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3 years ago

The given line segment has a midpoint at (−1, −2).

1 answer:

6 0

As the midpoint is a point on the perpendicular line, we can solve by eliminating the other equations, like this:
Check if the point (-1,-2) is on the line y=-4x-4:
-4(-1)-4=0 is differente from -2.
-4(-1)-6=-2, so (-1,-2) is on the line y=-4x-6.
We check also for the other equation like this:
Third equation: (-1)-4 = -5 not equal to -2.
Fourth equation: (-1)-6=-7 not equal to -2.
So the correct answer is the line y=-4x-6

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What is the product?

aliya0001 [1]

Answer:

20s³ + 50s² + 32s + 6

Step-by-step explanation:

( 4s + 2 ) ( 5s² + 10s + 3)

use the distribution property:

20s³ + 40s² + 12s + 10s² + 20s + 6

combine like terms:

20s³ + 50s² + 32s + 6

therefore

the expand value = 20s³ + 50s² + 32s + 6

3 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

9 months ago

Add. What is the sum (2 * 10 exponent 16) + ( 7 * 10 exponent 16) in scientific notation

Roman55 [17]

Answer:

9 * 10^16

Step-by-step explanation:

2 * 10^16

+7 * 10^16

9 * 10^16

5 0

3 years ago

If the area of a rectangular garden is 45x2−12x square meters, and it has a length of 3x meters, what is its width?

nikitadnepr [17]

Answer:

(9x -4) meters

Step-by-step explanation:

The relationship between area, length, and width is ...

area = length × width

Dividing by length gives ...

width = area/length

Filling in the given expressions, we have ...

$w=\dfrac{45x^2-12x}{3x}=\dfrac{45x^2}{3x}-\dfrac{12x}{3x}=9x-4$

The width of the garden is 9x-4 meters.

4 0

2 years ago

Read 2 more answers

A retail outlet's sales for January were $45,000. In February, these sales decreased 10%; but from February to March, sales incr

likoan [24]

Step-by-step explanation:

$44,550

7 0

2 years ago