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2 years ago

6 times 1/5 less than 1 , equal to 1, or greater than 1?

Mathematics

1 answer:

denpristay [2]2 years ago

3 0

6 times 1/5 (which is also 0.2) is greater than 1

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a sphere and a cone have the same volume. each figure has a radius of 3 inches. what is the height of cone

jok3333 [9.3K]

(4/3) Pi r^3 = (1/3) Pi r^2 h
Solving for h = 4r

8 0

2 years ago

8a to the power of 6 times a to the power of 4 over 4a to the power of -3

svetlana [45]

Answer:

(8a)⁻¹⁸

1 / (8a)¹⁸

Step-by-step explanation:

(((8a)^6a)^4/4a))^-3

((8a)^6a*4/4a)^-3

((8a)^6)^-3

(8a)^6*(-3)

(8a)⁻¹⁸

1 / (8a)¹⁸

4 0

2 years ago

Solve this system of equations. <br><br> 12x − 18y = 27 <br> 4x − 6y = 10

kifflom [539]

Answer:

no solution

Step-by-step explanation:

12x − 18y = 27

4x − 6y = 10

Multiply the second equation by -3 so we can eliminate x

-3(4x − 6y) = 10*-3

-12x +18y = -30

Add the first equation to this new equation

12x − 18y = 27

-12x +18y = -30

------------------------

0 = -3

This is not a true equation, so there is no solution

7 0

2 years ago

Any 10th grader solve it <br>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

2 years ago

Need this very badly

NeTakaya

((5x5)/2)x2=50 square ft
(8x5)x2= 80 square ft
8x7= 56 square ft
50+80+56=186 square ft
D) 186 square ft

4 0

2 years ago