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3 years ago

If f(x) = x^2 and h(x) = 2-x, then hf (-3) =

Mathematics

1 answer:

zheka24 [161]3 years ago

7 0

Answer:

hf(-3) = -7

Step-by-step explanation:

f(x) = x²

h(x) = 2 - x

To find hf(x), we will substitute f(x) function as x into the h(x) function.

h(x) = 2 - x

hf(x) = 2 - (x)²

hf(x) = 2 - x²

How that we have the hf(x) function, to find hf(-3), we substitute x = -3 into the new function.

hf(x) = 2 - x²

hf(-3) = 2 - (-3)² = 2 - 9 = -7

You might be interested in

translate the following sentence into an equation. the sum of twice a number x and 13 is two less than three times x

sergiy2304 [10]

I’m pretty sure the equation would be

2x + 13 < 3x - 2

4 0

3 years ago

In order to remove water from a flooded basement, two pumps, each rated at 40 gallons per minute, are used. After half an hour,

Scrat [10]

Answer:

3600 gallons of water was removed from the basement.

Step-by-step explanation:

Given:

Rate of each pump = 40 gallons/min

We need to find find the number of gallons of water removed from the basement.

Solution:

Now Given:

After half an hour, the one pump burns out.

Now we know that;

1 hour = 60 mins

$\frac12$ hour = $\frac{60}{2}=30\ mins$

Now for 30 mins both the pumps were working and removing the water.

So we can say that;

Water remove for first 30 mins = $40\times2\times30=2400\ gallons$

Also Given:

the second pump finishes removing the water half an hour later.

So we can say that;

Water removed for next 30 mins = $40\times 30\times1 =1200\ gallons$

Now we can say that;

Total water removed from the basement is equal to sum of Water remove for first 30 mins and Water removed for next 30 mins .

framing in equation form we get;

Total water removed from the basement = $2400+1200=3600\ gallons.$

Hence 3600 gallons of water was removed from the basement.

6 0

3 years ago

G(x) = -2x^3 – 15x^2 + 36x

shusha [124]

Consider the function $G(x) = -2x^3 - 15x^2 + 36x.$ First, factor it:

$G(x) = -2x^3 - 15x^2 + 36x=-x(2x^2+15x-36)=\\ \\=-x\cdot 2\cdot \left(x-\dfrac{-15-\sqrt{513}}{4}\right)\cdot \left(x-\dfrac{-15+\sqrt{513}}{4}\right).$

The x-intercepts are at points $\left(\dfrac{-15-\sqrt{513} }{4},0\right),\ (0,0),\ \left(\dfrac{-15+\sqrt{513} }{4},0\right).$

1. From the attached graph you can see that

function is positive for $x\in \left(-\infrty, \dfrac{-15-\sqrt{513} }{4}\right)\cup \left(0,\dfrac{-15+\sqrt{513} }{4}\right);$
function is negative for $x\in \left(\dfrac{-15-\sqrt{513} }{4},0\right)\cup \left(\dfrac{-15+\sqrt{513} }{4},\infty\right).$