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2 years ago

10

What is the slope of y=-x perpendicular to the equation

1 answer:

tensa zangetsu [6.8K]2 years ago

7 0

$\frac{1}{2} x$

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The polygons are similar. The area of one polygon is given. Find the area of the other polygon.

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Answer:

Hello! answer: area is 6 cm2

Step-by-step explanation:

Triangle area formula: base × height ÷ 2 or base × height × 1/2 so...

12 × 16 = 192 192 ÷ 2 = 96

12 ÷ 4 = 3

16 ÷ 4 = 4

3 × 4 = 12 12 ÷ 2 = 6 therefore the area is 6 hope that helps!

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1 year ago

On Gavin's math test, 3/8 of the questions are division problems. Of the division problems, 4/9 of the problems he got incorrect

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Something is strange in the wording of the question. "Of the division problems, 4/9 of the problems he got incorrect. What fraction of the division problems did Gavin get incorrect?" :) Well 4/9 of course. One statement answers another. Maybe they wanted the fraction of those wrong answers were in relation to the entire test???

$\frac {3}{8} * \frac {4}{9} = \frac {12}{72} = \frac {1}{6}$

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2 years ago

Geometry 8.6.3 - Equations of Parallel and Perpendicular Lines and Proofs

Pepsi [2]

Option B: $\frac{1}{25}$

<u>Solution:</u>

<u>Slope of the perpendicular lines:</u>

If two lines are perpendicular then their slopes are negative reciprocals of each other.

$$m_1=\frac{-1}{ m_2}$

<u>Slope of the parallel lines:</u>

If two lines are parallel then they have the same slope.

i.e their slopes are equal.

$m_1=m_2$

Given line m and p are parallel.

Slope of the line m = $\frac{1}{25}$

Using slope of the parallel lines definition,

Slope of the line p = slope of the line m

Slope of the line p = $\frac{1}{25}$

Option B is the correct aswer.

Hence the slope of the of line p is $\frac{1}{25}$ .

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3 years ago

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the re

alukav5142 [94]

Answer:

Step-by-step explanation:

$\text{Given that:}$

$y = 1+ sec(x) \ \ y =3$

$\text{we draw the graph and the curves intersect at:}$

$x = - \dfrac{\pi}{3} \ and \ x = \dfrac{\pi}{3}$

$\text{Applying washer method;}$

$f(x) _{outer} - g(x) _{inner} --- (1)$

$V= \int ^b_a A(x) \ dx --- (2)$

$\text{outer radius = 3 - 1 = 2}$

$\text{inner radius =}$ $( 1 + sec(x) ) - 1 = sec (x)$

$A(x) = \pi ((2)^2 -(sec(x)^2) \\ \\ A(x) = \pi (4 - sec^2 (x)) ---- (3)$

$\text{The volume V =}\int ^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ A(x) \ dx$

$V = \int ^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ \pi (4- sec^2 (x) ) \ dx$

$V = 2 \pi \int ^{\dfrac{\pi}{3}}_{0}( 4 - sec^2 (x)) \ dx$

$V = 2 \pi \int ^{\pi/3}_{0} 4 . \ dx - 2 \pi \int ^{\pi/3}_{0} sec^2 (x) \ dx$

$V = 2 \pi(4) \int ^{\pi/3}_{0} 1 . \ dx - 2 \pi \Big( tan (x)\Big )^{\dfrac{\pi}{3}}_{0}$

$V = 8 \pi(x)^{\dfrac{\pi}{3}}_{0} - 2 \pi \Big( tan \dfrac{\pi}{3} -tan (0)\Big )$

$V = 8 \pi({\dfrac{\pi}{3}}-{0}) - 2 \pi \Big( tan \sqrt{3}-(0)\Big )$

$V = 8 \pi({\dfrac{\pi}{3}}) - 2 \pi \Big( \sqrt{3}\Big )$

$\mathbf{V = 2 \pi \Big(\dfrac{4\pi}{3}- \sqrt{3} \Big)}$

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2 years ago