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3 years ago

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What’s the square root of 81

2 answers:

dalvyx [7]3 years ago

8 0

Answer:

The square root of 81 is 9

Wittaler [7]3 years ago

5 0

The Square root of 81 would be 9.

Hope this helps, have a great day!

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I need the answers to this please and thank you:)

Liono4ka [1.6K]

Q # 1

Explanation

Given the parabola

$f\left(x\right)=\left(x-3\right)^2-1$

Openness

It OPENS UP, as 'a=1' is positive.

Finding Vertex

The vertex of an up-down facing parabola of the form

$y=ax^2+bx+c\:\mathrm{is}\:x_v=-\frac{b}{2a}$

$\mathrm{Rewrite}\:y=\left(x-3\right)^2-1\:\mathrm{in\:the\:form}\:y=ax^2+bx+c$

$y=x^2-6x+8$

$a=1,\:b=-6,\:c=8$

$x_v=-\frac{\left(-6\right)}{2\cdot \:1}$

$x_v=3$

Finding $y_v$

$y_v=3^2-6\cdot \:3+8$

$y_v=-1$

So vertex is:

$\left(3,\:-1\right)$

Horizontal Translation

$y=\left(x-3\right)^2$ moves the graph RIGHT 3 units.

Vertical Translation

$f\left(x\right)=\left(x-3\right)^2-1$ moves the graph DOWN 1 unit.

Stretch or Compress Vertically

As $a = 1$ , so it does not affect the stretchiness or compression.

Q # 2

Explanation:

$f\left(x\right)=-\left(x+1\right)^2-2$

Openness

It OPENS DOWN, as 'a=-1' is negative.

Vertex

$\mathrm{Rewrite}\:y=-\left(x+1\right)^2-2\:\mathrm{in\:the\:form}\:y=ax^2+bx+c$

$y=-x^2-2x-3$

$a=-1,\:b=-2,\:c=-3$

$x_v=-\frac{\left(-2\right)}{2\left(-1\right)}$

$x_v=-1$

$\mathrm{Plug\:in}\:\:x_v=-1\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}$

$y_v=-2$

So vertex is:

$\left(-1,\:-2\right)$

Horizontal Translation

$y=\left(x+1\right)^2$ moves the graph LEFT 1 unit.

Vertical Translation

$f\left(x\right)=\left(x+1\right)^2-2$ moves the graph DOWN 2 unit.

Stretch or Compress Vertically

As $a = -1$ < 0, so it is either stretched or compressed.

Q # 3

Explanation:

$f\left(x\right)=\frac{1}{3}\left(x-4\right)^2+6$

It OPENS UP, as 'a=1/3' is positive.

Vertex

$\mathrm{Rewrite}\:y=\frac{1}{3}\left(x-4\right)^2+6\:\mathrm{in\:the\:form}\:y=ax^2+bx+c$

$y=\frac{1\cdot \:x^2}{3}-\frac{8x}{3}+\frac{34}{3}$

$a=\frac{1}{3},\:b=-\frac{8}{3},\:c=\frac{34}{3}$

$x_v=-\frac{\left(-\frac{8}{3}\right)}{2\left(\frac{1}{3}\right)}$

$x_v=4$

Finding $y_v$

$y_v=\frac{1\cdot \:4^2}{3}-\frac{8\cdot \:4}{3}+\frac{34}{3}$

$y_v=6$

So vertex is:

$\left(4,\:6\right)$

Horizontal Translation

$f\left(x\right)=\left(x-4\right)^2$ moves the graph RIGHT 4 units.

Vertical Translation

$f\left(x\right)=}\left(x-4\right)^2+6$ moves the graph UP 6 unit.

Stretch or Compress Vertically

As $a=\frac{1}{3}$ , so it the graph is vertically compressed by a factor of 1/3.

Check the attached comparison graphs.

Q # 4

Explanation:

Given the function

$f\left(x\right)=-\left(x+3\right)^2$

It OPENS DOWN, as 'a=-1' is negative.

Vertex

The vertex of an up-down facing parabola of the form $y=a\left(x-m\right)\left(x-n\right)$

is the average of the zeros $x_v=\frac{m+n}{2}$

$y=-\left(x+3\right)^2$

$a=-1,\:m=-3,\:n=-3$

$x_v=\frac{m+n}{2}$

$x_v=\frac{\left(-3\right)+\left(-3\right)}{2}$

$x_v=-3$

Finding $y_v$

$y_v=-\left(-3+3\right)^2$

$y_v=0$

So vertex is:

$\left(-3,\:0\right)$

Horizontal Translation

$y=\left(x+3\right)^2$ moves the graph LEFT 3 units.

Vertical Translation

$y=\left(x+3\right)^2$ does not move the graph vertically.

Stretch or Compress Vertically

As $a=-1$ , so it the graph is either vertically stretched or compressed.

Q # 5

Explanation:

$f\left(x\right)=\left(x+5\right)^2-3$

Openness

It OPENS UP, as 'a=1' is positive.

Vertex

$\mathrm{Rewrite}\:y=\left(x+5\right)^2-3\:\mathrm{in\:the\:form}\:y=ax^2+bx+c$

$y=x^2+10x+22$

$a=1,\:b=10,\:c=22$

$x_v=-\frac{10}{2\cdot \:1}$

$x_v=-5$

Finding $y_v$

$y_v=\left(-5\right)^2+10\left(-5\right)+22$

So vertex is:

$\left(-5,\:-3\right)$

Horizontal Translation

$f\left(x\right)=\left(x+5\right)^2$ moves the graph LEFT 5 units.

Vertical Translation

$f\left(x\right)=\left(x+5\right)^2-3$ moves the graph DOWN 3 unit.

Stretch or Compress Vertically

As $a = 1$ , so it does not affect the stretchiness or compression.

Check the attached comparison graphs.

Q # 6

THE DETAILS OF COMPLETE SOLUTION OF QUESTION 6 IS ATTACHED IN THE DIAGRAM AS THE 5000 CHARACTERS WERE ALREADY FILLED. SO, I solved via the attached figure.

SO, PLEASE CHECK THE LAST FIGURE TO FIND THE COMPLETE SOLUTION OF THE Q#6.

6 0

3 years ago

I am honestly the worst at geometry and this is my last proof for my class could someone help me, I don't wanna fail this proof.

Phantasy [73]

Why cant you ask your teacher thats what they are for

5 0

3 years ago

How do you solve <br> 6⋅1/3x+6⋅1/2=2x+3

pogonyaev

Answer:

x = -3/2, 1

Step-by-step explanation:

1. Use Rule of One

2. Take out the constants

3. Simplify 6/3 to 2

4. Use the product rule

5. Simplify

6. Move all terms to one side

7. Simplify

8. Split the second term

9. Factor out the common terms

10. Factor out the common term (again)

11. Solve for x

5 0

3 years ago

I am not good at algebra so I REALLY NEED HELP WITH THE PROBLEM 8n+4n^2-8n!

NeX [460]

Answer:

4n^2

Step-by-step explanation:

8n+4n^2-8n

The first 8n + the other 8n is 0 because you combine like terms

6 0

3 years ago

Use the pythagorean theorem to solve the problem. Round your answer to the nearest tenth if necessary.

lyudmila [28]

4^2 + b^2 = 12^2
16 + b^2 = 144
B^2 = 144-16
B^2 = 128
B= 11.3

7 0

3 years ago