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horrorfan [7]
3 years ago
11

Enter the slope-intercept equation for the image of line ℓ after a clockwise rotation of 90°. (Hint: To find the image of line ℓ

, choose two or more points on the line and find the images of the points.)

Mathematics
1 answer:
frutty [35]3 years ago
7 0

Answer:

y=-3x-3

Step-by-step explanation:

Rotate 90 degrees

find slope

find y intercept

write equation

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PLEASE I NEED THE ANSWER DON'T IGNORE MY QUESTION!!!
kenny6666 [7]

Answer:

x=7.8

Step-by-step explanation:

take 49 degree as reference angle

using cos rule

cos 49=adjacent/hypotenuse

0.65=x/12

0.65*12=x

7.8=x

3 0
2 years ago
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HELP PLEASE for brainliest!!
marta [7]
-3 • (5x + 2y = 7)
-15x - 6y = -21
-2x + 6y = 9
__________
-17x = -12
x = 12/17
5(12/17) + 2y = 7
2y = 7 - 60/17
2y = 3.47
y = 1.735
3 0
3 years ago
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What is 10x-10y=0 -20x-5y=-25
Arada [10]

-10x-15-(-25)

-5(2x+3-5)

-5(2x-2)

-5*2 (x-1)

= -10(x-1)

6 0
3 years ago
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Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10
zvonat [6]

The approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

<h3>What is depreciation?</h3>

Depreciation is to decrease in the value of a product in a period of time. This can be given as,

FV=P\left(1-\dfrac{r}{100}\right)^n

Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.

Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is n_1. Thus, by the above formula for the first car,

0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}

Take log both the sides as,

\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85

Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.

Thus, the depreciation price of the car is 0.6y. Let the number of year is n_2. Thus, by the above formula for the second car,

0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}

Take log both the sides as,

\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14

The difference in the ages of the two cars is,

d=4.85-3.14\\d=1.71\rm years

Thus, the approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

Learn more about the depreciation here;

brainly.com/question/25297296

4 0
2 years ago
Laurie says summer is 1 over 4 of the year. Maria says summer is 3 over 12 of the year. Who is correct? explain
nasty-shy [4]

Answer:


Step-by-ste

Oh, oh! This is such a tricky question...

Guess what!

Both of them are correct. Why?

Because if we simplify 3/12, we get 1/4th!

(How do you simplify?)

3/12 = 1/4

Because if we divide both "3" and "12" in 3/12 by 3, we get 1 instead of 3 and 4 instead of 12.


I hope you've got it! ;)


4 0
2 years ago
Read 2 more answers
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