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tankabanditka [31]
3 years ago
6

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

Mathematics
1 answer:
VLD [36.1K]3 years ago
4 0
<span>f '(x) = [(-40x +11)(7x - 9) - 7(-4x +3)(5x + 1)]/(7x - 9)2</span> 
<span>= [(-280x2<span> + 360x + 77x - 99) - 7(-20x</span>2<span> - 4x + 15x + 3)]/(7x - 9)</span>2</span> 
<span>= [(-280x2<span> + 437x - 99) + (140x</span>2<span> + 28x - 105x - 21)]/(7x - 9)</span>2</span> 
<span>= (-140x2<span> +360x - 120)/(7x - 9)</span><span>2
</span></span>
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}

<em>Hence the limit of the function </em>\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
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