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3 years ago

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14. NUTRITION A cookie contains 9 grams of fat. If you eat no fewer than 4 and no more than 7 cookies, how many grams of fat wil

l you consume?a cookie contains many grams of fat if you eat no fewer than four and no more than seven cookies how many grams of fat will you consume

1 answer:

ra1l [238]3 years ago

8 0

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What is the product of -2x^3+x-5 and x^3-3x-4 ?

murzikaleks [220]

To find the product of <span>-2x^3+x-5 and x^3-3x-4, we need to multiply each term in the first polynomial by the second polynomial. (So, x^3 - 3x - 4) times ....
-2x^3 = -2x^6 + 6x^4 + 8x^3
x = x^4 - 3x^2 - 4x
-5 = -5x^3 + 15x + 20
If we add all these together, we get (-2x^6 + 7x^4 + 3x^3 - 3x^2 + 11x + 20)</span>

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3 years ago

Four years ago, Jane was twice as old as Sam. Four years on from now, Sam will be 3/4 of Jane's age. How old is Jane now?

Sever21 [200]

How old is sam? We need that first

8 0

3 years ago

Read 2 more answers

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

Jane earns a 5.5% commission on the selling price of each home she sells. She earned $9,020 in commission on the sale of a home.

ratelena [41]

Answer:

"$164000" is the appropriate answer.

Step-by-step explanation:

Let the selling price be "s".

As per the question,

If Jane take 5.5% of selling price i.e.,

= $0.055\times s$

Given commission,

= $9,020

then,

The selling price will be:

⇒ $0.055\times s=9020$

$s=\frac{9020}{0.055}$

$=164000$ ($)

6 0

2 years ago

Will give brainliest to whoever answers first and correct

aliina [53]

Answer:

A is f(t)=-(t-4)^2+50

B is 50. the top of the graph is at 50

C is at 4. that is the x value when y=50

If you need the graph put the answer from question A into Desmos

4 0

3 years ago