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Yuki888 [10]
3 years ago
12

PLEASE ANSWER ASAP FOR BRAINLEST!!!!!!!!!!!!!!!

Mathematics
1 answer:
IrinaK [193]3 years ago
7 0

Step-by-step explanation:

i think u had done something wrong in equation

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can someone help me out with this?? I am God awful at basic math such as this...apologies if I sound absurdly stupid;;​
GenaCL600 [577]

For graphs, always start with line <em>x</em> then do line <em>y</em>.

4 0
3 years ago
Identify the relationship (complementary, linear pair/supplementary, or vertical) and find the measure of angle b in the image b
iren2701 [21]

Answer:

complementary

b = 45 deg

Step-by-step explanation:

Angles b and 45-deg are complementary since their measures ad to 90 deg.

45 + b = 90

b = 45

6 0
3 years ago
Read 2 more answers
2. Casey says, "I pick a number.
victus00 [196]

Step-by-step explanation:

7,14,21,28,35,42,49,56,63,70

18,36,54,72,90,108,126,144,162,180

4 0
3 years ago
Find the area of the parallelogram whose vertices are given below. ​A(negative 2​,2​) ​B(2​,0​) ​C(10​,3​) ​D(6​,5​)
swat32

Answer: Area of Parallelogram is 28 units squared.

Step-by-step explanation:

Knowing, any two vector sides of a parallelogram sharing the same initial point, we can find the area of parallelogram that two vectors are shaping, using the cross product of these two vectors a×b, where the area is given by

Area= |a×b| ___ (1)

From the given points,we may choose any three of them such and find a two vector and expresses the sides of parallelogram " or one side and a diagonal of parallelogram" ,we may choose for example B ,C and D.

Moreover, we may choose point B, to be common point for the two sides " the initial point of two vector" thus we need to find vector BD and vector BC .

Calculations are given in picture.

5 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
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