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3 years ago

What is 8.1% of 252?

Mathematics

1 answer:

VMariaS [17]3 years ago

7 0

Answer:

20.412

Step by step explanation:

8.1% of 252 = 0.081 × 252 = 20.412

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Find the GCF of the terms of the polynomial. 26x2 + 34x4

Ivanshal [37]

Answer:

D) 2x2

Step-by-step explanation:

i got i right on the test

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4 years ago

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

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

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

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

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

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

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

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

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

$\neq 0$

ie., $RHS\neq 0$

Therefore $LHS\neq RHS$

ie., $\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0$

Hence proved

5 0

3 years ago

Find (fºg)(1) f(x) = x² - 1 g(x) = √x A.0 B.1 C.4 √2 D. √8

Savatey [412]

Given:

The function are

$f(x)=x^2-1$

$g(x)=\sqrt{x}$

To find:

The value of $(f\circ g)(1)$ .

Solution:

We have,

$f(x)=x^2-1$

$g(x)=\sqrt{x}$

We need to find the value of $(f\circ g)(1)$ .

$(f\circ g)(1)=f(g(1))$ $[\because (f\circ g)(x)=f(g(x))]$

$(f\circ g)(1)=f(\sqrt{1})$ $[\because g(x)=\sqrt{x}]$

$(f\circ g)(1)=f(1)$

$(f\circ g)(1)=(1)^2-1$ $[\because f(x)=x^2-1]$

$(f\circ g)(1)=1-1$

$(f\circ g)(1)=0$

The value of $(f\circ g)(1)$ is 0.

Therefore, the correct option is A.

4 0

3 years ago

A company wants to construct an open rectangular box with a volume of 525 in3 so that the length of its base is 3 times its widt

vfiekz [6]

The length of the base 3 times the width. And we need to express the surface. I'm assuming the surface of the base. So, to get the area of the base, multiply w * l.

5 0

3 years ago

Read 2 more answers

Is anyone good at math (geometry)? I need some help!

wel

Yea me what do you need help with

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3 years ago

What is 8.1% of 252?​

What is 8.1% of 252?