       60+90+ y = 180

       150+ y =180

        y=180-150

        y=30°

Thus, the angles for the given triangle are:

From the trigonometric ratio sinФ, you can find the hypotenuse the triangle 2. See below.

[tex]sin \theta =\frac{opposite\ leg }{hypotenuse} \\ \\ sin 45 =\frac{hypotenuse \ triangle 2 }{hypotenuse \ triangle 1} \\ \\ \frac{\sqrt{2} }{2}=\frac{hypotenuse \ triangle 2 }{10\sqrt{2} }\\ \\ 2*hypotenuse \ triangle 2 = 10*2\\ \\ hypotenuse \ triangle 2 = \frac{10*2}{2} \\ \\ hypotenuse \ triangle 2 =10[/tex]

If the hypotenuse of triangle 2 is 10, you can apply again the trigonometric ratio sinФ for finding x

[tex]sin \theta =\frac{opposite\ leg }{hypotenuse} \\ \\ sin 60=\frac{x }{hypotenuse \ triangle 2} \\ \\ \frac{\sqrt{3} }{2}=\frac{x }{10}\\ \\2x=10 \sqrt3\\ \\ x=5\sqrt3[/tex]

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This relationship is not a function because more than one amount of plant food remains each day.

An expression, rule, or law that defines a relationship between one variable (the independent variable) and another variable (the dependent variable).

Given that, Ryan did an experiment by observing the effects of plant food and the growth of the sprouting bean plant. He made a note of the height of the bean plant and the amount of food remaining in the container for the plant after feeding each day.

Thus, he observed that the plant food in the container decreases by an equal amount every day as the bean plant grew. Thus, this states that it is not a  function between the two, as there are more than one plant food remaining each day for the plant.

Hence, This relationship is not a function because more than one amount of plant food remains each day.

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Answer:

32 books

Step-by-step explanation:

Well if each have 8 classes and 2 books per class then to get the total of books you'd multiply 8x2 and you get 16 for one alone. Since its two people sharing a single locker than add 16 with another 16 and you get 32!

sorry I can't physically show the work but writing down a 8x2 on the side and then adding 16 to that answer should show it off! :]

Answer:

[tex]9b + 10u > 1400[/tex]

Using simple mathematical operations, the missing values are as follows:

(B) Dijon Mustard: 18g

(C) Barbeque sauce: 70g

(D) Worcestershire sauce: 10.35g

(E) Kosher salt: 1.5g

(F) Ground black pepper: 2.3g

(G) Crushed red pepper: 1.15g

So, we can calculate missing values as follows:

(B) Dijon Mustard: 18g

(C) Barbeque sauce: 70g

(D) Worcestershire sauce: 10.35g

(E) Kosher salt: 1.5g

(F) Ground black pepper: 2.3g

(G) Crushed red pepper: 1.15g

Therefore, using simple mathematical operations, the missing values are as follows:

(B) Dijon Mustard: 18g

(C) Barbeque sauce: 70g

(D) Worcestershire sauce: 10.35g

(E) Kosher salt: 1.5g

(F) Ground black pepper: 2.3g

(G) Crushed red pepper: 1.15g

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The complement of set A is { 3, 6, 13, 14, 17, 19}

The complement of a set is the set that includes all the elements of the universal set that are not present in the given set. Let's say A is a set of all coins which is a subset of a universal set that contains all coins and notes, so the complement of set A is a set of notes (which do not includes coins).

U = {2,3,6,9,10,13,14,17,19}, A =  {2,9,10}

set A complement are all the elements in the Universal set U not contained in set A. Those elements are in set A' =  { 3, 6, 13, 14, 17, 19}

In conclusion the complement of set A is  { 3, 6, 13, 14, 17, 19}

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The a < 6 or a>-9 compound inequality is different than the rest as it involves one interval instead of a reunion of two intervals.

A phrase having two inequality statements connected by the words "or" or "and" is referred to as a compound inequality. The conjunction "and" denotes the simultaneous truth of both truths in the compound phrase. It is when the solution sets for the various assertions to cross over or intersect.

When the word and is used to connect two inequalities, the compound inequality is solved when both inequalities are true at the same time. It is the intersection or overlap of each inequality's solutions.

1)(−∞,−3)∪(4,∞)

2)(−∞,−2)∪(8,∞)

3)(−∞,−5)∪(7,∞)

4)(−∞,6)∪(−9,∞)=(−9,6)

The fourth compound inequality is different than the rest as it involves one interval instead of a reunion of two intervals.

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The length of AC is 17.3 cm

acc to our question,

Hence,The length of AC is 17.3 cm

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In linear equation, 27/2 pancakes are in the box .

What in mathematics is a linear equation?

Number of b pancake mix  to be need  =  1½ cups

Number of box has pancake mix   = 9 cups

Therefore, in order to get the total number of  batches of pancakes  needed for the batches will be calculated thus

             =  1½  * 9

             = 3/2 * 9

             = 27/2

therefore 27/2 pancakes are in the box .

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Answer:

3*3*3 or 3^3

Step-by-step explanation:

27/3=9/3=3

so 3*3*3 or 3^3

Answer:

-40

Step-by-step explanation:

The Concept here is Complex Numbers

where [tex]\sqrt{-1}=i \ and \ i^{2} = -1[/tex]

Then [tex]\sqrt{-4}=\sqrt{4i^{2} } =\sqrt{(2i)^{2} } =2i[/tex]

Also [tex]\sqrt{-25}=\sqrt{25i^{2} } =\sqrt{(5i)^{2} } =5i[/tex]

Then [tex](\sqrt{-4} )(4\sqrt{-25} )=(2i)(4*5i)=40i^{2} =40*-1=-40[/tex]

Glad to Help:)

Answer: a) = 4

b) = 30

Step-by-step explanation:

a) They are asking for the range of the marks so you do 10-6 as 10 is the highest mark and 6 is the lowest

b) You find this out by adding the frequency up 10+4+7+4+5 = 30

The linear inequality representing the number of apps and games that Kumiko can buy is [tex]1.25a+2.5g\leq50[/tex].

Expressions using linear inequalities compare any two values using inequality symbols like "," ">," "," or "."

The linear inequality representing the number of apps and games that Kumiko can buy is [tex]1.25a+2.5g\leq50[/tex].

Expressions using linear inequalities compare any two values using inequality symbols like "," ">," "," or "." These values might be either numerical, algebraic, or both.

The number of apps Kumiko buy is a and the number of games she can buy is g.

The gift card has 50$.

Total amount of apps = 1.25a

Total amount of games = 2.5g

So, the linear equality will be [tex]1.25a+2.5g\leq50[/tex].

She can can only games and apps for amount less than and equal to $50.

Therefore, The linear inequality representing the number of apps and games that Kumiko can buy is [tex]1.25a+2.5g\leq50[/tex].

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The given question is Incomplete. Here is the complete question:

Kumiko has a $50 gift card for a Web site that sells apps and games. Games cost $2.50 each, and apps cost $1.25 each. Write an inequality that represents the number of apps a and the number of games g that Kumiko can buy and describe any constraints.

Answer: z=-2

Step-by-step explanation:

3z+6z-30=-48

9z-30=-48

9z-30=-48+30

9z=-18

z=-2

Answer:

see explanation

Step-by-step explanation:

there is a common difference between consecutive terms in the sequence, that is

8 - 13 = 3 - 8 = - 2 - 3 = - 7 - (- 2) = - 5

this indicates the sequence is arithmetic with nth term

[tex]a_{n}[/tex] = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

here a₁ = 13 and d = - 5 , then

[tex]a_{n}[/tex] = 13 - 5(n - 1) = 13 - 5n + 5 = 18 - 5n

then

a₁₀₀ = 18 - 5(100) = 18 - 500 = - 482

The area of the circle on the paper is 4.5 cm when traced around the edge of the quarter on a sheet of paper.

The area of the circle is equal to the product of the square of the radius of the circle and pi.

A = πr²

where 'r' is the radius of the circle

The diameter of a quarter is 2.4 cm. You trace around the edge of the quarter on a sheet of paper.

The area of the circle on the paper = πr²

Here r = (diameter of a quarter)/2 = 2.4/2 = 1.2 cm

The area of the circle on the paper = π(1.2)²

The area of the circle on the paper = π × 1.44

The area of the circle on the paper = 4.5238

Round the answer to the nearest tenth.

The area of the circle on the paper = 4.5 cm

Thus, the area of the circle on the paper is 4.5 cm

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Answer:its 4.5

Step-by-step explanation:

The equation of the line that passes through the points  (9,2) and (3,4) is x + 3y = 15  and the graph  is shown below .

In the question ,

it is given that

the required line passes through the points (9,2) and (3,4)

in order to find the equation of  line we first need to find slope .

so , the slope(m) will be

m = (4-2)/(3-9)

m = 2/(-6)

m = -1/3

The equation of the line passing through the point (3,4) having slope -1/3 is

y - 4 = (-1/3)(x - 3)

3(y - 4) = -1(x - 3)

3y - 12 = -x + 3

x + 3y = 12 + 3

x + 3y = 15

the graph of the line is drawn below .

Therefore ,The equation of the line that passes through the points  (9,2) and (3,4) is x + 3y = 15  .

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7. The composite functions are given as follows:

a. f(g(x)) = [tex]\frac{3x}{-x^2 + 3x - 1}[/tex].

b. g(f(x)) = [tex]\frac{3x(x - 1)}{9x^2 + x - 1}[/tex]

c. (f(g(3)) = -9.

8. The inverse function is given as follows:

[tex]f^{-1}(x) = \frac{4 + x}{3 - x}[/tex]

The composite functions are obtained using the inner function as the input to the outer function.

In this problem, the functions are defined as follows:

The composite of f and g is:

[tex]f(g(x)) = f\left(\frac{3x}{x^2 + 1}\right) = \frac{\frac{3x}{x^2+1}}{\frac{3x}{x^2+1} - 1} = \frac{\frac{3x}{x^2+1}}{\frac{-x^2 + 3x - 1}{x^2 + 1}} = \frac{3x}{-x^2 + 3x - 1}[/tex]

At x = 3, the numeric value of the composite function is given as follows:

f(g(3)) = 9/(-9 + 9 - 1) = -9.

The composite function of g and f is given as follows:

[tex]g(f(x)) = g\left(\frac{x}{x - 1}\right) = \frac{\frac{3x}{x - 1}}{\left(\frac{3x}{x - 1}\right)^2 + 1} = \frac{3x}{x - 1} \times \frac{(x - 1)^2}{9x^2 + x - 1} = \frac{3x(x - 1)}{9x^2 + x - 1}[/tex]

The function is given as follows:

y = (3x + 4)/(x - 1)

The find the inverse, the variables x and y are exchanged, and then y is isolated, as follows:

x = (3y + 4)(y - 1)

xy - x = 3y + 4

3y - xy = 4 + x

y(3 - x) = 4 + x

y = (4 + x)/(3 - x)

[tex]f^{-1}(x) = \frac{4 + x}{3 - x}[/tex]

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The absolute maximum value is f(0,±2) = 4 and the absolute minimum value is f(1,0) = -1.

The given function is  F(x,y) = [tex]x^{2}[/tex] + [tex]y^{2}[/tex] - [tex]2x[/tex]   → 1

D is the closed triangular region with the vertices (0,0), (0,2) and (0,-2).

Differentiate 1 partially with respect to x:

df/dx =  [tex]x^{2}[/tex] + [tex]y^{2}[/tex] - [tex]2x[/tex]

[tex]f^{'}[/tex](x) = 2x - 2

Differentiate 1 partially with respect to y:

df/dy = [tex]x^{2}[/tex] + [tex]y^{2}[/tex] - [tex]2x[/tex]

[tex]f^{'}[/tex](y)  = 2y

Consider [tex]f^{'}[/tex](x) = 0 and [tex]f^{'}[/tex](y) = 0 to find the critical points.

[tex]f^{'}[/tex](x) ⇒ 2x -2 =0

x = 1

[tex]f^{'}[/tex](y) ⇒ 2y

y = 0

Thus, the critical point is (x,y) = (1,0).

Calculate the values of f(x,y) at (0,0), (0,2), (0,-2) and (1,0).

At (x,y) = (0,0)

f(0,0) ⇒ 0 + 0 - 0 = 0

At (x,y) = (0,2)

f(0,2) ⇒ 0 + 4 - 0 = 4

At (x,y) = (0,-2)

f(0,-2) ⇒ 0 + 4 - 0 = 4

At (x,y) = (1,0)

f(1,0) ⇒ 1 + 0 - 2 = -1

Therefore, the absolute maximum is f(0,±2) = 4 and the absolute minimum is f(1,0) = -1.

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Answer:

[tex](x + 4)(2x + 5) - (x + 3)(x + 2) = 79 \\ 2x {}^{2} + 5x + 8x + 20 - {x}^{2} - 2x - 3x - 6 = 79 \\ {x}^{2} + 8x + 14 = 79 \\ {x}^{2} + 8x - 65 = 0 \\ {x}^{2} + 13x - 5x - 65 = 0 \\ x(x + 13) - 5( x + 13) = 0 \\ (x - 5)(x + 13) = 0[/tex]

x-5=0

x=5

the length of smaller rectangle= 8 m

and breadth of s. rectangle= 7 m

l x b = 8 x 7

[tex]56 {m}^{2} [/tex]

hopefully this will u

The area of the largest rectangle is 25 square units.

It is given that the radius of the semicircle is [tex]4[/tex].

It is required to inscribe a rectangle of maximum area.

Consider a rectangle of sides [tex]x[/tex], [tex]y[/tex] as shown in the given figure.

From the attached figure,[tex]AB=x[/tex], [tex]OB=r[/tex] and [tex]OB=\frac{y}{2}[/tex].

Now, in the triangle OAB, apply Pythagoras' theorem as,

[tex](OA)^{2}=(OB)^{2}+(AB)^2[/tex]

[tex]r^2=(\frac{y}{2} )^2+x^2\\\\[/tex]

[tex]16=\frac{y}{4}+x^2[/tex]

[tex]y^2=4(16-x^2)[/tex]

[tex]y=2\sqrt{16-x^2}[/tex]

So, the area of the rectangle will be,

Area,

[tex]A = 2x \times y[/tex]

= [tex]2x \times \sqrt{(16 - x^2)}[/tex]

Differentiate the area with respect to ,

[tex]A' = 2 \times \sqrt{(16 - x^2)} - 2x^2/(16 - x^2)[/tex]

Differentiate the area twice as,

When[tex]x = 0, y = 5[/tex] and when [tex]x = 5, y = 0[/tex], [tex]area = 0[/tex].

It implies that area is maximum when the value of x lies between 0 and 5.

This will occur where [tex]A'= 0[/tex].

⇒ [tex]2 \times \sqrt{(16 - x2) }- 2x^2/(16 - x^2) = 0[/tex]

⇒[tex]2 \times \sqrt{(16 - x^2)} = 2x^2/(16 - x^2)[/tex]

On simplification, we get,

⇒ [tex]2 \times (16 - x^2) = 2x^2[/tex]

⇒ [tex]16 - x^2 = x^2[/tex]

⇒ [tex]2x^2= 16[/tex]

⇒ [tex]x^2 = 16/2[/tex]

⇒ [tex]x = \sqrt{8}[/tex]

Now, [tex]y = \sqrt{(16 - x^2)}[/tex] becomes

⇒[tex]y = \sqrt{(16 - (\sqrt8)^2)}[/tex]

⇒ [tex]y = \sqrt{(16 - 8)}[/tex]

[tex]y =\sqrt{8}[/tex]

Maximum area = 2xy

[tex]=2(\sqrt{8)}(\sqrt{8})[/tex]

[tex]=16[/tex]

Therefore, the area of the largest rectangle is 16 square units.

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Step-by-step explanation:

9 × 1/2 = 4.5

8 × 1/2 = 4

hope it's helpful

Answer:

V<4400

Step-by-step explanation:

Answer:

the answer is V<4,400

if you are using khan academy, do not add the comma for 4,400, otherwise it will not work.

khan academy answer: V<4400

Step-by-step explanation:

if the air leaked out, which is V, that means V would be less than 4,400 because 4,400 is what we have started with.

Therefore, the inequality would be V<4,400 or for khan academy, V=4400

Answer:

see explanation

Step-by-step explanation:

- 1, 2, 5, 8

(a)

the common difference

d = a₂ - a₁ = 2 - (- 1) = 2 + 1 = 3

(b)

to find a term in the sequence add d = 3 to the previous term

a₅ = = 8 + 3 = 11

a₆ = 11 + 3 = 14

a₇ = 14 + 3 = 17

a₈ = 17 + 3 = 20

(c)

the recursive rule is

[tex]a_{n}[/tex] = [tex]a_{n-1}[/tex] + 3 : a₁ = - 1

(d)

the explicit rule for an arithmetic sequence is

[tex]a_{n}[/tex] = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

here a₁ = - 1 and d = 3 , then

[tex]a_{n}[/tex] = - 1 + 3(n - 1) = - 1 + 3n - 3 = 3n - 4

[tex]\fbox{90}[/tex]

In order to find 36 is 40% of what number, we will need to use the following formula:

[tex]number\times100/Percent[/tex]

[tex]\frac{36\times100}{40}[/tex]

[tex]36\times100=3,600[/tex]

[tex]3,600\div40[/tex]

[tex]=90[/tex]

The solution of the number in the expression ''40% of a number is 36'' is 90.

Given that,

To find the solution for 40% of that number is 36.

Used the concpet of percentage which stated that,

A number or ratio that can be expressed as a fraction of 100 or a relative value indicating the hundredth part of any quantity is called a percentage.

Let us assume that the solution of the given condition is x.

So, it can be written as,

[tex]40\% \times x = 36[/tex]

[tex]\dfrac{40}{100} \times x = 36[/tex]

Multiply both sides by 100,

[tex]40x = 36 \times 100[/tex]

[tex]40x = 3600[/tex]

Divide both sides by 40,

[tex]x = \dfrac{3600}{40}[/tex]

[tex]x = 90[/tex]

So, the solution for the number is 90.

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The perimeter of the composite figure would be = 21 (to the nearest tenth).

Perimeter is defined as the quantity that is used to evaluate the total length of the sides of an object.

From the given diagram, the composite figure is made up of 6 sides. The addition of these sides is the figures perimeter.

The sides of the composite figure are:

a = 6; b = 12; c = 9; d = 9; e = -9; f = -6.

The perimeter = a + b + c + d + e + f.

= 6 + 12 + 9 +9+ (-9) + (-6)

= 27 - 6

= 21 ( to the nearest tenth)

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Answer: always, always

Step-by-step explanation:

Every circle is similar to every other circle.

Answer:

Step 6 is wrong

Using percentages we calculate in the beginning there were 125 jellybeans in the jar.

Let the number of jellybeans in the jar was x.

Now student ate 20% of the jellybeans.

Number of jellybeans students ate = 20% of x = 0.2x

Number of jellybeans left = x - 0.2x = 0.8x

But the number of jellybeans is given as 100.

Therefore 0.8x = 100

or, x = 100 /0.8 = 125

Hence there were 125 jellybeans in the jar.

The English word "percentage" derives from the Latin phrase "per centum," which means "by the hundred." Percentages are fractions with a denominator of 100.

In other words, that is a relationship where it is always assumed that the value of the whole is 100.

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