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3 years ago

What are the domain and range of the function f(x)=square root x-7 +9?

Mathematics

2 answers:

MArishka [77]3 years ago

8 0

Answer:

Domain is all real numbers greater than or equal to 7. Range is all real numbers greater than or equal to 9.

Step-by-step explanation:

The domain is possible x-values while the range is possible y-values. This question look like

$\sqrt{x - 7} + 9$

The domain of square root function cant have a the radical inside being a negative number because we can't take the sqr root of a negative number and graph it on a cartisean plane We can take the sqr root of 0 . So we set x-7=0. And which x=7 so the domain is all real numbers that are equal to or greater than 7. The range of a sqr root function can only yield any positive y-values. So we must find the lowest point of it. Since the lowest x value we can possible use is 7 we plug it in

$\sqrt{7- 7} + 9$

which equals 9. So the range is all real numbers greater than or equal to 9.

amid [387]3 years ago

4 0

Answer:

I think that the short answer will be C) on edge.

Step-by-step explanation:

Edge 2021.

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11Alexandr11 [23.1K]

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A.Angle
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A line segment is part of a line that is bounded by 2 distinct end points. In this case, the end points are the edges of the index cards.

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2 years ago

Chandra has 15 feet of ribbon to make bows for 6 teams how much ribbon can she use for each bow if she cuts the ribbon into equa

SIZIF [17.4K]

Answer:

2.5 feet for each bow

Step-by-step explanation:

15/6= 2.5

6 0

2 years ago

I can't figure out how to do (i + j) x (i x j)for vector calc

Vinil7 [7]

In three dimensions, the cross product of two vectors is defined as shown below

$\begin{gathered} \vec{A}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k} \\ \vec{B}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \\ \Rightarrow\vec{A}\times\vec{B}=\det (\begin{bmatrix}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {a_1} & {a_2} & {a_3} \\ {b_1} & {b_2} & {b_3}\end{bmatrix}) \end{gathered}$

Then, solving the determinant

$\Rightarrow\vec{A}\times\vec{B}=(a_2b_3-b_2a_3)\hat{i}+(b_1a_3+a_1b_3)\hat{j}+(a_1b_2-b_1a_2)\hat{k}$

In our case,

$\begin{gathered} (\hat{i}+\hat{j})=1\hat{i}+1\hat{j}+0\hat{k} \\ \text{and} \\ (\hat{i}\times\hat{j})=(1,0,0)\times(0,1,0)=(0)\hat{i}+(0)\hat{j}+(1-0)\hat{k}=\hat{k} \\ \Rightarrow(\hat{i}\times\hat{j})=\hat{k} \end{gathered}$

Where we used the formula for AxB to calculate ixj.

Finally,

$\begin{gathered} (\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=(1,1,0)\times(0,0,1) \\ =(1\cdot1-0\cdot0)\hat{i}+(0\cdot0-1\cdot1)\hat{j}+(1\cdot0-0\cdot1)\hat{k} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=1\hat{i}-1\hat{j} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=\hat{i}-\hat{j} \end{gathered}$

Thus, (i+j)x(ixj)=i-j

8 0

11 months ago

I need this done it’s due tmr and I need each problem simplified in the lowest fraction

Ivenika [448]

Answer:

photomath

Step-by-step explanation:

theres an app where you can take a pic of your problem and it would solve it for you,including fractions. I'm pretty sure it will simplify your answers too.The app is called photomath

8 0

3 years ago

Need ASAP Evaluate h^-2 g for h = 3 and g = 27. 9 3 1/3

tangare [24]

Answer:

Step-by-step explanation:

h^-2 g

h= 3 and g = 27

(3) ^-2 (27)

1/3^2 (27)

1/9 *27

27/9

5 0

3 years ago