All categories

2 years ago

What is the value of f[g(3)] for the functions f(x) =x+2 and g(x) = 5x - 1

Mathematics

1 answer:

alexdok [17]2 years ago

7 0

Answer:

Step-by-step explanation:

In this case we know the composition of f and g is

$f(g(x))=f(5x-1)=5x-1+2=5x+1$

Thus if we change x by 3 we get

$f(g(3))=5(3)+1=16$

You might be interested in

The circumference of a circle is 9 pi.What is the area,in square inches,of a circle?Express your answer in terms of pi.

Alla [95]

Answer:

20.25π

Step-by-step explanation:

The circumference (C) of a circle is calculated using the formula

C = 2πr ← r is the radius

given C = 9π, then

2πr = 9π ( divide both sides by 2π )

r = $\frac{9\pi }{2\pi }$ ( cancel the π on numerator/denominator )

= 4.5

The area (A) of a circle is calculated using the formula

A = πr² = π × 4.5² = 20.25π

5 0

3 years ago

Find the values of x, y, and z. The diagram is not to scale.

Kruka [31]

we know that

The sum of the internal angles in the triangle must be $180$ degrees

see the attached figure with letters to better understand the problem

Step $1$

Find the measure of the angle x

In the triangle ABC

$63\°+36\°+x\°=180\°$

solve for x

$99\°+x\°=180\°$

$x\°=180\°-99\°$

$x=81\°$

therefore

the answer Part a) is

the measure of angle x is $81\°$

Step $2$

Find the measure of the angle z

we know that

$x+z=180\°$ --------> by supplementary angles

substitute the value of x

$81\°+z=180\°$

$z=180\°-81\°$

$z=99\°$

therefore

the answer Part b) is

the measure of angle z is $99\°$

Step $3$

Find the measure of the angle y

In the triangle ACD

$13\°+z\°+y\°=180\°$

solve for y

$13\°+99\°+y\°=180\°$

$112\°+y\°=180\°$

$y\°=180\°-112\°$

$y\°=68\°$

therefore

the answer Part c) is

the measure of angle y is $68\°$

5 0

3 years ago

Read 2 more answers

WILL CHOOSE BRAINLIEST

fredd [130]

The volume of the first cube is (5h^2)^3, while the volume of the second cube is (3k)^3, so their total volume is (5h^2)^3 + (3k)^3. We can use the special formula for factoring a sum of two cubes:
x^3 + y^3 = (x + y)(x^2 - xy + y^2)
(5h^2)^3 + (3k)^3 = (5h^2 + 3k)((5h^2)^2 - (5h^2)(3k) + (3k)^2)
= (5h^2 + 3k)(25h^4 - 15(h^2)(k) + 9k^2)
This is the second of the given choices.

5 0

3 years ago

Read 2 more answers

2. Tessa used the computer every day. Her mother said Tessa could only use the

Vlada [557]

Answer:

3/4 of an hour

Step-by-step explanation:

To find the leftover time, subtract the total allotted time by the time used.

First make sure they have common denominators.

1 1/2 = 3/2 = 6/4

6/4 - 3/4 = 3/4

Tessa has 3/4 of an hour left.

7 0

2 years ago

Determine if the columns of the matrix form a linearly independent set. Justify your answer. [Start 3 By 4 Matrix 1st Row 1st Co

Volgvan

Answer:

Linearly Dependent for not all scalars are null.

Step-by-step explanation:

Hi there!

1)When we have vectors like $v_{1},v_{2},v_{3}, ...$ we call them linearly dependent if we have scalars $a_{1},a_{2},a_{3},...$ as scalar coefficients of those vectors, and not all are null and their sum is equal to zero.

$a_{1}\vec{v_{1}}+a_{2}\vec{v_{2}}+a_{3}\vec{v_{3}}+...a_{m}\vec{v_{m}}=0$

When all scalar coefficients are equal to zero, we can call them linearly independent

2) Now let's examine the Matrix given:

$\begin{bmatrix}1 &-2 &2 &3 \\ -2 & 4 & -4 &3 \\ 0&1 &-1 & 4\end{bmatrix}$

So each column of this Matrix is a vector. So we can write them as:

$\vec{v_{1}}=\left \langle 1,-2,1 \right \rangle,\vec{v_{2}}=\left \langle -2,4,-1 \right \rangle,\vec{v_{3}}=\left \langle 2,-4,4 \right \rangle\vec{v_{4}}=\left \langle 3,3,4 \right \rangle$ Or

Now let's rewrite it as a system of equations:

$a_{1}\begin{bmatrix}1\\ -2\\ 0\end{bmatrix}+a_{2}\begin{bmatrix}-2\\ 4\\ 1\end{bmatrix}+a_{3}\begin{bmatrix}2\\ -4\\ -1\end{bmatrix}+a_{4}\begin{bmatrix}3\\ 3\\ 4\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}$

2.1) Since we want to try whether they are linearly independent, or dependent we'll rewrite as a Linear system so that we can find their scalar coefficients, whether all or not all are null.

Using the Gaussian Elimination Method, augmenting the matrix, then proceeding the calculations, we can see that not all scalars are equal to zero. Then it is Linearly Dependent.

$\left ( \left.\begin{matrix}1 &-2 &2 &3 \\ -2 &4 &-4 &3 \\ 0 & 1 &-1 &4 \\ \end{matrix}\right|\begin{matrix}0\\ 0\\ 0\end{matrix} \right )R_{1}\times2 +R_{2}\rightarrow R_{2}\left ( \left.\begin{matrix}1 &-2 &2 &3 \\ 0 &0 &9 &0\\ 0 & 1 &-1 &4 \\ \end{matrix}\right|\begin{matrix}0\\ 0\\ 0\end{matrix} \right )\ R_{2}\Leftrightarrow R_{3}\left ( \left.\begin{matrix}1 &-2 &2 &3 \\ 0 &1 &-1 &4 \\ 0 &0 &9 &0 \\ \end{matrix}\right|\begin{matrix}0\\ 0\\ 0\end{matrix} \right )$ $\left\{\begin{matrix}1a_{1} &-2a_{2} &+2a_{3} &+3a_{4} &=0 \\ &1a_{2} &-1a_{3} &+4a_{4} &=0 \\ & & &9a_{4} &=0 \end{matrix}\right.\Rightarrow a_{1}=0, a_{2}=a_{3},a_{4}=0$

$S=\begin{bmatrix}0\\ a_{3}\\ a_{3}\\ 0\end{bmatrix}$

3 0

3 years ago