Im one paragraph, summarize the purpose of the graphic and its key findings

Answer:

32

Step-by-step explanation:

Opposite angles are congruent, so Angle BAC is 58 degrees.

Line KL and FG are perpendicular (forms a 90 degree angle), so Angle ABC is 90 degrees.

Points ABC form a triangle, and the interior angles of a triangle must add up to 180 degrees.

Thus, 58 + 90 + Angle BCA = 180. Solving for BCA, we get 32 degrees. Since Angle GCJ is opposite to BCA, x is 32 degrees.

ANSWER: The third statement appears to be incorrect.

Considering the area of the rectangle, we have that the length is of 8 units and the width is of 10 units.

The area of a rectangle of length l and width w is given as follows:

A = lw.

In this problem, the area is of 80 square units, while the length and the width are consecutive even integers, hence:

Then:

l(l + 2) = 80

l² + 2l - 80 = 0

(l + 10)(l - 8) = 0.

We need the positive measure, hence:

l -  8= 0 -> l = 8 units.

w = l + 2 = 10 units.

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The solutions of the equations are (0,4.5), (1, 1), (2, -2.5) & (3, -6) and (0,0), (1, -0.4), (2, -0.8) & (3, -1.2)

Equation 1

7x + 2y = 9

Make y the subject

2y = 9- 7x

Divide by 2

y = 4.5 - 3.5x

Let x = 0, 1, 2 and 3

y = 4.5 - 3.5 * 0 = 4.5

y = 4.5 - 3.5 * 1 = 1

y = 4.5 - 3.5 * 2 = -2.5

y = 4.5 - 3.5 * 3 = -6

Hence, the solutions of the equation are (0,4.5), (1, 1), (2, -2.5) and (3, -6)

Equation 2

2x = -5y​

Make y the subject

y = -0.4x

Let x = 0, 1, 2 and 3

y = -0.4 * 0 = 0

y = -0.4 * 1 = -0.4

y = -0.4 * 2 = -0.8

y = -0.4 * 3 = -1.2

Hence, the solutions of the equation are (0,0), (1, -0.4), (2, -0.8) and (3, -1.2)

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The angle of m∠5 is 116 degrees.

When parallel lines are cut by a transversal, angle relationships are formed. This include corresponding angles, alternate angles , vertical angles etc.

Therefore, line m and n are parallel lines cut by the a transversal.

Hence,

m∠2 = 64 degrees

Therefore,

m∠5 = m∠4 (alternate angles)

Therefore,

m∠5  + m∠2 = 180

m∠5  = 180 - 64

m∠5  = 116 degrees.

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The degree measure of the angle labelled 5; m<5 = 116°.

The angle measure of the angle labelled 5 as in the complete task content in the attached image can be determined as follows;

It follows from the complete task content that; that the measure of angle 2 is; 64°.

On this note, it follows from the sum of angle measures on a straight line theorem that; m<4 = 180° - 64° = 116°.

Additionally, it follows from the congruence of alternate angles theorem that <4 and <5 are congruent and hence, their measures are equal.

Therefore, m<5 = 116°.

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Answer:

1/xyz^3

Step-by-step explanation:

exponents are subtracted when divided

x^0 is 1

negative exponents are positive in the denominator

(4x^0y^-2z^-3) / (4xy-1)

(4xy^-2z^-3) / (4xy^-1)

x-1y^-1z^-3

1/xyz^3

Answer:

x = 6

Step-by-step explanation:

y = -8

x = 8

-x = -8 = y

-x = y

now, when y is -6

-x = y = -6

-x = -6

x = 6

Answer:

20m

Step-by-step explanation:

write a equation duh

x * 4x = 100

u can split 100 into 5 x 20

x * 4x = 5 * 20

x = 5

and 4x = 20

The values of x that will give the quadratic equation no real number solutions are -5 and 9

Real roots of a quadratic equation are values of x, whose numbers can be represented on a number line.

Analysis:

We have -[tex]x^{2}[/tex] +3x +c = 0

[tex]b^{2}[/tex] - 4ac  is called the discriminant of a quadratic equation.

if [tex]b^{2}[/tex] - 4ac  < 0, then the quadratic equation has no real root.

where a = -1, b = 3 c = c

[tex](3)^{2}[/tex] -4(-1)c < 0

9 + 4c < 0

4c < -9

c < -9/2

Which means values lower than -9/2 would give the solution no real root.

From, the options,  only -5 would give no real solution, since it is lower than -9/2.

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Answer: (-2, 1)

Step-by-step explanation:

       We will graph the given equations. The point of intersection is the solution to the system of equations. See attached.

       They intersect at the point (-2, 1) giving us an answer of x = -2 and y = 1, but this question asks for the answer as an ordered pair so we leave it as is.

[tex]\frac{3}{4}[/tex] gallons of water fill the bin.

Given: Nancy left a bin outside in her garden to collect rainwater.

She notices that 1 over 2 gallons of water fills 2 over 3 of the bin.

To Find: Write and solve an expression to find the amount of water that will fill the entire bin.

Let's say the amount of  Water to  fill the bin  = W gallon

Equate both:

[tex]\frac{2W}{3} =\frac{1}{2}[/tex]  is the expression to find the amount of water that will fill the entire bin.

[tex]\frac{2W}{3} =\frac{1}{2} \\2W=\frac{3}{2} \\W=\frac{3}{4}[/tex]

Therefore, [tex]\frac{3}{4}[/tex] gallons of water fill the bin.

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The total area of the figure shown is 320 square units.

We can see that we have 4 triangular faces and one square face.

The area of a square of side length S is:

A = S²

Here we can see that the side length of the square is 10 units, then the area of the base is:

a = (10)² = 100

For a triangle of base B and height H, the area is:

[tex]A = B*H/2[/tex]

Here we can see that the base measures 10 units and the height 11 units, so the area of each triangular face is:

A' = (10)*(11)/2 = 55

Then the total area is:

area = 100 + 4*55 = 320

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Answer:

7x  + 9/10  ft

Step-by-step explanation:

To find the perimeter of a triangle, add the lengths of the three sides

P = 2x + 1/5  +  4x + 1/2  + x+ 1/ 5

Combine like terms

P = 7x + 2/5 +1/2

Get a common denominator

P = 7x + 2/5(2/2) + 1/2 * (5/5)

   = 7x + 4/10 + 5/10

   = 7x  + 9/10

The solutions of x² - 3x + 2 ≡ mod 5 does not exist.

We are given the Polynomial;

x² - 3x + 2 ≡ mod 5

Finding the roots of the equation gives;

x² - 2x - x + 2 = 0(mod 5)

x(x - 2) - 1(x - 2) = 0(mod 5)

Thus, RHS = 0(mod 5) = 0

The roots of the equation are;

x = 1 and x = 2

Thus, the solutions of x² - 3x + 2 ≡ mod 5 does not exist.

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Answer:Answer: true

Step-by-step explanation: f(x) is a function because each x value (left oval) has one y value (right oval)!

Answer:

4:10

Step-by-step explanation:

4:10 = 4/10 = 2/5

12:20 = 12/20 = 6/10 = 3/5.

So 4:10 is smaller than 12:20 because 2/5 is less than 3/5.

Answer:

4:10

Step-by-step explanation:

First, I wrote each ratio as a fraction. Then I found a common denominator of 20. The fraction 4/10 is 8/20. Comparing numerators, 8 is less than 12, so the ratio 4:10 is smaller.

The piecewise function is

f(x) = 2, 0 < x ≤ 1/2

f(x) = 2 + 0.7x, x > 1/2

From the question, we have:

Fares = $2.00 ⇒ up to (and including) 1/2 miles

Fares = $0.70 increment ⇒ greater than 1/2 miles

Let x represent the number of miles, and f(x) the fares.

So, we have:

f(x) = 2, 0 < x ≤ 1/2

f(x) = 2 + 0.7x, x > 1/2

The above represents the piecewise function

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Last option. The correct statement about the graph is g(x) is an odd degree polynomial with a positive leading coefficient.

We have the function as

[tex]f(x) = x^2 - 2x + 1[/tex]

The polynomial that we have here is of an even degree. The highest degree we have here is 2. The coefficient that has the highest degree is 1.

Hence all of the options that have f(x) are incorrect.

The answer to the question is option 4.

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Answer:

1 +10=11 (since 0 is neither negative or positive the first and second positive integerswith 1 or 0

The height h of the triangle is 5.42 ft

The height of the triangle can be found using the formula below:

Area = 1 / 2 bh

where

Therefore,

A = √s(s - a)(s - b)(s - c)

Hence,

s = 14 + 7 + 11 / 2 = 32 / 2 = 16

A = √16(16 - 11)(16 - 7)(16 - 14)

A = √1440

A = 37.947331922

A = 37.95 ft²

37.95 = 1 / 2 × 14 × h

h = 37.95 / 7

h = 5.42104741743

h = 5.42 ft

Therefore, the height h of the triangle is 5.42 ft

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The solutions are: θ = 90º, 270º, 30º, and 150º

Given,

sin2θ = cosθ

sin2 θ = 2sinθ cosθ

2sinθ cosθ = cosθ

2sinθ cosθ - cosθ = 0

factor

cosθ(2sinθ-1) = 0

cosθ = 0 or 2sinθ - 1 = 0

cosθ = 0 when θ=90º or θ=270º (look on the unit circle)

2sinθ - 1 = 0

2sinθ = 1

sinθ = 1/2

look at the unit circle and see that sinθ=1/2 when θ=30º

sin is positive in Quadrant2 as well as Quadrant 1

180º-30º=150º

θ=150º

answers: θ=90º, 270º, 30º, and 150º

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The Bulldogs' distribution is symmetric, but the Wolverines' distribution is negatively skewed.

A box plot is used to study the distribution and level of a set of scores. On the box, the first line to the left represents the lower (first) quartile. The next line on the box represents the median. The third line on the box represents the upper (third) quartile.

A box plot is symmetric when the median is at the center of the box and negatively skewed if the median is closer to the upper quartile.

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Answer is C as all other choices are even numbers therefore make the equation non-integer.

Answer:

C)7

Step-by-step explanation:

Due to the fact that all other options are even numbers so it Wii make the equation non-integer.

Answer:

x = 34

Step-by-step explanation:

The three angles form a straight line and are thus supplementary angles.  The sum of a supplementary angles is always 180, so we can use the equation 180 = x + 12 + 100 + x to find x:

180 = 2x + 112

68 = 2x

x = 34

The correct statements are Neither was correct and Joe used the height and radius to calculate the slant height.

The square inches of paper needed A = surface area of cone +  area of overlap

The surface area of the cone is given by A = πr[r + √(h² + r²)] where

So, A = πr[r + √(h² + r²)]

A = π × 2 × [2 + √(6² + 2²)]

A = π × 2 × [2 + √(36 + 4)]

A = π × 2 × [2 + √40]

A = π × 2 × [2 + 2√10]

A = 2π[2 + 2√10]

A = 4π + 4π√10

The area of overlap A' = wL where

So, A' = wL

A' = w[√(h² + r²)]

A' =  1/2[√(6² + 2²)]

A' =  1/2[√(36 + 4)]

A' =  1/2[√40]

A' =  1/2 × 2√10

A' = √10

So, the number of square inches of paper needed is A" = A + A'

=  4π + 8π√10 + √10

Since the number of square inches of paper needed is 4π + 4π√10 + √10

So, the correct statements are Neither was correct and Joe used the height and radius to calculate the slant height.

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Based on the given variation, y does not vary directly with x and the constant of variation are 8, 3.2 and 1.25 respectively.

y = k × x

where,

k = constant of proportionality

y = -40

x = -5

y = k × x

-40 = k × -5

-40 = -5k

k = -40/-5

k = 8

when,

y = 8 and x = 2.5

y = k × x

8 = k × 2.5

8 = 2.5k

k = 8/2.5

k = 3.2

when,

y = 5 and x = 4

y = k × x

5 = k × 4

5 = 4k

k = 5/4

k = 1.25

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1. The TRUE statement about the graph, which is represented here by the data table, is D. The average salary increased by some irregular amount each year.

2. The average salary per year between 2001 and 2006 is $3,600?

An average number is the mean of a total value.

The average is computed as the total value of observations divided by the number of observations.

Year       Salary

2001     $1,500

2002     2,600

2003     3,200

2004      4,100

2005    5,000

2006    5,200

Total  $21,600

Average = $3,600 ($21,600/6)

Thus, the graph or the data table shows that D. The average salary increased by some irregular amount each year.

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1. Option D: The average salary increased by some regular amount each year.

2, What is the average salary per year between 2001 and 2006?

Answer: C-The average salary increased by the same amount each year.

NEXT ANSWER-B

Step-by-step explanation:

Answer:

  24

Step-by-step explanation:

The semi-axis lengths are 'a' and 'b' in the equation ...

  (y -k)²/a² -(x -h)²/b² = 1

The area of the central rectangle will be ...

  A = LW

The axis lengths are 2a and 2b, so the area is ...

  A = (2a)(2b) = 4ab

In terms of the numbers in the given equation, this is ...

  A = 4√(a²b²) = 4√(3×12)

  A = 4×6 = 24

The area of the central rectangle is 24 square units.

Answer:

2.5

Step-by-step explanation:

IVE BEEN WATCHING YOUR ACCOUNT STOP STEALING POINTS OR YOUR ACCOUNT WILL BE TERMINATED.

1) [tex]\overline{AB} \perp \overline{BD}[/tex], [tex]\overline{AB} \cong \overline{DB}[/tex], [tex]\overline{AC} \cong \overline{DE}[/tex] (given)

2) [tex]\angle ABD[/tex] is a right angle (perpendicular lines form right angles)

3) [tex]\triangle ABD[/tex] and [tex]\triangle DEB[/tex] are right triangles (a triangle with a right angle is a right angle)

4) [tex]\triangle ABD \cong \triangle DEB[/tex] (HL)

5) [tex]\angle ACB \cong \angle DEB[/tex] (CPCTC)