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solmaris [256]
3 years ago
15

WCLN PCMath1l - Rev. Sept/2018

Mathematics
1 answer:
Verdich [7]3 years ago
4 0

Answer:

\frac{p + 1}{p - 8}

Step-by-step explanation:

The  \: slope: \frac{2p + 1 - p}{p - 5 - 3}  =  \frac{p + 1}{p - 8}

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What is the equation of the graph below?<br>​
Sonja [21]

The equation of the graph is y= -(x+3)^2 +1

Choice A is the correct answer

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3 years ago
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Heather finished the 12-kilometer race in 2 hours. If Heather kept her pace constant, what was her rate? A. 24 km/hr. B. 14 km/h
andriy [413]

Answer: D, 6km/hr.

Step-by-step explanation:

Heather can finish a 12-kilometer race in 2 hours, and now we have to find how many kilometers she can ride/run at in 1 hour.

How many hours can Heather run in 1 hour? To solve that, we can use the equation 12 ÷ 2.

12 ÷ 2 = 6.

Therefore, if Heather keeps her pace constant, then her rate will be 6km/hr, or D.

7 0
3 years ago
Solve for x open image to see
Natasha2012 [34]

Answer:

x=15

Step-by-step explanation:

Since both lines are parallel, and one line goes through them, angle (3x+2) and 47 are congruent

3x+2=47

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x=15

6 0
3 years ago
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20 POINTS!!!!
nevsk [136]

Answer:

Step 1: Remove parentheses by multiplying factors.

= (x * x) + (1 * x) + (2 * x) + (2 * 1)

Step 2: Combine like terms by adding coefficients.

(x * x) = x2

(1 * x) = 1x

(2* x) = 2x

Step 3: Combine the constants.

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= x2 + 3x + 2.

3 0
2 years ago
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A fastball is hit straight up over home plate. The ball's height, h (in feet), from the ground is modeled by ℎ = −16푡 ଶ+103푡+5 w
Liono4ka [1.6K]

Question:

A fastball is hit straight up over home plate. The ball's height, h (in feet), from the ground is modeled by h(t)=-16t^2+80t+5, where t is measured in seconds.  Write an equation to determine how long it will take for the ball to reach the ground.

Answer:

t = 5.0625

Step-by-step explanation:

Given

h(t)=-16t^2+80t+5

Required

Find t when the ball hits the ground

This implies that h(t) = 0

So, we have:

0=-16t^2+80t+5

Reorder

-16t^2+80t+5 = 0

Using quadratic formula, we have:

t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}

Where

a = -16      b =80      c = 5

So, we have:

t = \frac{-80\±\sqrt{80^2 - 4*-16*5}}{2*-16}

t = \frac{-80\±\sqrt{6400 +320}}{-32}

t = \frac{-80\±\sqrt{6720}}{-32}

t = \frac{-80\±82.0}{-32}

This gives:

t = \frac{-80+82.0}{-32} or t = \frac{-80-82.0}{-32}

t = \frac{2}{-32} or t = \frac{-162}{-32}

t = -\frac{2}{32} or t = \frac{162}{32}

But time can not be negative.

So, we have:

t = \frac{162}{32}

t = 5.0625

<em>Hence, time to hit the ground is 5.0625 seconds</em>

3 0
3 years ago
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