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2 years ago

Help needed here please help

Mathematics

1 answer:

bazaltina [42]2 years ago

4 0

Answer:

g(f(x))= (x+3)(x+3)

Step-by-step explanation:

f(x) =x² + 6x +7

g(x)= x + 2

1st Step: Substitute the x in g(x) =x + 2 by the value of f(x)

g(f(x))= (x² + 6x +7) + 2

g(f(x))= x² + 6x + 9

2nd Step: Simplify

g(f(x))= x² + 6x + 9

g(f(x))= (x+3)(x+3)

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Select all of the following that are quadratic equations.

kifflom [539]

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3 years ago

Simplify the expression 6h +(-7.1d) – 17 + 4d – 2.4h

Alexxx [7]

Answer:

11.1d + 3.2h - 15

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Explain how to find surface area of a rectangular prism. Be specific and use examples.

Leya [2.2K]

As you can see in the picture above, there are six faces of a rectangular prism; two are formed with dimensions width and height, two are formed by the dimensions length and width, and two are formed by the dimensions length and height. So, if you know the length, width, and height of the rectangular prism, then the formula for the surface area is

=(2⋅ℎ⋅ℎ)+(2⋅ℎ⋅ℎℎ)+(2⋅ℎ⋅ℎℎ)

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3 years ago

Please help me! Find the number x such that f(x) =1

STatiana [176]

Answer:

Step-by-step explanation:

We have the piecewise function:

$f(x) = \left\{ \begin{array}{ll} -\frac{1}{2}x-1 & \quad x \leq -2 \\ x & \quad x > -2 \end{array} \right.$

And we want to find x such that f(x)=1.

So, let's substitute 1 for f(x):

$1 = \left\{ \begin{array}{ll} -\frac{1}{2}x-1 & \quad x \leq -2 \\ x & \quad x > -2 \end{array} \right.$

This has two equations. So, we can separate them into two separate cases. Namely:

$1=-\frac{1}{2}x-1\text{ or } 1=x$

Let's solve for x in each case.

Case I:

We have:

$1=-\frac{1}{2}x-1$

Add 1 to both sides:

$2=-\frac{1}{2}x$

Let's cancel out the fraction by multiplying both sides by -2. So:

$2(-2)=(-2)\frac{-1}{2}x$

The right side cancels:

$-4=x\\$

Flip:

$x=-4$

So, x is -4.

Case II:

We have:

$1=x$

Flip:

$x=1$

This is the solution for our second case.

So, we have:

$x_1=-4\text{ or } x_2=1$

Now, can check to see if we have to to remove solution(s) that don't work.

Note that x=-4 is the solution to our first equation.

The first equation is defined only if x is less than -2.

-4 is less than -2. So, x=-4 is indeed a solution.

x=1 is the solution to our second equation.

The second equation is defined only if x is greater than or equal to -2.

1 is greater than or equal to -2. So, x=1 is also a solution.

Therefore, our two solutions are:

$x_1=-4\text{ or } x_2=1$

Out of our answer choices, we can pick D.

And we're done!

7 0

3 years ago

The area of a mirror is 225 square inches, and its width is 13 3/4 inches. Will the mirror fit in a space that is 15 inches by 1

gavmur [86]

If you multiply length times width to get area, it's the same with division. You do area divided by length or width. The equation would be 225 divided by 13 3/4. 225/13 3/4= 16.36 inches. 16.36 > 16 inches, so the answer would be no. (The mirrors length and width are 13 3/4 inches by 16 9/25)

7 0

3 years ago