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3 years ago

10

Can someone plz plz help me

1 answer:

VikaD [51]3 years ago

6 0

Answer:

I have know clue

Step-by-step explanation:

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A container holds 13 quarts of water. How much is this in gallons?

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Answer:

3 1/4 gallons or 3.25 gallons

Step-by-step explanation:

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3 years ago

You only have to get this second answer right (30 points)

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Recall the equation that modeled the volume of the raised flower bed, y, in terms of the width of the box, y = x3 + 11x2 − 312x.

lakkis [162]

Answer:

48 cubic feet of mulch

Step-by-step explanation:

12 feet long x 4 feet wide x 1 foot deep of mulch= 48 cubic feet of mulch

8 0

3 years ago

Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2

Verdich [7]

Answer:

(a) $f'(x)=-\frac{2}{x^3}$

(b) $y=-0.25x+0.75$

Step-by-step explanation:

The given function is

$f(x)=\frac{1}{x^2}$ .... (1)

According to the first principle of the derivative,

$f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}$

Cancel out common factors.

$f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}$

By applying limit, we get

$f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}$

$f'(x)=\frac{-2x)}{x^4}$

$f'(x)=\frac{-2)}{x^3}$ .... (2)

Therefore $f'(x)=-\frac{2}{x^3}$ .

(b)

Put x=2, to find the y-coordinate of point of tangency.

$f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25$

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

$m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}$

Substitute x=2 in equation 2.

$f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25$

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

$y-y_1=m(x-x_1)$

$y-0.25=-0.25(x-2)$

$y-0.25=-0.25x+0.5$

$y=-0.25x+0.5+0.25$

$y=-0.25x+0.75$

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

5 0

3 years ago

What is the product?

Sophie [7]

Answer:

Step-by-step explanation:

In exponent multiplication, if base are same just add the powers.

$(-6a^{3}b+2ab^{2})(5a^{2}-2ab^{2}-b)=(-6a^{3}b)*(5a^{2}-2ab^{2}-b)+2ab^{2}*(5a^{2}-2ab^{2}-b)\\\\$

$=(-6a^{3}b)*5a^{2}-(-6a^{3}b)*2ab^{2}-(-6a^{3}b)*b+ 2ab^{2}*5a^{2}-2ab^{2}*2ab^{2}-2ab^{2}*b\\\\=-30a^{3+2}b+12a^{3+1}b^{1+2}+6a^{3}b^{1+1}+10a^{1+2}b^{2}-4a^{1+1}b^{2+2}-2ab^{2+1}\\\\=-30a^{5}b+12a^{4}b^{3}+6a^{3}b^{2}+10a^{3}b^{2}-4a^{2}b^{4}-2ab^{3}\\$

6 0

3 years ago