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seraphim [82]
3 years ago
12

Identify the x-intercept and y-intercept of the line 4x-2y=-12

Mathematics
1 answer:
Alex73 [517]3 years ago
7 0
To find x-intercept, put the Y value = 0;

4x - 2y = -12

-> 4x -2.0 = -12
-> 4x = -12
-> x = -12/4
-> x = -3

X-intercept: (-3,0)

Now do the reverse to find the y-intercept, X = 0;

4x - 2y = -12

-> 4.0 - 2y = -12
-> -2y = -12 x(-1)
-> 2y = 12
-> y = 12/2
-> y = 6

Y-intercept: (0, 6)
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Which line contains the point (3,6)
spayn [35]
Many lines do. One of them would be y = 2x
6 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

4 0
2 years ago
Simplify:<br> 49 - 4 x 49 ÷ 7<br><br><br> 21<br> 34<br> 36
enot [183]

Answer:

21

Step-by-step explanation:

49 - 4 x 49 / 7 =

49 - 196 / 7 =

49 - 28 =

21

<em>Hope that helps!</em>

5 0
2 years ago
Read 2 more answers
Help me please ASAP
Mashutka [201]

Answer: \frac{5}{35}

Step-by-step explanation:

Since, the total number of contestant = 7,

Thus, the probability that out of 3 contestant, me and my friend is chosen

= \frac{2_C_2\times 5_C_1}{7_C_3}

= \frac{\frac{2!}{2!0!}\times \frac{5!}{1!\times 4!}}{\frac{7!}{4!\times 3!}}

= \frac{1\times 5}{\frac{7\times 6\times 5\times 4!}{4!\times 6}}

=  \frac{5}{35}

⇒ Third Option is correct.

4 0
3 years ago
What is the solution to the division problem below x^3-x^2-11x+3/ x+3
evablogger [386]
I believe the answer is x^2-4x+1

(I did the problem in my head but it seems correct)
3 0
3 years ago
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