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rodikova [14]
2 years ago
9

Ian walks down a sloped path. After 6 minutes, he is 3 m above street level. Write the equation in slope-intercept form of the l

ine that represents Ian’s height above street level, y, over time x, if he knows the path is parallel to another path with equation y = -2/3x + 12
Mathematics
2 answers:
Georgia [21]2 years ago
8 0

Answer:

y=-2/3x+7

Step-by-step explanation:

Since you know the path is parallel to the equation y+-2/3x+12, you already know the slope is -2/3.

It says y is the height above street level, so that is 3. The problem says x is time, which is 6, so all you have to solve for is the y intercept.

3 (y, or the height above street level) = (-2/3) 6 (x, or time) + b (unknown y intercept)

3 = -4 + b

7 = b

So, y = -2/3x + 7

zubka84 [21]2 years ago
3 0

Answer:

it 5y-4x=(334-444)

Step-by-step explanation:

is the right answer i fu.cking promise bro fkkkkk

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The monthly profit for a small company that makes long-sleeve T-shirts depends on the price per shirt. If the price is too high,
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Answer:

A. $18

B. $200

C. p\in (16,20)

Step-by-step explanation:

Function:

f(p) = -50p^2 + 1,800p - 16,000

<u>Parts A and B:</u>

The price that generates the maximum profit is ate vertex of parabola. Find the coordinates of the vertex:

p_v=\dfrac{-b}{2a}=\dfrac{-1,800}{2\cdot (-50)}=\dfrac{1,800}{100}=18\\ \\f(p_v)=-50\cdot 18^2+1,800\cdot 18-16,000=200

The price that generates the maximum profit is $18

The maximum profit is $200

<u>Part C:</u>

The company breaks even when the profit is positive. From the graph of the function you can see that the graph of the function is over p-axis for all p\in (16,20), so the positive profit is for p\in (16,20)

In p=16 and p=20, the profit is 0 and when p<16 and p>20, there will be a loss.

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2 years ago
This form will be sent to Lily by the end of January. She will use this W-2 form to...
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1 year ago
The vertices of ΔRST are R(–1,–1), S(–1,11) and T(4,11). Which could be the side lengths of a triangle that is similar but not c
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The side lengths could be 10, 24 and 26 units.

We must first find the side lengths.  We use the distance formula to do this.

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d=\sqrt{(11--1)^2+(-1--1)^2}&#10;\\=\sqrt{(11+1)^2+(-1+1)^2}&#10;\\=\sqrt{12^2+0^2}=\sqrt{144}=12

For ST:
d=\sqrt{(11-11)^2+(4--1)^2}&#10;\\=\sqrt{0^2+(4+1)^2}=\sqrt{5^2}=\sqrt{25}=5

For TR:
d=\sqrt{(11--1)^2+(4--1)^2}&#10;\\=\sqrt{(11+1)^2+(4+1)^2}=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13

Our side lengths, from least to greatest, are 5, 12 and 13.

To be similar but not congruent, the side lengths must have the same ratio between corresponding sides but not be the same length.  10, 24 and 26 are all 2x the original side lengths, so this works.
5 0
2 years ago
ΔABC underwent a sequence of rigid transformations to give ΔA′B′C′. Which transformations might have taken place?
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Answer: The correct option is second, a rotation 90^{\circ} clockwise about the origin followed by a reflection across the x-axis.

Explanation:

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If a point P(x,y) reflection across the x-axis followed by a reflection across the y-axis, then the image of point after transformation will be P'(-x,-y), therefore it is not the correct option.

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