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3 years ago

Which equation has the graph shown?

Mathematics

1 answer:

kykrilka [37]3 years ago

7 0

Answer:

2nd answer choice

Step-by-step explanation:

Graph is increasing by 1

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ice sculpture originally has a height of 74 1/4 inches. The ice sculpture begins to melt and after several hours, the height has

Tomtit [17]

The height of the sculpture now is 68.8125

7 0

3 years ago

Please solve all and work needs to be shown ❤️❤️

Alex73 [517]

1).

a=bh

=(13)(9)

=117in^2

2).

A=1/2h(b1+b2)

=1/2(14)(15+9)

=168cm^2

3).

A=1/2d1d2

=1/2(18)(11)

=99in^2

4 0

3 years ago

Integrala x la a treia ori ln la a doua dx va rog

Studentka2010 [4]

I don't speak Romanian, but the closest translation for this suggests you're trying to compute

$\displaystyle \int x^3 \ln(x)^2 \, dx$

Integrate by parts:

$\displaystyle \int x^3 \ln(x)^2 \, dx = uv - \int v \, du$

where

u = ln(x)² ⇒ du = 2 ln(x)/x dx

dv = x³ dx ⇒ v = 1/4 x⁴

$\implies \displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \int x^3 \ln(x) \, dx$

Integrate by parts again:

$\displaystyle \int x^3 \ln(x) \, dx = u'v' - \int v' du'$

where

u' = ln(x) ⇒ du' = dx/x

dv' = x³ dx ⇒ v' = 1/4 x⁴

$\implies \displaystyle \int x^3 \ln(x) \, dx = \frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx$

So, we have

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \left(\frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx \right)$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \int x^3 \, dx$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \left(\frac14 x^4\right) + C$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac1{32} x^4 + C$

$\boxed{\displaystyle \int x^3 \ln(x)^2 \, dx = \frac1{32} x^4 \left(8\ln(x)^2 - 4\ln(x) + 1\right) + C}$

3 0

2 years ago

Plzz help will mark brainiest Don't do it for the points

UNO [17]

Answer:

I think the answer is B

Step-by-step explanation:

6 0

3 years ago

how do you do this? i don't need it solved i just don't know how to do it. i know the note on the side says how to but i'm still

olchik [2.2K]

Answer:

Perimeter of triangle J= 18.0 units (3 s.f.)

Perimeter of triangle K= 19.0 units (3 s.f.)

Step-by-step explanation:

The length of each side of the triangle can be found using the distance formula below:

$\boxed{d= \sqrt{(y1- y2)^{2} + (x1 - x2)^{2} } }$

Please see the attached pictures for full solution.

5 0

3 years ago