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2 years ago

Can somebody tell me which one it is? options are (BE, CF, DG, GF)

Mathematics

1 answer:

mixas84 [53]2 years ago

7 0

Answer:

Step-by-step explanation:

Mark as brainliest please

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Girardo is using the model below to solve the equation 3x+1=4x+(-4)

V125BC [204]

Answer:

x=5

Step-by-step explanation:

3x+1= 4x+(-4)

-4×+1= -4x-1 combine like terms

-1x= -5 divided by -1

x= 5

5 0

2 years ago

Can someone answer this?

sesenic [268]

C.. third one

Hope it helps you

6 0

3 years ago

Find the 12th term of thsi sequence 5 12 19

enyata [817]

It's adding 7 each time. So...

5, 12, 19, 26, 33, 40, 47, 54, 61, 68, 75, and

82

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3 years ago

The graph shows the relationship between the number of apples and their cost. For the point (1, r), what is the value of r?

3241004551 [841]

Answer:

Step-by-step explanation:

When you check the point that is on X, it is a little over halfway to the point 1,2. It would be like 1,1.5 or 1,1.6.

5/3 is the equivalent to 1.6 so that answer would make the most sense.

4 0

2 years ago

Read 2 more answers

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

5 0

1 year ago