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3 years ago

1. (05.06 MC)

Mathematics

1 answer:

adoni [48]3 years ago

8 0

Part A: it is linear because it is not curving and it consists of straight lines.

Part B: in side A it is increasing because it has a positive slope. In side b it is constant because the slope is 0 since it is straight. Finally, side C is decreasing because the slope is negative.

Part C: during side A the ant is crawling out of the hole in 2 seconds. After that, the ant stops for 2 more seconds as shown in side B. Then, he crawls back into the hole as shown by the decrease in distance due to the slope.

Hope this helps!!!

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How I writing this in a expresión

Anon25 [30]

Answer:

2.75x+122.5

Step-by-step explanation:

Horse=x

Stork=x+70

Whale=3/4(x+70) = 0.75(x+70) = 0.75x+52.5

x+x=70+0.75x+52.5

2.75x+122.5

4 0

3 years ago

Read 2 more answers

√secA+tanA/√secA-tanA × √cosecA-1/√cosecA+1=1

Snowcat [4.5K]

$Use:\\\\\sec A=\dfrac{1}{\cos A}\\\\\tan A=\dfrac{\sin A}{\cos A}\\\\\csc A=\dfrac{1}{\sin A}\\\\\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}\\---------------------------------\\\\\sec A+\tan A=\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}=\dfrac{1+\sin A}{\cos A}\\\\\sec A-\tan A=\dfrac{1-\sin A}{\cos A}\\\\\csc A-1=\dfrac{1}{\sin A}-\dfrac{\sin A}{\sin A}=\dfrac{1-\sin A}{\sin A}\\\\\csc A+1=\dfrac{1+\sin A}{\sin A}$

$\dfrac{\sqrt{\sec A+\tan A}}{\sqrt{\sec A-\tan A}}\cdot\dfrac{\sqrt{\cos A-1}}{\sqrt{\cos A+1}}=1\\\\L_s=\sqrt{\dfrac{\sec A+\tan A}{\sec A-\tan A}\cdot\dfrac{\cos A-1}{\cos A+1}}=\sqrt{\dfrac{\frac{1+\sin A}{\cos A}}{\frac{1-\sin A}{\cos A}}\cdot\dfrac{\frac{1-\sin A}{\sin A}}{\frac{1+\sin A}{\sin A}}}\\\\=\sqrt{\dfrac{1+\sin A}{\cos A}\cdot\dfrac{\cos A}{1-\sin A}\cdot\dfrac{1-\sin A}{\sin A}\cdot\dfrac{\sin A}{1+\sin A}}\\\\\text{Everything are simplified}\\\\=\sqrt{1}=1=R_s$

7 0

3 years ago

Math geometry show work Thanks

aev [14]

Problem 1, part (a)

<h3>Answer: False</h3>

For instance, 200 feet in real life can be reduced to scale down to say 2 inches on paper. So we have a reduction going on, and not an enlargement.

====================================================

Problem 1, part (b)

<h3>Answer: true</h3>

This is because a scale drawing involves similar polygons. This is true whenever any dilation is applied.

====================================================

Problem 2

I'm not sure how your teacher wanted you to answer this question. S/he didn't give you any numbers for the side lengths of the polygon. The angle measures are missing as well.

4 0

3 years ago

A pie company arranges its three-flavor cheesecake sampler into equal-sized slices as shown, where S represents strawberry, C re

Citrus2011 [14]

Answer:

$37.68$ square inches

Step-by-step explanation:

Given:

A pie company arranges its three-flavor cheesecake sampler into equal-sized slices: S represents strawberry, C represents chocolate, and P represents plain.

Diameter of the pie = 12 inches

To find: total area of the strawberry portion in the pie

Solution:

Diameter of the pie = 12 inches

Radius of the pie (r) = Diameter / 2 = $\frac{12}{2}=6$ inches

Area of the pie = $\pi r^2=3.14(6)^2=113.04$ square inches

Ratio of the strawberry portion in the pie = $\frac{2}{6}=\frac{1}{3}$

Therefore,

total area of the strawberry portion in the pie = $\frac{1}{3}(113.04)=37.68$ square inches

6 0

3 years ago

What value is added to both sides of the equation x2 − 4x = −2 in order to solve by completing the square? 4, 2 , −2 , −4

Fiesta28 [93]

X^2 - 4 x = -2
Add 4 to both sides:
x^2 - 4 x + 4 = 2
Write the left-hand side as a square:
(x - 2)^2 = 2
Take the square root of both sides:
x - 2 = sqrt(2) or x - 2 = -sqrt(2)
Add 2 to both sides:
x = 2 + sqrt(2) or x - 2 = -sqrt(2)
Add 2 to both sides:
Answer: x = 2 + sqrt(2) or x = 2 - sqrt(2)

6 0

3 years ago