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2 years ago

Write an equation in slope-intercept form for the line that satisfies the following condition.

Mathematics

1 answer:

skelet666 [1.2K]2 years ago

4 0

Answer:

The equation of the line we want to write is 4y = -x + 22

Step-by-step explanation:

Here, we want to write the equation of a line

The standard equation of a straight line is given as:

y = mx + c

where m is the slope and c represents the y-intercept

Now, let’s look at the line y = 4x + 16

The slope of this line is 4

Now, the equation of the line we want to write is perpendicular to this line

When two lines are perpendicular, the product of their slopes = -1

Hence;

m * 4 = -1

m = -1/4

So the slope of the line we want to write is -1/4

Now, using the point-slope form for the new equation;

y-y1/x-x1 = m

From the point given (x1,y1) = (10,3)

Thus;

y-3/x-10 = -1/4

4(y-3) = -1(x - 10)

4y - 12 = -x + 10

4y = - x + 10 + 12

4y = -x + 22

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X3 - 3y when x=6 and y= -3

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Answer:

Step-by-step explanation:

6×3=18 3×3=9 9+18

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2 years ago

"A pieces of licorice is cut into 10 equal-sized pieces. If the length of the piece of licorice is 2/3 yard, how long will each

jok3333 [9.3K]

Each piece of licorice will be 2.4 inches

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3 years ago

A transformation that maps the vertex A( 3 , 4 ) to A’ ( -4 , 3 ) is

Katen [24]

Answer:

"A translation of 7 units to the left followed by a translation of 1 unit down".

Step-by-step explanation:

There are multiple transformations that map one point into another, here is one example that works particularly for translations, which are the simplest (and usually the most used) transformations.

Suppose that we have the point (a, b) which is transformed into (a', b')

Then we have a horizontal translation of (a' - a) units followed by a vertical translation of (b' - b) units.

(the order of the translations does not matter, is the same having first the vertical translation and then the horizontal one).

Here we have the point A (3, 4) transformed into (-4, 3)

Then we have a horizontal translation of ((-4) - 3) = -7 units followed by a vertical translation of (3 - 4) = -1 units.

Where a horizontal translation of -7 units is a translation of 7 units to the left, and a vertical translation of -1 unit is a translation of 1 unit down.

Then we can write this transformation as:

"A translation of 7 units to the left followed by a translation of 1 unit down".

5 0

2 years ago

I need help with math

frutty [35]

Answer:

I think it may be C? I used desmos to graph it but I'm not 100%

4 0

2 years ago

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

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2 years ago