Ask question

Ask question

All categories

iogann1982 [59]

3 years ago

9

How to simplify 24yz^5/32y^2z

1 answer:

Maurinko [17]3 years ago

8 0

24y*z^5/32y^2*z

24/32 = 3/4
y/y^2 = 1/y
z^5/z = z^4/1

Soo now that they are all simplified the answer is 3z^4/4y

You might be interested in

A composition of reflections across two parallel lines is a _____

S_A_V [24]

It's transilation (if that's how you spell it)

8 0

3 years ago

Read 2 more answers

Find the angle measure.<br><br> PLEASE EXPLAIN I NEED TO SHOW MY WORK

oksian1 [2.3K]

Answer:

135

Step-by-step explanation:

Because the two vertical lines are parallel

that answer your question?

3 0

3 years ago

Bu cevaplar mısın acil?

Arisa [49]

Answer:

çok üzgünüm ama biz daha bu konuya geçmedik

6 0

2 years ago

What is the solution set of the equation using the quadratic formula?

oee [108]

(1)

we are given

$x^2+6x+10=0$

we can use quadratic formula

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

now, we can compare and find a , b and c

we get

$a=1,b=6,c=10$

now, we can plug these values into quadratic formula

$x=\frac{-6\pm \sqrt{6^2-4(1)(10)}}{2(1)}$

$x=\frac{-6+\sqrt{6^2-4\cdot \:1\cdot \:10}}{2\cdot \:1}$

we can simplify it

$x=-3+i$

$x=\frac{-6-\sqrt{6^2-4\cdot \:1\cdot \:10}}{2\cdot \:1}$

$x=-3-i$

so, we will get solution

{−3+i, −3−i}.........Answer

(2)

we are given equation as

$x^2-8x+14=0$

Since, Jamal solve this equation by completing square

so, firstly we will move constant term on right side

so, subtract both sides by 14

$x^2-8x+14-14=0-14$

$x^2-8x=-14$

we can write

$-8x=-2\times 4\times x$

so, we will add both sides by 4^2

$x^2-8x+4^2=-14+4^2$

we get

$x^2-8x+16=-14+16$ ..............Answer

4 0

2 years ago

Find the solution of the differential equation that satisfies the given initial condition. dy/dx = xe^y, y(0) = 0

Anna [14]

$\dfrac{\mathrm dy}{\mathrm dx}=xe^y$
$e^{-y}\,\mathrm dy=x\,\mathrm dx$
$-e^{-y}=\dfrac{x^2}2+C$

$y(0)=0$
$\implies-1=C$
$\implies -e^{-y}=\dfrac{x^2}2-1$
$e^{-y}=1-\dfrac{x^2}2$
$-y=\ln\left(1-\dfrac{x^2}2\right)$
$y=-\ln\left(1-\dfrac{x^2}2\right)$

3 0

3 years ago