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3 years ago

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If two lines are perpendicular, which statement must be true? A) The product of their slopes is 1. B) Their slopes are opposites

. C) Their slopes are equal. D) The product of their slopes is -1.

1 answer:

castortr0y [4]3 years ago

4 0

If two lines are perpendicular, the statement that must be true is : D. The product of their slopes is -1

Lets say that a line has a slope of 2, the line which perpendicular to that line would have a slope of -1/2

Which means, the product would be 2(-1/2) = -1

hope this helps

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A)
145 - 5d
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b)
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145-95 = 50

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3 years ago

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Each person in a certain group supports only one of the two candidates R and T. Of the people in the group, 45 percent support C

aleksley [76]

Missing Information:

Of the people in the group who support Candidate R, 58 percent are in favor of flat tax
Of the people in the group who support Candidate T, 22 are in favor of a flat tax.

Answer:

$Total = 26.1\%N + 22$

Step-by-step explanation:

Given

$R = 45\%$ -- $Percentage$ $of\ R$ in the total population

$Favor\ Flat\ Tax = 58\%$ in group R

$Favor\ Flat\ Tax = 22$ in group T

Required

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First, the proportion of those that supports flax rate in R needs to be calculated.

$Proportion = 45\% * 58\%$

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So, those that favor flat tax rate in group R are:

$Favor\ Flat\ Tax = 26.1\%N$

Where N represents the total population.

Total number of people that supports the flax tax rate is gotten by adding people that support flat tax rate in each group.

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3 years ago

3. Two cars are parked, 150m apart, and on opposite sides of a building. A camera on the top of the building rotates and can vie

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Answer:

<em>The building is 61.5 m tall</em>

Step-by-step explanation:

The image below is a diagram where all the given distances and angles are shown. We have additionally added some variables:

h = height of the building

a, b = internal angles of each triangle

x = base of each triangle

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a = 90° - 37° = 53°

b = 90° - 42° = 48°

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$\displaystyle \tan a=\frac{\text{opposite leg}}{\text{adjacent leg}}$

$\displaystyle \tan 53^\circ=\frac{150-x}{h}$

$\displaystyle \tan 48^\circ=\frac{x}{h}$

From the last equation:

$x=h.\tan 48^\circ$

Substituting into the first equation:

$\displaystyle \tan 53^\circ=\frac{150-h.\tan 48^\circ}{h}$

Operating on the right side:

$\displaystyle \tan 53^\circ=\frac{150}{h}-\tan 48^\circ$

Rearranging:

$\displaystyle \tan 53^\circ+\tan 48^\circ=\frac{150}{h}$

Solving for h:

$\displaystyle h=\frac{150}{\tan 53^\circ+\tan 48^\circ}$

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