All categories

3 years ago

Find the inverse of function f. f(x)=9x+7 A. B. C. D.

Mathematics

1 answer:

9966 [12]3 years ago

8 0

Answer:

y = (x-7)/9

Step-by-step explanation:

y = 9x + 7

x = 9y + 7

x - 7 = 9y

y = (x-7)/9

You might be interested in

Please help. Geometry.

tia_tia [17]

And what's the problem with Geometry?

4 0

3 years ago

What is the approximate distance between two points with coordinates (3, 5) and (-4, -8)? Round your answer to the nearest hundr

Anika [276]

There is a problem,in order to solve this u need a cordanite bride for them to go on.

3 0

3 years ago

Read 2 more answers

What are the key characteristics of 2d figures?

Dafna1 [17]

Flat and idk the other one

3 0

3 years ago

Isaiah brought two dozen donuts to a class party. Half of the donuts had vanilla frosting with sprinkles, 2 had strawberry frost

enot [183]

Answer:

22/24

ALL TOGETHER there is 24 donuts.

take half of 24, or 24 divided by 2, and you get 12.

so we know 12 of the donuts are vanilla with sprinkles.

then 2 had strawberry frosting and the rest had chocolate and sprinkles.

we took 12 away so we only have twelve left and if 2 of those 12 are strawberry with no sprinkles and the rest are chocolate WITH sprinkles, this means 10 are chocolate with sprinkles.

so 10 chocolate and sprinkles +12 vanilla and sprinkles=22 donuts with sprinkles out of 24 donuts total

7 0

3 years ago

Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"

vampirchik [111]

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h$

Employ a standard trick used in proving the chain rule:

$\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h$

The limit of a product is the product of limits, i.e. we can write

$\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)$

The rightmost limit is an exercise in differentiating $\sqrt x$ using the definition, which you probably already know is $\dfrac1{2\sqrt x}$ .

For the leftmost limit, we make a substitution $y=\sqrt x$ . Now, if we make a slight change to $x$ by adding a small number $h$ , this propagates a similar small change in $y$ that we'll call $h'$ , so that we can set $y+h'=\sqrt{x+h}$ . Then as $h\to0$ , we see that it's also the case that $h'\to0$ (since we fix $y=\sqrt x$ ). So we can write the remaining limit as

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}$

which in turn is the derivative of $\tan y$ , another limit you probably already know how to compute. We'd end up with $\sec^2y$ , or $\sec^2\sqrt x$ .

So we find that

$\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$

7 0

3 years ago