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3 years ago

How do you reflect over y axis

2 answers:

5 0

Think of a point not on the y-axis.
For example, think of the point (5, 5).
This point is in quadrant 1, right of the y-axis and above the x-axis.
To reflect it over the y-axis, the reflection will be in quadrant 2, still above the x-axis, but now left of the y-axis. The distance the reflection is from the y-axis is the same as the original point is from the y-axis.

The reflection of (5, 5) over the y-axis is (-5, 5).

To reflect any point over the y-axis, replace the x-coordinate with the additive inverse of the x-coordinate.

aleksley [76]3 years ago

3 0

If you want to reflect over the y axis then your answer would want to be going either from the left side to the right side, or right side over to the left side. If the figure was on the top or bottom you would go over the x axis.

So to reflect over the y axis you must find the points that each angle or corner of the figure your reflecting. You'll want to remember those locations of the points, then you'll transfer them over to the other side on the same points but on the other side. I'll make an example to make it more clear

For the first number, before the coma, if the number is negative you'll go left, if the number is positive you go right
For the second number, after the coma, you go up if it's positive or down if its negative.

If A is located at (-3, +2)
Angle B at (-1, +2)
Angle C at (-2, +0)

To transfer them over the y axis the numbers being positive or negative will change
For example

Angle A will become (+3, +2)
Angle B will become (+1, +2)
And angle C will become (+2, 0)

If the number is zero then it stays on the default line.

Hope this helped,reply if you need more help!!

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Please help! Its for my big test tomorrow!

Travka [436]

Answer:

# The solution x = -5

# The solution is x = 1

# The solution is x = 6.4

# The solution is x = 4

# The solution is 1.7427

# The solution is 0.190757

Step-by-step explanation:

* Lets revise some rules of the exponents and the logarithmic equation

# Exponent rules:

1- b^m × b^n = b^(m + n) ⇒ in multiplication if they have same base

we add the power

2- b^m ÷ b^n = b^(m – n) ⇒ in division if they have same base we

subtract the power

3- (b^m)^n = b^(mn) ⇒ if we have power over power we multiply

them

4- a^m × b^m = (ab)^m ⇒ if we multiply different bases with same

power then we multiply them ad put over the answer the power

5- b^(-m) = 1/(b^m) (for all nonzero real numbers b) ⇒ If we have

negative power we reciprocal the base to get positive power

6- If a^m = a^n , then m = n ⇒ equal bases get equal powers

7- If a^m = b^m , then a = b or m = 0

# Logarithmic rules:

1- $log_{a}b=n-----a^{n}=b$

2- $loga_{1}=0---log_{a}a=1---ln(e)=1$

3- $log_{a}q+log_{a}p=log_{a}qp$

4- $log_{a}q-log_{a}p=log_{a}\frac{q}{p}$

5- $log_{a}q^{n}=nlog_{a}q$

* Now lets solve the problems

# $3^{x+1}=9^{x+3}$

- Change the base 9 to 3²

∴ $9^{x+3}=3^{2(x+3)}=3^{2x+6}$

∴ $3^{x+1}=3^{2x+6}$

- Same bases have equal powers

∴ x + 1 = 2x + 6 ⇒ subtract x and 6 from both sides

∴ 1 - 6 = 2x - x

∴ -5 = x

* The solution x = -5

# ㏒(9x - 2) = ㏒(4x + 3)

- If ㏒(a) = ㏒(b), then a = b

∴ 9x - 2 = 4x + 3 ⇒ subtract 4x from both sides and add 2 to both sides

∴ 5x = 5 ⇒ divide both sides by 5

∴ x = 1

* The solution is x = 1

# $log_{6}(5x+4)=2$

- Use the 1st rule in the logarithmic equation

∴ 6² = 5x + 4

∴ 36 = 5x + 4 ⇒ subtract 4 from both sides

∴ 32 = 5x ⇒ divide both sides by 5

∴ 6.4 = x

* The solution is x = 6.4

# $log_{2}x+log_{2}(x-3)=2$

- Use the rule 3 in the logarithmic equation

∴ $log_{2}x(x-3)=2$

- Use the 1st rule in the logarithmic equation

∴ 2² = x(x - 3) ⇒ simplify

∴ 4 = x² - 3x ⇒ subtract 4 from both sides

∴ x² - 3x - 4 = 0 ⇒ factorize it into two brackets

∴ (x - 4)(x + 1) = 0 ⇒ equate each bract by 0

∴ x - 4 = 0 ⇒ add 4 to both sides

∴ x = 4

∵ x + 1 = 0 ⇒ subtract 1 from both sides

∴ x = -1

- We will reject this answer because when we substitute the value

of x in the given equation we will find $log_{2}(-1)$ and this

value is undefined, there is no logarithm for negative number

* The solution is x = 4

# $log_{4}11.2=x$

- You can use the calculator directly to find x

∴ x = 1.7427

* The solution is 1.7427

# $2e^{8x}=9.2$ ⇒ divide the both sides by 2

∴ $e^{8x}=4.6$

- Insert ln for both sides

∴ $lne^{8x}=ln(4.6)$

- Use the rule $ln(e^{n})=nln(e)$ ⇒ ln(e) = 1

∴ 8x = ln(4.6) ⇒ divide both sides by 8

∴ x = ln(4.6)/8 = 0.190757

* The solution is 0.190757

7 0

3 years ago

Read 2 more answers

For an annual membership fee of $500,Mr.Bailey can join a country club that would allow him to play a round of golf for $35. Wit

IRISSAK [1]

Answer:

25 rounds

Step-by-step explanation:

Let

x -----> the number of rounds of golf

y ---> total charges to play

The linear equation in slope intercept form is equal to

$y=mx+b$

where

m is the slope or unit rate

b is the y-intercept or initial value

<em>Membership</em>

$y=35x+500$ ----> equation A

The slope is m=$35 per round

The y-intercept b=$500 (annual membership fee)

<em>Non-Membership</em>

$y=55x$ ----> equation B

The slope is m=$55 per round

The y-intercept b=$0

Equate equation A and equation B

$55x=35x+500$

solve for x

$55x-35x=500$

$20x=500$

$x=25\ rounds$

For x=25 rounds the cost to be the same with and without a membership

8 0

3 years ago

Read 2 more answers

A short-wave radio antenna is supported by two guy wires, 150 ft and 180 ft long. Each wire is attached to the top of the antenn

Dafna1 [17]

The distance between the anchor points of the two guy wires holding radio antenna is 181 feet (to the nearest foot)

<h3>How to determine the distance between the two anchor points of the guy wire</h3>

The problem will be solved using SOH CAH TOA

let the distance between the 150 ft long guy wire and the radio antenna be x

let the distance between the 180 ft long guy wire and the radio antenna be y

cos 65° = x / 150

x = cos 65° * 150

x = 63.39 ft

The height of the antenna

sin 65° = height of antenna / 150

height of antenna = sin 65 * 150

height of antenna = 135.95

using Pythagoras theorem

(length of guy wire)² = (height of the antenna)² + (anchor distance)²

(anchor distance)² = 180² - 135.95²

anchor distance = √(180² - 135.95²)

anchor distance = 117.97

The anchor points distance apart

= 63.39 + 117.97

= 181 (to the nearest foot)

Learn more on Pythagoras theorem here:

brainly.com/question/29241066

#SPJ1

4 0

1 year ago

Pls help me and show work pls

ElenaW [278]

3/6 cause there are 3 evens and 3 odds on a dice

3 0

3 years ago

ANSWER QUICK PLEASEEE

Sliva [168]

Answer:

I think the answer is 664....but I'm nt sure. Plz don't blame me if I'm wrong lol

Step-by-step explanation:

12x8x7= 672

672- 8= 664

3 0

3 years ago