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3 years ago

What is the inverse of the function f(x) =1/9x + 2?

Mathematics

1 answer:

stira [4]3 years ago

7 0

f^-1 (x)=9x-18 or f(x)=9x+18

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Solve n+m/p = q for the variable n

konstantin123 [22]

N+m/p=q
Multiply by p
Np+m=pq
Subtract m
Np=pq-m
Divide by p
N=(pq-m)/p

8 0

3 years ago

Read 2 more answers

4. Molly had $21 to spend on two notebooks. After buying them he had $13. How much did each notebook cost?

otez555 [7]

21-13=8
8/2=4
Each notebook costs $4
Hope this helps :)

8 0

3 years ago

Read 2 more answers

Find f(0), if not supplied, and then find the equation of the given linear function.

My name is Ann [436]

Answer:

Part 1) $f(0)=2$

Part 2) $f(x)=3x+2$

Step-by-step explanation:

Part 1) Find f(0)

we know that

Looking at the data in the table

When the value of x increases by 1 unit, the value of y increases by 3 units

therefore

When the value of x decreases by 1 unit from 1 to 0, the value of y would decreases 3 units from 5 to 2

thus

$f(0)=2$

This value represent the y-intercept

Part 2) Find the equation of the given linear

Find the slope

take the points (1,5) and (2,8)

$m=(8-5)/(2-1)=3$

The equation of the line in slope intercept form is equal to

$f(x)=mx+b$

we have

$m=3\\b=2$

substitute

$f(x)=3x+2$

3 0

3 years ago

Please can someone help.

enot [183]

Step-by-step explanation:

w = 3(2a + b) - 4

w = 6a + 3b - 4

6a = w - 3b + 4

a = 1/6 * (w - 3b + 4).

6 0

3 years ago

Please solve this, will rate 5 stars and mark as STAR!

Nina [5.8K]

Answer:

$\boxed{5 \cdot \sqrt{2} \cdot \sqrt[6]{5} }$

Step-by-step explanation:

$\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }$

$\sqrt{\sqrt[3]{10} } \implies (10^\frac{1}{3} )^\frac{1}{2} =10^\frac{1}{6} =\sqrt[6]{10}$

$\therefore \sqrt{\sqrt[3]{10} }=\sqrt[6]{10}$

$\text{Solving }\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }$

$250=2 \cdot 5^3$

$\sqrt[3]{250}=\sqrt[3]{2\cdot 5^3}=5 \sqrt[3]{2}$

Once

$\sqrt[6]{2} \cdot \sqrt[6]{5} = \sqrt[6]{10}$

We have

$5 \sqrt[3]{2} \cdot \sqrt[6]{2} \cdot \sqrt[6]{5}$

We can proceed considering the common base of exponentials

$\sqrt[3]{2} \cdot \sqrt[6]{2} = 2^{\frac{1}{3}} \cdot 2^{\frac{1}{6} } = 2^{\frac{3}{6} } = 2^{\frac{1}{2} }=\sqrt{2}$

Therefore,

$5 \sqrt[3]{2} \cdot \sqrt[6]{2} \cdot \sqrt[6]{5} = 5 \cdot \sqrt{2} \cdot \sqrt[6]{5}$

7 0

3 years ago