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3 years ago

15

30 POINTS!!! Hello!! Marie here! Im giving BRAINLIEST and THANKS and FIVE STARS to the person who answers this question and expl

ains it!
How much miles is race track if the racers did 500 miles in 200 laps?

1 answer:

9966 [12]3 years ago

5 0

<h2>.....Question:-)</h2>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>How </em><em>much </em><em>miles </em><em>is </em><em>race </em><em>track </em><em>if </em><em>the </em><em>racers </em><em>did </em><em>5</em><em>0</em><em>0</em><em>m</em><em>i</em><em>l</em><em>e</em><em>s</em><em> </em><em>in </em><em>2</em><em>0</em><em>0</em><em> </em><em>laps?</em><em>?</em>

<h2><em>Answer</em><em> (✷‿✷)</em></h2>

<em> </em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>2</em><em>.</em><em>5</em><em> </em><em>miles </em><em>(</em><em>4</em><em>k</em><em>m</em><em>)</em><em> </em><em>is </em><em>race </em><em>track </em><em>if </em><em>the </em><em>racers </em><em>did </em><em>5</em><em>0</em><em>0</em><em> </em><em>miles </em><em>in </em><em>2</em><em>0</em><em>0</em><em> </em><em>laps.</em><em>.</em><em>.</em><em>.</em>

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Find an equation of the parabola y = ax2 + bx + c that passes through (0, 1) and is tangent to the line y = 4x − 4 at (1, 0).

larisa86 [58]

We are given the following:
- parabola passes to both (1,0) and (0,1)
<span> - slope at x = 1 is 4 from the equation of the tangent line </span>

<span>First, we figure out the value of c or the y intercept, we use the second point (0, 1) and substitute to the equation of the parabola. W</span><span>hen x = 0, y = 1. So, c should be equal to 1. The</span><span> parabola is y = ax^2 + bx + 1 </span>

<span>Now, we can substitute the point (1,0) into the equation,
</span>0 = a(1)^2 + b(1) + 1
<span>0 = a + b + 1
a + b = -1 </span>

<span>The slope at x = 1 is equal to 4 which is equal to the first derivative of the equation.</span>
<span>We take the derivative of the equation ,
y = ax^2 + bx + 1</span>
<span>y' = 2ax + b
</span>
<span>x = 1, y' = 2
</span>4 = 2a(1) + b
<span>4 = 2a + b </span>

So, we have two equations and two unknowns,<span> </span>
<span>2a + b = 4 </span>
<span>a + b = -1
</span><span>
Solving simultaneously,
a = 5 </span>
<span>b = -6</span>

<span>Therefore, the eqution of the parabola is y = 5x^2 - 6x + 1 .</span>

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2 years ago

Write a linear equation in slope intercept form to model a tree 4 feet tall the grows 3 inches per year

Hoochie [10]

Y = 3x + 48 where x is your number of years. I got 48 by multiplying 4 and 12 since there are 12 inches in a foot. (:

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3 years ago

Please help ME I BEG YOU

VMariaS [17]

Step-by-step explanation:

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2 years ago

Read 2 more answers

A box with a rectangular base and open top must have a volume of 128 f t 3 . The length of the base is twice the width of base.

noname [10]

Answer:

$Width = 4ft$

$Height = 4ft$

$Length = 8ft$

Step-by-step explanation:

Given

$Volume = 128ft^3$

$L = 2W$

$Base\ Cost = \$9/ft^2$

$Sides\ Cost = \$6/ft^2$

Required

The dimension that minimizes the cost

The volume is:

$Volume = LWH$

This gives:

$128 = LWH$

Substitute $L = 2W$

$128 = 2W * WH$

$128 = 2W^2H$

Make H the subject

$H = \frac{128}{2W^2}$

$H = \frac{64}{W^2}$

The surface area is:

Area = Area of Bottom + Area of Sides

So, we have:

$A = LW + 2(WH + LH)$

The cost is:

$Cost = 9 * LW + 6 * 2(WH + LH)$

$Cost = 9 * LW + 12(WH + LH)$

$Cost = 9 * LW + 12H(W + L)$

Substitute: $H = \frac{64}{W^2}$ and $L = 2W$

$Cost =9*2W*W + 12 * \frac{64}{W^2}(W + 2W)$

$Cost =18W^2 + \frac{768}{W^2}*3W$

$Cost =18W^2 + \frac{2304}{W}$

To minimize the cost, we differentiate

$C' =2*18W + -1 * 2304W^{-2}$

Then set to 0

$2*18W + -1 * 2304W^{-2} =0$

$36W - 2304W^{-2} =0$

Rewrite as:

$36W = 2304W^{-2}$

Divide both sides by W

$36 = 2304W^{-3}$

Rewrite as:

$36 = \frac{2304}{W^3}$

Solve for $W^3$

$W^3 = \frac{2304}{36}$

$W^3 = 64$

Take cube roots

$W = 4$

Recall that:

$L = 2W$

$L = 2 * 4$

$L = 8$

$H = \frac{64}{W^2}$

$H = \frac{64}{4^2}$

$H = \frac{64}{16}$

$H = 4$

Hence, the dimension that minimizes the cost is:

$Width = 4ft$

$Height = 4ft$

$Length = 8ft$

8 0

2 years ago

Pentagon RSTUV is circumscribed about a circle. What is the value of x if RS = 8, ST = 12, TU = 15, UV = 9, and VR = 12?

ivanzaharov [21]

Answer:

The value of x is 7 ⇒ 1st answer

Step-by-step explanation:

* Lets revise a fact in the circle

- The two tangents drawn from a point out side the circle are equal

∵ RSTUV is circumscribed about a circle

∴ Each side of the pentagon is a tangent to the circle

- Look to the attached figure to know how we will solve the problem

- Each tangent divided into two parts

# RS = x + y

∵ RS = 8

∴ x + y = 8 ⇒ (1)

# RV = x + n

∵ RV = 12

∴ x + n = 12 ⇒ (2)

- Subtract (2) from (1)

∴ y - n = -4 ⇒ (3)

# ST = y + z

∵ ST = 12

∴ y + z = 12 ⇒ (4)

# TU = z + m

∵ TU = 15

∴ z + m = 15 ⇒ (5)

- Subtract (5) from (4)

∴ y - m = -3 ⇒ (6)

# UV = m + n

∵ UV = 9

∴ m + n = 9 ⇒ (7)

- Add (6) and (7)

∴ y + n = 6 ⇒ (8)

- Lets solve equation (3) and equation (8) to find y

∵ y - n = -4 ⇒ (3)

∵ y + n = 6 ⇒ (8)

- Add (3) and (8)

∴ 2y = 2 ⇒ divide two sises by 2

∴ y = 1

- Lets substitute the value of y in equation (1)

∵ x + y = 8 ⇒ (1)

∵ y = 1

∴ x + 1 = 8 ⇒ subtract (1) from both sides

∴ x = 7

* The value of x is 7

3 0

3 years ago