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3 years ago

How do I solve this I’m dumb

Mathematics

1 answer:

laila [671]3 years ago

8 0

-1/4 + f/8 = 1/2.
The LCD is 8. Rewrite this equation as -8/4 + 8f/8 = 8(1/2) and then reduce the result:

-2 + f = =4

Then f = 6.

You should check this result.

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Y = 2f(g(x))

IrinaK [193]

Yes you used the chain rule properly to follow the correct steps to get the right answer. Great job.

If you wanted, you can come up with examples for f(x) and g(x) to help confirm the answer. A quick way to do this is to use something like GeoGebra to help graph the two expressions and you'll notice that the curves match up perfectly (indicating equivalent expressions). Note: GeoGebra can handle derivatives through the Derivative[] comand or you can type the function in the input bar with a tickmark after it to tell GeoGebra to derive the function.

7 0

4 years ago

Solve for x.<br>8 + 6x<br>4*+2

zheka24 [161]

Answer:

x = 0

Step-by-step explanation:

8 + 6x = 4 * 2

If this is the equation then start with multiplying 4 by two and subtracting 8 from both sides

6x = 0

Divide 6 from both sides

x = 0

To test this, insert the value of x which is 0

8 + 6(0) = 4 * 2

Multiply 6 and 0

8 + 0 = 4 * 2

Multiply 4 and 2

8 + 0 = 8

Add

8 = 8

This equation is true as you can see so x is equal to 0.

Hope this helps.

- Que

5 0

4 years ago

Read 2 more answers

Integrate dx/3sinx+4cosx

german

$\displaystyle\int\frac{\mathrm dx}{3\sin x+4\cos x}$

A standard approach would be the tangent half-angle substitution:

$t=\tan\dfrac x2\implies\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx$

Then

$\sin x=2\sin\dfrac x2\cos\dfrac x2\implies\sin x=\dfrac{2t}{1+t^2}$

$\cos x=\cos^2\dfrac x2-\sin^2\dfrac x2\implies\cos x=\dfrac{1-t^2}{1+t^2}$

from which we get

$\mathrm dx=\dfrac2{1+t^2}\,\mathrm dt$

So the integral becomes

$\displaystyle\int\frac{\frac2{1+t^2}}{\frac{6t}{1+t^2}+\frac{4(1-t^2)}{1+t^2}}\,\mathrm dt=\int\frac{\mathrm dt}{3t+2(1-t^2)}=-\int\frac{\mathrm dt}{2t^2-3t-2}$

Rewrite the denominator as

$2t^2-3t-2=(2t+1)(t-2)$

and expand the integrand into its partial fractions:

$\dfrac1{2t^2-3t-2}=\dfrac15\left(\dfrac1{t-2}-\dfrac2{2t+1}\right)$

We have

$\displaystyle-\frac15\int\frac1{t-2}-\frac2{2t+1}\,\mathrm dt=-\frac15(\ln|t-2|-\ln|2t+1|)+C$

$=\dfrac15\ln\left|\dfrac{2t+1}{t-2}\right|+C$

$=\dfrac15\ln\left|\dfrac{2\tan\frac x2+1}{\tan\frac x2-2}\right|+C$

6 0

3 years ago

Any answers? My last question. ;-;

faltersainse [42]

? 42
sorry if i’m wrong

7 0

3 years ago

Read 2 more answers

7 2\3+ (5 1\6-1 5\9)

lorasvet [3.4K]

Start off with: 5 1/6 - 1 5/9
turn it into an entire fraction: 31/6 - 14/9
make a common denominator (which is 18): 93/18 - 28/18 = 93 - 28 = 65/18
65/18 + 7 2/3 is the new equation
7 2/3 = 23/3
23/3 = 138/18
65/18 + 138/18 = 203/18
Simplify: 203/18 = 11 5/18
Answer: 11 5/18

3 0

3 years ago

Read 2 more answers