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2 years ago

Please answer this question!

Mathematics

1 answer:

bogdanovich [222]2 years ago

5 0

it's answer is 82 but I think

it will wrong

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Which expression is equivalent to 2x2 + (4x - 6x2) + 9 - (6x + 3)?

Troyanec [42]

Answer:

-4 $x^2$ -10x+6

Step-by-step explanation:

I am going to assume the (Something)x2 means (Something) $x^2$ . Then, the expression is 2 $x^2$ -6 $x^2\\$ +4x-6x+9-3=-4 $x^2$ -10x+6

4 0

3 years ago

Read 2 more answers

Lan parents asked him to create a budget For his 1000 monthly income he determines That he would like to save the remaining amou

LiRa [457]

Answer:

for an equal income to everyone you can do 250 to each person adding up to 750 then 250 to save

Step-by-step explanation:

6 0

3 years ago

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

2 years ago

Can someone help me ?

Alex Ar [27]

All you need to do is divide 2.25 by 2!

The answer is 1.125!

I hope this helps! :)

5 0

3 years ago

Cosx - 4 = sinx -4<br> (0,2pi)<br>Find solutions

lutik1710 [3]

$\bf cos(x)-4=sin(x)-4\implies cos(x)=sin(x)\implies \cfrac{cos(x)}{sin(x)}=0
\\\\\\
cot(x)=0\implies \measuredangle x=cot^{-1}(0)\implies \measuredangle x=\frac{\pi }{2}~~,~~\frac{3\pi }{2}$

8 0

3 years ago