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3 years ago

If f(x) and f–1(x) are inverse functions of each other and f(x)=2x+5, what is f–1(8)?

Mathematics

2 answers:

yarga [219]3 years ago

6 0

f(x)=2x+5

first find inverse

write it without f(x)

y = 2x + 5

x = 2y + 5 swap the variables, now solve for y

- 5 -5

x - 5 = 2y

/2 /2 divide both sides by 2

(x-5)/2 = y

so f-1(x) = (x-5)/2

f-1(8) = (8-5)/2

= (3)/2

f-1(8) = 3/2

Roman55 [17]3 years ago

4 0

Answer:

3/2

Step-by-step explanation:

f(x)=2x+5

first find inverse (write it without f(x))

y = 2x + 5

x = 2y + 5 (swap the variables, now solve for y)

- 5 -5

x - 5 = 2y

/2 /2 (divide both sides by 2)

(x-5)/2 = y

so f-1(x) = (x-5)/2

f-1(8) = (8-5)/2

= (3)/2

f-1(8) = 3/2

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2 years ago

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

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

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rth term of given arithmetic progression is c

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

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

Now to prove LHS=RHS

Now take LHS

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

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

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

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

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

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

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

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

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ie., $RHS\neq 0$

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

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3 years ago

Plz help

Semenov [28]

This is a classic example of Scientific Notation.
5.3 x 10^5 actually equals 530,000
3.8 x 10^4 actually equals 38,000
Since it is asking for the sum of the numbers you are adding.
Your new equation look like this:
530,000 + 38,000
= 568,000
now all you have to do is count the decimal points. You start from the end zero and count the spaces in between until you get to the tenths place. (Right behind the first number) There are 5
So your answer will be:
B. 5.68 x 10^5.

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3 years ago