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4 years ago

I don’t know what do do help

Mathematics

1 answer:

astraxan [27]4 years ago

7 0

The degree of a polynomial is the greatest exponent. In this case, the degree is 8.

The polynomial is a trinomial because it has 3 terms.

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L'Shonda purchased a container of juice

cupoosta [38]

L'Shonda purchased a container of juice. 80 calories are from carbohydrates.

<u>Solution:</u>

Given that, LShonda purchased a container of juice that contains 200 calories.

The label says that 40% of the calories are from carbohydrates.

We have to find how many calories are from carbohydrates?

Now, according to the given information,

<em>Number of calories from carbohydrates = 40% of the total calories of the juice </em>

Then, calories from carbohydrates = 40% of 200 calories

40% can be written in fraction as $\frac{40}{100}$

$\text { Calories from carbohydrates }=\frac{40}{100} \times 200=40 \times 2=80 \text { calories }$

Hence, 80 calories are from carbohydrates.

5 0

3 years ago

What are 3 points to the solution of the equation 2x-y=4

Sidana [21]

your solution to this problem would be
Y= -2x+4

8 0

3 years ago

How to write a product of its prime factors

matrenka [14]

Write 24 as a product as its prime factor , now think of all the small numbers that can fit into it.that divides into 12, now we can use 2 and write the 12 as 2 x 6 to give 2 2 2 and 3 are all prime numbers so we have our answer.

dose this help?

4 0

4 years ago

Evaluate the integral

cluponka [151]

The value of the integration is

$\int {\frac{-7x^{3} }{x^{3}+1} } \, dx =7x - \frac{7}{3} \ln(x+1) + \frac{7}{6} ln (x^{2} -x+1)-\frac{7\sqrt{3} }{3} tan^{-1} (\frac{2x-1}{\sqrt{3} })+C$ , where C is an integrating constant.

We know that finding the area of the curve's undersurface is the process of integration. To do this, cover the area with as many tiny rectangles as possible, then add up their areas. The sum gets closer to a limit that corresponds to the area under a function's curve. Finding an antiderivative of a function is the process of integration. If a function can be integrated and its integral over the domain is finite with the given bounds, then the integration is definite.

Here, we have to determine the value of the given integral

$\int {\frac{-7x^{3} }{x^{3}+1} } \, dx$ .

So,

$\int {\frac{-7x^{3} }{x^{3}+1} } \, dx = 7 \int {\frac{1-x^{3}-1 }{x^{3}+1} } \, dx$ $= 7 \int [{1-\frac{1 }{x^{3}+1} } ]\, dx =7\int \, dx - 7 \int {\frac{1 }{x^{3}+1} } \, dx$ ...(1)

Now,

$\frac{1}{x^{3} +1} = \frac{1}{(x+1)(x^{2} -x+1)}$

We can write

$\frac{1}{(x+1)(x^{2} -x+1)}=\frac{A}{(x+1)} + \frac{Bx+C}{(x^{2} -x+1)}$

i.e. $1 = A(x^{2} -x+1)+(Bx+C)(x+1)$

i.e. $1 = Ax^{2} -Ax + A+Bx^{2} +Bx+Cx+C$

i.e. $1=(A+B)x^{2} +(-A+B+C)x +(A+C)$

Comparing both sides, we get

$A+B=0 \implies B = -A$ ...(2)

$-A+B+C = 0 \implies -A+(-A)+C=0 \implies -2A+C=0$ ...(3)

$A+C=1$ ...(4)

Subtracting (3) from (4),

$A+C+2A-C=1-0 \implies 3A=1 \implies A=\frac{1}{3}$

From (4), $C= 1-\frac{1}{3} =\frac{3-1}{3} =\frac{2}{3}$

From (2), $B =-\frac{1}{3}$

We get

$\frac{1}{x^{3}+1 } =\frac{1}{3} \frac{1}{(x+1)} - \frac{1}{3} \frac{x-2}{x^{2} -x+1}$

$=\frac{1}{3} \frac{1}{(x+1)} - \frac{1}{6} \frac{2x-4}{x^{2} -x+1}$

$=\frac{1}{3} \frac{1}{(x+1)} - \frac{1}{6} \frac{2x-1-3}{x^{2} -x+1}$

$=\frac{1}{3} \frac{1}{(x+1)} - \frac{1}{6} \frac{2x-1}{x^{2} -x+1}+\frac{3}{6} \frac{1}{x^{2} -x+1}$

$=\frac{1}{3} \frac{1}{(x+1)} - \frac{1}{6} \frac{2x-1}{x^{2} -x+1}+\frac{1}{2} \frac{1}{x^{2} -x+1}$

Integrate both sides,

$\int \frac{1}{x^{3}+1 }\, dx =\frac{1}{3} \int \frac{1}{(x+1)} \, dx - \frac{1}{6} \int \frac{2x-1}{x^{2} -x+1}\, dx+\frac{1}{2} \int \frac{1}{x^{2} -x+1}\, dx$

i.e. $\int \frac{1}{x^{3}+1 }\, dx =\frac{1}{3} \ln(x+1) - \frac{1}{6} ln (x^{2} -x+1)+\frac{\sqrt{3} }{3} tan^{-1} (\frac{2x-1}{\sqrt{3} })$

From (1),

$\int {\frac{-7x^{3} }{x^{3}+1} } \, dx =7x - \frac{7}{3} \ln(x+1) + \frac{7}{6} ln (x^{2} -x+1)-\frac{7\sqrt{3} }{3} tan^{-1} (\frac{2x-1}{\sqrt{3} })+C$

Therefore, the value of the integration is

$\int {\frac{-7x^{3} }{x^{3}+1} } \, dx =7x - \frac{7}{3} \ln(x+1) + \frac{7}{6} ln (x^{2} -x+1)-\frac{7\sqrt{3} }{3} tan^{-1} (\frac{2x-1}{\sqrt{3} })+C$ , where C is an integrating constant.

Learn more about integrations here -

brainly.com/question/7541827

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6 0

2 years ago

Please help me with this question

kompoz [17]

The answer is both suravi and tori hope this helps! :) ✨

5 0

3 years ago