Match each system of equations to its solution represented by an augmented matrix.

GEOMETRY | can some please help me? i'm really confused (20 POINTS + BRAINLIEST)

Mademuasel [1]

This is the last answer because if you reflect it 4 times over the x and y axis you will get back to the same spot

5 0

3 years ago

Please just give me a good explanation on how to reflect the x-axis and y-axis so I can learn it on my own as well as answering

MrMuchimi

Answer:

Alright! BTW, the answer to the question is D :)

Step-by-step explanation:

Reflecting across the x-axis, the x-coordinate ALWAYS remains the same, but the y-coordinate is transformed into its opposite. If you forget this, fold along the x line, and you should be able to figure it out. Reflecting across the y-axis, the y-coordinate ALWAYS remains the same, but the x-coordinate is transformed into its opposite. If you forget this, fold along the y line, and you should be able to figure it out.

Hope I helped!

6 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

A car is `160` inches long. <br><br><br><br> A truck is `7\%` longer than the car.

iragen [17]

Answer:

171.2 inches

Step-by-step explanation:

7% is equaled to 0.07 and multiplied by 160 equals 11.2 then I added that to 160 to get 171.2 inches. I hope im correct.

6 0

2 years ago

Someone Help I will give Branlist for whoever answers first!!!

Assoli18 [71]

Answer:

A its like, getting a plate before you make a sandwich, you have to get a plate first, or else you can't start making your sandwich.

Step-by-step explanation:

you don't have to give brainliest, im just helping lol

6 0

2 years ago

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