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1 year ago

Allen is 5 feet 9 inches tall, terrel is 6 feet 2 inches tall. what is the average height

1 answer:

4 0

The average height is $5feet$ and $11.5$ inches if Allen is $5 feet 9 inches$ tall and Terrel is $6 feet 2 inches$ tall.

How to find the average height ?

Allen height is $5feet ,9inches$

Terrel height is $6feet,2inches$

And we know that $1feet=12inches$

So Allen height is

$=5*12+9\\=60+9\\=69 inches$

And Terrel height is

$=6*12+2\\=72+2\\=74inches$

Average of the Allen and Terrel height is

$\frac{69+74}{2\\} \\=143/2\\=71.5inches$

And average height in feet is $5feet,11.5inches$

Learn more about the average height is here:

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klio [65]

Answer:

y = -1/4x -2

Step-by-step explanation:

the perpendicular slope would be -1/4

2 = -1/4 (-16) + b

b = -2

the equation should be y = -1/4x -2

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5 0

3 years ago

Pleeeeese help me as fast as you can!!!

Aleonysh [2.5K]

Answer:

A. $y=3x-10$

C. $y+6=3(x-15)$

Step-by-step explanation:

Given:

The given line is $6x+18y=5$

Express this in slope-intercept form $y=mx+b$ , where m is the slope and b is the y-intercept.

$6x+18y=5\\18y=-6x+5\\y=-\frac{6}{18}x+\frac{5}{18}\\y=-\frac{1}{3}x+\frac{5}{18}$

Therefore, the slope of the line is $m=-\frac{1}{3}$ .

Now, for perpendicular lines, the product of their slopes is equal to -1.

Let us find the slopes of each lines.

Option A:

$y=3x-10$

On comparing with the slope-intercept form, we get slope as $m_{A}=3$ .

Now, $m\times m_{A}=-\frac{1}{3}\times 3=-1$ . So, option A is perpendicular to the given line.

Option B:

For lines of the form $x=a$ , where, a is a constant, the slope is undefined. So, option B is incorrect.

Option C:

On comparing with the slope-point form, we get slope as $m_{C}=3$ .

Now, $m\times m_{C}=-\frac{1}{3}\times 3=-1$ . So, option C is perpendicular to the given line.

Option D:

$3x+9y=8\\9y=-3x+8\\y=-\frac{3}{9}x+\frac{8}{9}\\y=-\frac{1}{3}x+\frac{8}{9}$

On comparing with the slope-intercept form, we get slope as $m_{D}=-\frac{1}{3}$ .

Now, $m\times m_{D}=-\frac{1}{3}\times -\frac{1}{3}=\frac{1}{9}$ . So, option D is not perpendicular to the given line.

8 0

2 years ago

Write 23.061 in words.

Tasya [4]

23.061 written in words is:

twenty-three and sixty-one thousandths.

Note the decimal point.

hope this helps

4 0

3 years ago

F(3) = 8; f^ prime prime (3)=-4; g(3)=2,g^ prime (3)=-6 , find F(3) if F(x) = root(4, f(x) * g(x))

Marrrta [24]

Given:

$f(3)=8,f^{\prime}(3)=-4,g(3)=2,\text{ and }g^{\prime}(3)=-6$

Required:

$We\text{ need to find }F^{\prime}(3)\text{ if }F(x)=\sqrt[4]{f(x)g(x)}.$

Explanation:

Given equation is

$F(x)=\sqrt[4]{f(x)g(x)}.$ $F(x)=(f(x)g(x))^{\frac{1}{4}}$ $F(x)=f(x)^{\frac{1}{4}}g(x)^{\frac{1}{4}}$

Differentiate the given equation for x.

$Use\text{ }(uv)^{\prime}=uv^{\prime}+vu^{\prime}.\text{ Here u=}\sqrt[4]{f(x)}\text{ and v=}\sqrt[4]{g(x)}.$

$F^{\prime}(x)=f(x)^{\frac{1}{4}}(\frac{1}{4}g(x)^{\frac{1}{4}-1})g^{\prime}(x)+g(x)^{\frac{1}{4}}(\frac{1}{4}f(x)^{\frac{1}{4}-1})f^{\prime}(x)$ $=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{1}{4}-\frac{1\times4}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{1}{1}-\frac{1\times4}{4}}f^{\prime}(x)$ $=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{1-4}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{1-4}{4}}f^{\prime}(x)$ $F^{\prime}(x)=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{-3}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{-3}{4}}f^{\prime}(x)$

Replace x=3 in the equation.

$F^{\prime}(3)=\frac{1}{4}f(3)^{\frac{1}{4}}g(3)^{\frac{-3}{4}}g^{\prime}(3)+\frac{1}{4}g(3)^{\frac{1}{4}}f(3)^{\frac{-3}{4}}f^{\prime}(3)$ $Substitute\text{ }f(3)=8,f^{\prime}(3)=-4,g(3)=2,\text{ and }g^{\prime}(3)=-6\text{ in the equation.}$ $F^{\prime}(3)=\frac{1}{4}(8)^{\frac{1}{4}}(2)^{\frac{-3}{4}}(-6)+\frac{1}{4}(2)^{\frac{1}{4}}(8)^{\frac{-3}{4}}(-4)$ $F^{\prime}(3)=\frac{-6}{4}(8)^{\frac{1}{4}}(2^3)^{\frac{-1}{4}}+\frac{-4}{4}(2)^{\frac{1}{4}}(8^3)^{\frac{-1}{4}}$ $F^{\prime}(3)=\frac{-3}{2}(8)^{\frac{1}{4}}(8)^{\frac{-1}{4}}-(2)^{\frac{1}{4}}(8^3)^{\frac{-1}{4}}$ $F^{\prime}(3)=\frac{-3}{2}\frac{\sqrt[4]{8}}{\sqrt[4]{8}}-\frac{\sqrt[4]{2}}{\sqrt[4]{8^3}}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^9}}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^4(2)^4}(2)}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{4\sqrt[4]{}(2)}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{1}{4}$ $F^{\prime}(3)=\frac{-3\times2}{2\times2}-\frac{1}{4}$ $F^{\prime}(3)=\frac{-6-1}{4}$ $F^{\prime}(3)=\frac{-7}{4}$

Final answer:

$F^{\prime}(3)=\frac{-7}{4}$

8 0

11 months ago

What kind of triangle is this?

skad [1K]

Answer:

answer: obtuse

Step-by-step explanation:

Hope that helps!

7 0

2 years ago

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