All categories

3 years ago

The first one is it right or wrong?

Mathematics

1 answer:

gayaneshka [121]3 years ago

3 0

The first one is wrong

(do you need help solving it?)

You might be interested in

Dan thinks of a number

puteri [66]

Answer:

-1

Step-by-step explanation:

2 - 3 = -1

5 0

2 years ago

The middle school population is made up of 55% girls and 45% boys. Of the boys 40% have siblings what percent of the middle scho

tester [92]

Answer:2.5%

Step-by-step explanation:

2.5% of the school has sibloings

55+45=100

100÷40=2.5

8 0

2 years ago

Can someone help me?

Pie

Answer:

x = 12 efg = 132

Step-by-step explanation:

3 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

Giving everything. Plzzz help.

jarptica [38.1K]

Pretty sure it’s A sorry if that’s wrong

8 0

3 years ago

Read 2 more answers