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2 years ago

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Write an equation in slope-intercept form for the line that passes through (0, 1) and is perpendicular to the line whose equatio

n is y = 2x? Ay= -2x + 1 B y = 2x + 1 C y = x + 1 Dy=-%2x + 1

1 answer:

loris [4]2 years ago

8 0

Answer:

y= -1/2 x + 1

Step-by-step explanation:

The line passes through point (0, 1) and is perpendicular to line with equation :

y=2x

Perpendicular lines are those that give -1 when the product of their slopes is calculated.

This means m₁ *m₂= -1

In this case, m₁=2, finding m₂ as;

m₁ * m₂ = -1

2 * m₂ = -1

m₂ = -1/2

Using point (0, 1),m₂ = -1/2, and imaginary line (x, y) then the equation of the line in slope-intercept form y= mx + b will be;

-1/2 = y-1/x-0

-1/2 x = y-1

1-1/2x = y

The equation will be ;

y= -1/2 x + 1

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3 years ago

Find the sum of the zeros in the following equation: (2x-1)(x+8)(2x-4)(x-1)

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Step-by-step explanation:

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3 years ago

The caffeine content (in mg) was examined for a random sample of 50 cups of black coffee dispensed by a new machine. The mean an

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Answer:

We are 98% confident interval for the mean caffeine content for cups dispensed by the machine between 107.66 and 112.34 mg .

Step-by-step explanation:

Given -

The sample size is large then we can use central limit theorem

n = 50 ,

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98% confidence interval for the mean caffeine content for cups dispensed by the machine = $\overline{(y)}\pm z_{\frac{\alpha}{2}}\frac{\sigma}\sqrt{n}$

= $110\pm z_{.01}\frac{7.1}\sqrt{50}$

= $110\pm 2.33\frac{7.1}\sqrt{50}$

First we take + sign

$110 + 2.33\frac{7.1}\sqrt{50}$ = 112.34

now we take - sign

$110 - 2.33\frac{7.1}\sqrt{50}$ = 107.66

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3 years ago

If figure A is similar to figure B, what is the relationship between their perimeters?

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8 0

3 years ago

HELP PLEASE ASAP

Pie

$\bold{\huge{\underline{ Solution }}}$

<u>We </u><u>have</u><u>, </u>

Line segment AB
The coordinates of the midpoint of line segment AB is ( -8 , 8 )
Coordinates of one of the end point of the line segment is (-2,20)

Let the coordinates of the end point of the line segment AB be ( x1 , y1 ) and (x2 , y2)

<u>Also</u><u>, </u>

Let the coordinates of midpoint of the line segment AB be ( x, y)

<u>We </u><u>know </u><u>that</u><u>, </u>

For finding the midpoints of line segment we use formula :-

$\bold{\purple{ M( x, y) = }}{\bold{\purple{\dfrac{(x1 +x2)}{2}}}}{\bold{\purple{,}}}{\bold{\purple{\dfrac{(y1 + y2)}{2}}}}$

<u>According </u><u>to </u><u>the </u><u>question</u><u>, </u>

The coordinates of midpoint and one of the end point of line segment AB are ( -8,8) and (-2,-20) .

<u>For </u><u>x </u><u>coordinates </u><u>:</u><u>-</u>

$\sf{ -8 = }{\sf{\dfrac{(- 2 +x2)}{2}}}$

$\sf{2}{\sf{\times{ -8 = - 2 + x2 }}}$

$\sf{ - 16 = - 2 + x2 }$

$\sf{ x2 = -16 + 2 }$

$\bold{ x2 = -14 }$

<h3><u>Now</u><u>, </u></h3>

<u>For </u><u>y </u><u>coordinates </u><u>:</u><u>-</u>

$\sf{ 8 = }{\sf{\dfrac{(- 20 +x2)}{2}}}$

$\sf{2}{\sf{\times{ 8 = - 20 + x2 }}}$

$\sf{ 16 = - 20 + x2 }$

$\sf{ y2 = 16 + 20 }$

$\bold{ y2 = 36 }$

Thus, The coordinates of another end points of line segment AB is ( -14 , 36)

Hence, Option A is correct answer

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2 years ago