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2 years ago

Please give me answer please

Mathematics

1 answer:

777dan777 [17]2 years ago

7 0

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A^3b^2 divided by a^-1b^-3

Naya [18.7K]

Answer:

$\frac{a^3 b^2}{\frac{1}{a} \frac{1}{b^3}}$

And simplifying we got:

$a^3 b^2 a b^3$

$a^3 a b^2 b^3 = a^{3+1} b^{2+3} = a^4 b^5$

Step-by-step explanation:

We want to simplify the following expression:

$\frac{a^3 b^2}{a^{-1} b^{-3}}$

And we can rewrite this expression using this property for any number a:

$a^{-1}= \frac{1}{a}$

And using this property we have:

$\frac{a^3 b^2}{\frac{1}{a} \frac{1}{b^3}}$

And simplifying we got:

$a^3 b^2 a b^3$

$a^3 a b^2 b^3 = a^{3+1} b^{2+3} = a^4 b^5$

3 0

3 years ago

What are the zeros of the polynomial function f(x) = x3 – 4x2 – 12x

AleksAgata [21]

fx=x3−4x2−12x.‌fx/x‌=‌x3−4x2−12x/x‌

f=x2−4x−12

4 0

3 years ago

Read 2 more answers

Franco is reading a book with 482 pages he reads 30 pages each night. how many pages will he read in 7 nights? show your work

Sergeu [11.5K]

The number of pages in the book is actually irrelevant.

Franco reads 30 pages in 1 night. To find how many pages he reads in 7 nights, just multiply by 7.

30*7=210

Franco reads 210 pages in 7 nights.

Solving with Equivalent Ratios:

The ratio of pages he reads per night is 30:1. We need to figure out the ratio x:7, where the variable x is the number of pages he reads in 7 nights.

30:1=x:7

To find x we need to see the similarity between both ratios. This is found by dividing two of the equal sides, here it would be 7 and 1.

7/1=7

This means that the second ratio is 7 times the first ration. To find x, just multiply 30 by 7.

30*7=210

Franco reads 210 pages every 7 nights.

4 0

3 years ago

Read 2 more answers

Any 10th grader solve it for 50 points

kkurt [141]

Answer:

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0$ is proved for the sum of pth, qth and rth terms of an arithmetic progression are a, b,and c respectively.

Step-by-step explanation:

Given that the sum of pth, qth and rth terms of an arithmetic progression are a, b and c respectively.

First term of given arithmetic progression is A

and common difference is D

ie., $a_{1}=A$ and common difference=D

The nth term can be written as

$a_{n}=A+(n-1)D$

pth term of given arithmetic progression is a

$a_{p}=A+(p-1)D=a$

qth term of given arithmetic progression is b

$a_{q}=A+(q-1)D=b$ and

rth term of given arithmetic progression is c

$a_{r}=A+(r-1)D=c$

We have to prove that

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)=0$

Now to prove LHS=RHS

Now take LHS

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)$

$=\frac{A+(p-1)D}{p}\times (q-r)+\frac{A+(q-1)D}{q}\times (r-p)+\frac{A+(r-1)D}{r}\times (p-q)$

$=\frac{A+pD-D}{p}\times (q-r)+\frac{A+qD-D}{q}\times (r-p)+\frac{A+rD-D}{r}\times (p-q)$

$=\frac{Aq+pqD-Dq-Ar-prD+rD}{p}+\frac{Ar+rqD-Dr-Ap-pqD+pD}{q}+\frac{Ap+prD-Dp-Aq-qrD+qD}{r}$

$=\frac{[Aq+pqD-Dq-Ar-prD+rD]\times qr+[Ar+rqD-Dr-Ap-pqD+pD]\times pr+[Ap+prD-Dp-Aq-qrD+qD]\times pq}{pqr}$

$=\frac{Arq^{2}+pq^{2} rD-Dq^{2} r-Aqr^{2}-pqr^{2} D+qr^{2} D+Apr^{2}+pr^{2} qD-pDr^{2} -Ap^{2}r-p^{2} rqD+p^{2} rD+Ap^{2} q+p^{2} qrD-Dp^{2} q-Aq^{2} p-q^{2} prD+q^{2}pD}{pqr}$