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3 years ago

F(x)=( x^4 + 5x^3 - 2x^2 + 5x - 3) d(x)=( x^2 + 1) Divide f(x) by d(x)

1 answer:

4 0

f '(x) = [(-40x +11)(7x - 9) - 7(-4x +3)(5x + 1)]/(7x - 9)2
= [(-280x2 + 360x + 77x - 99) - 7(-20x2 - 4x + 15x + 3)]/(7x - 9)2
= [(-280x2 + 437x - 99) + (140x2 + 28x - 105x - 21)]/(7x - 9)2
= (-140x2 +360x - 120)/(7x - 9)2

i think thats how you would solve it
hope this helps tho:)

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Abby and Alex were practicing free throws. Abby attempted 40 shots and made 36 of them. Alex attempted 60 shots and made 54 of t

ZanzabumX [31]

Answer:

Abby is the answer

Step-by-step explanation:

When you calculate Abby’s free throws you should realize

40 - 36 = 4

However with Alex’s free throws you calculate

60 - 54 = 6

When you compare the two Abby has a higher percentage of attempts because she was more successfu!

3 0

3 years ago

Read 2 more answers

Evaluate the limit with either L'Hôpital's rule or previously learned methods.lim Sin(x)- Tan(x)/ x^3x → 0

Vsevolod [243]

Answer:

$\dfrac{-1}{6}$

Step-by-step explanation:

Given the limit of a function expressed as $\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3}$ , to evaluate the following steps must be carried out.

Step 1: substitute x = 0 into the function

$= \dfrac{sin(0)-tan(0)}{0^3}\\= \frac{0}{0} (indeterminate)$

Step 2: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the function

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ sin(x)-tan(x)]}{\frac{d}{dx} (x^3)}\\= \lim_{ x\to \ 0} \dfrac{cos(x)-sec^2(x)}{3x^2}\\$

Step 3: substitute x = 0 into the resulting function

$= \dfrac{cos(0)-sec^2(0)}{3(0)^2}\\= \frac{1-1}{0}\\= \frac{0}{0} (ind)$

Step 4: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 2

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ cos(x)-sec^2(x)]}{\frac{d}{dx} (3x^2)}\\= \lim_{ x\to \ 0} \dfrac{-sin(x)-2sec^2(x)tan(x)}{6x}\\$

$= \dfrac{-sin(0)-2sec^2(0)tan(0)}{6(0)}\\= \frac{0}{0} (ind)$

Step 6: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 4

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ -sin(x)-2sec^2(x)tan(x)]}{\frac{d}{dx} (6x)}\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^2(x)sec^2(x)+2sec^2(x)tan(x)tan(x)]}{6}\\\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^4(x)+2sec^2(x)tan^2(x)]}{6}\\$

Step 7: substitute x = 0 into the resulting function in step 6

$= \dfrac{[ -cos(0)-2(sec^4(0)+2sec^2(0)tan^2(0)]}{6}\\\\= \dfrac{-1-2(0)}{6} \\= \dfrac{-1}{6}$

Hence the limit of the function $\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3} \ is \ \dfrac{-1}{6}$ .

3 0

3 years ago

Sandra used partial products to find the product of 438 × 17 by multiplying 438 by 1 and 438 by 7 to get 3,066. Find the product

boyakko [2]

Let us determine the product of 438 and 17 by partial products.

Consider $438 = 400+30+8$

and $17 = 10+7$

So, $438 \times 17 = (400+30+8) \times (10+7)$

$438 \times 17 = (400\times 10)+(30 \times 10)+(8 \times 10) +(400 \times 7)+(30 \times 7)+(8 \times 7)$

= $438 \times 17 = (4000+300+80+2800+210+56)$

= 7446

Therefore, the product of 438 and 17 is 7446.

No, Sandra's answer is not correct.

Because she should have expressed 17 as (10+7), then if she multiplied (438 by 10) and (438 by 7). And, then added the results.Then her answer would be correct.

6 0

3 years ago

2 2/3 x 2 2/3 please

Ivan

Hello
Here the answer goes
7 1/9
Good luck

5 0

2 years ago

you have to bake a lot of brownies for your 6th grade fund raiser! Although you have plenty of flour, cocoa, milk, and oil, you

Mrac [35]

You can make 11 batches of brownies. Divide 8.5 (8 1/2) by .75(3/4)

4 0

2 years ago

Read 2 more answers