All categories

3 years ago

What is the range of h

Mathematics

1 answer:

soldi70 [24.7K]3 years ago

3 0

Answer: Choice D) The h(x) values -4, -2, 2, 4, and 6

Explanation:

The range is the set of all possible y outputs, or h(x) outputs.

We simply list the y coordinate of each blue point shown on the graph.

We cannot say something like -4 ≤ h(x) ≤ 6 because there are gaps where the function is not defined. For instance, h(x) = 0 is not possible, so 0 is not in the range. Put another way: there is no point with y coordinate 0 on the graph, so that's why 0 is not part of the range.

You might be interested in

I need help with 3 and 4 pls

vladimir2022 [97]

Answer U cant answer 4 without answering 3 good luck

Step-by-step explanation:

7 0

3 years ago

Solve the matrix and prove that it is equal 0

Art [367]

Step-by-step explanation:

$\underline{ \underline{ \text{Given}}} :$

$\tt{ {A}^{T} = \begin{bmatrix} 2 & - 4 \\ 4 & 3 \\ \end{bmatrix}}$

$\underline{ \underline { \text{To \: Find}}} :$

$\sf{ {A}^{2} - 5A+ 22I= 0}$

$\underline{ \underline{ \text{Solution}}} :$

The new matrix obtained from a given matrix by interchanging it's rows and columns is called the transposition of matrix. It is denoted by $\sf{ {A}^{T}}$ . Again , Interchange it's rows and columns in order to find ' A '.

$\tt{A = \begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix}}$

Now , LEFT HAND SIDE ( L.H.S )

$\tt{ {A}^{2} - 5A+ 22I}$

Here, I refers to identity matrix. A diagonal matrix in which all the elements of leading diagonal are 1 ( unit ) is called unit or identity matrix.

⟼ $\begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix} \times \begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix} - 5 \times \begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix} + 22 \times \begin{bmatrix} 1 & 0 \\ 0 & 1\\ \end{bmatrix}$

⟼ $\begin{bmatrix} 2 \times 2 + 4 \times ( - 4)& 2 \times 4 + 4 \times 3 \\ - 4 \times 2 + 3 \times ( - 4) & - 4 \times 4 + 3 \times 3 \\ \end{bmatrix} - \begin{bmatrix} 10 & 20 \\ - 20& 15 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}$

⟼ $\begin{bmatrix} 4 + ( - 16) & 8 + 12 \\ - 8 + ( - 12) & - 16 + 9 \\ \end{bmatrix} - \begin{bmatrix} 10 & 20 \\ - 20 & 15 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}$

⟼ $\begin{bmatrix} - 12 & 20\\ - 20& - 7 \\ \end{bmatrix} - \begin{bmatrix} 10 & 20 \\ - 20 & 15 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}$

⟼ $\begin{bmatrix} - 22 & 0 \\ 0& - 22 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}$

⟼ $\begin{bmatrix} - 22 + 22 & 0 + 0 \\ 0 + 0 & - 22 + 22 \\ \end{bmatrix}$

⟼ $\begin{bmatrix} 0 & 0\\ 0 & 0 \\ \end{bmatrix}$

⟼ $\sf{0}$

RIGHT HAND SIDE ( R.H.S ) : 0

L.H.S = R.H.S [ Hence , proved ! ]

Hope I helped ! ♡

Have a wonderful day / night ! ツ

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

4 0

3 years ago

A candy store wants to create a mix of two hard candies. One is priced at $5.45 per pound, and the other is priced a $7.33 per p

Art [367]

$x-\ amount\ of\ candy\ which\ cost\ 5,45\$\\\\y-\ amount\ of\ candy\ which\ cost\ 7,33\$\\\\ 7,33*y+11*5,45=6,14(11+y)\\\\
7,33y+59,95=67,54+6,14y\ \ \ \ | subtract\ 6,14y\\\\
1,19y=67,54-59,95\\\\
1,19y=7,59\ \ \ \ | divide\ by\ 1,19\\\\y=6,38\ pound\\\\
6,38\ pounds\ should\ be\ mixed\ with\ those\ wich\ cost\ 5,45.$

3 0

3 years ago

Jordan wants to solve the following system using the elimination method:

sergeinik [125]

Answer:

see explanation

Step-by-step explanation:

Given the 2 equations

2x + 3y = 10 → (1)

x + y = 7 → (2)

Multiplying (2) by - 3 and adding to (1) will eliminate y

- 3x - 3y = - 21 → (3)

Add (1) and (3) term by term to eliminate y, that is

- x = - 11 ( multiply both sides by - 1 )

x = 11

Substitute x = 11 into either of the 2 equations and evaluate for y

Substituting into (2)

11 + y = 7 ( subtract 11 from both sides )

y = - 4

Solution is (11, - 4 )

6 0

4 years ago

How many terms are in this expression? 6m + 8 + m - 2n

Ne4ueva [31]

Answer:

There are 4 terms in this expression.

Step-by-step explanation:

A term can be a signed number, a variable, or a constant multiplied by a variable or variables. This document can help you remember what is what in expressions. https://www.usatestprep.com/modules/gallery/file.php?id=35913

Hope this helps!

5 0

3 years ago