All categories

3 years ago

2.8 to 3.6 Find the ratio to the first # To the 2nd #

Mathematics

1 answer:

Jlenok [28]3 years ago

3 0

Answer:

1:4 & 1:2

Step-by-step explanation:

divide first set by 2 a piece and the second by 3

You might be interested in

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

3 years ago

Please explain with working!!! Find the set of values of x that satisfy the inequality 9x^2-15x<6

Oksi-84 [34.3K]

Answer:

$-1/3$

Step-by-step explanation:

When solving a quadratic inequality, first solve it normally like you would for a normal quadratic equation. We have:

$9x^2-15x$

Ignore the less than sign and replace it with an equal sign and solve the quadratic for its zeros:

$9x^2-15x=6$

Subtract 6 from both sides:

$9x^2-15x-6=0$

Divide everything by 3:

$3x^2-5x-2=0$

Factor. Find two numbers that equal (3)(-2)=-6 that add up to -5.

-6 and 1 works. Thus:

$3x^2-6x+x-2=0\\3x(x-2)+1(x-2)=0\\(3x+1)(x-2)=0$

Find the x using the Zero Product Property:

$3x+1=0 \text{ or }x-2=0\\x=-1/3\text{ or }x=2$

Now, we need to replace the equal signs with symbols again. To do so, we need to test which symbol to place. Let's do the first zero first.

So, the first zero is:

$x=-1/3$

Assume that the correct symbol is >. Thus,

$x>-1/3$

Now, pick any number that is greater than -1/3. I'll pick 0 since it's the easiest. Now, plug 0 back into the very original inequality. If it works, then the sign is correct, if it doesn't, then simply use the opposite one. Therefore:

$9x^2-15x$