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3 years ago

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soldier1979 [14.2K]3 years ago

4 0

there is nothing attached?

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The slope f′(x) at each point (x,y) on a curve y=f(x) is given, along with a point (a,b) on the curve. Use this information to f

Montano1993 [528]

$f'(x)=\dfrac{4x}{1+7x^2}$

Integrating gives

$f(x)=\displaystyle\int\frac{4x}{1+7x^2}\,\mathrm dx$

To compute the integral, substitute $u=1+7x^2$ , so that $\frac27\,\mathrm du=4x\,\mathrm dx$ . Then

$f(x)=\displaystyle\frac27\int\frac{\mathrm du}u=\frac27\ln|u|+C$

Since $u=1+7x^2>0$ for all $x$ , we can drop the absolute value, so we end up with

$f(x)=\dfrac27\ln(1+7x^2)+C$

Given that $f(0)=10$ , we have

$10=\dfrac27\ln1+C\implies C=10$

so that

$\boxed{f(x)=\dfrac27\ln(1+7x^2)+10}$

7 0

3 years ago

Use the LCD to rewrite 34, 58, and 110 with the same denominator pls help

galben [10]

Answer: 30/40, 25/40, 4/40

Step-by-step explanation: Using the Least Common Denominator, I looked at each multiple of each number to determine 40 as the least common denominator, and since we did 4 x 10 to get 40 in the first one, we do 3 x 10 in the first one as well making it 30/40. The second one we did 8 x 5 to get 40, making the numerator (top number) the product of 5 x 5 which is 25 making it 25/40. For the last one, I used 10 x 4 to get 40 meaning I would have to do 1 x 4 to get the numerator (top number) making it 4/40

4 0

3 years ago

Please help ASAP don’t mind the scribbles

hjlf

U = 1

L = 0

R = -2

I = 24

M = 17

H = -3

E = 100

S = -439

O = 11

B = -6

T = -41

D = 19

C = 206

<em>Hope this helped ! :)</em>

8 0

3 years ago

I dont understand this question.Help?

Maurinko [17]

the first step Andre made was

2x+6=3x-8

or, -x+6=-8

And the first step Diego made was

2x+6=3x-8

or, +6=3x-2x

or, 6=x-8

...

4 0

2 years ago

Simplify 1- cos^2 Θ / cos^2 Θ<br><br> Answers<br><br> sin2 θ<br><br> cos2 θ<br><br> tan2 θ<br><br> 1

Tems11 [23]

Answer:

<h3><u /></h3>

<u>Trigonometric identities</u>

<u />

$\large \begin{aligned}\cos^2 \theta + \sin^2 \theta & = 1\\\implies \quad \quad \sin^2 \theta & = 1 - \cos^2 \theta\\\\\tan \theta=\dfrac{\sin \theta}{\cos \theta} \end{aligned}$

Therefore, using the <u>identities</u>, the given expression can be <em>simplified</em> as follows:

$\large\begin{aligned}\implies \dfrac{1- \cos^2 \theta}{\cos^2 \theta} & = \dfrac{\sin^2 \theta}{\cos^2 \theta}\\\\& = \tan^2 \theta\end{aligned}$

3 0

2 years ago

Read 2 more answers