All categories

3 years ago

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

You might be interested in

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