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kotegsom [21]
4 years ago
9

Find the solution to the system of equations. Select all that apply. Y=x^2-2x-3. Y=x-3

Mathematics
1 answer:
sukhopar [10]4 years ago
6 0

Answer:

The solution to the system of equations are (0 , -3) ⇒ answer B and

(3 , 0) ⇒ answer D

Step-by-step explanation:

* Lets explain the problem

- The quadratic equation represented by parabola

- The linear equation represented by line

- If the two equation solved together, then we have 3 types of solutions

# They intersected in two different points (2 solution)

# They intersected in one point (1 solution)

# They did not intersect in any points (No solution)

* Now lets solve the problem

∵ The quadratic equation is y = x² - 2x - 3

- It represented by parabola (red graph)

∵ The linear equation is y = x - 3

- It represented by line (blue graph)

- The two graphs intersect each other at points (3 , 0) and (0 , -3)

∵ The solution of the system of the equation is the intersection points

  between the two graphs

∵ The two graphs intersect each other at points (3 , 0) and (0 , -3)

∴ The solution to the system of equations are (3 , 0) and (0 , -3)

* The solutions are B and D

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You increase the size of a page by 30%. Then you decrease it by 30%. What is the size of the page now?
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Douglas invests money in two simple interest accounts. He invests three times as much in an account paying 14% as he does in an
Rainbow [258]

Douglas invested $ 1300 altogether

<em><u>Solution:</u></em>

Let x represent the amount invested in the account paying 14% interest.

Let y represent the amount invested in the account paying 5% interest

He invests three times as much in an account paying 14% as he does in an account paying 5%

Which means,

x = 3y

<em><u>The simple interest is given by formula:</u></em>

S.I = \frac{p \times n \times r}{100}

Where,

"p" is the principal

"n" is the number of years

"r" is the rate of interest

<em><u>He earns $152.75 in interest in one year from both accounts combined</u></em>

Therefore,

Combined S.I = 152.75

n = 1 year

<em><u>Considering the account earning 14% interest:</u></em>

S.I = \frac{3y \times 14 \times 1}{100}\\\\S.I = 0.42y

<em><u>Considering the account earning 5% interest:</u></em>

S.I = \frac{y \times 5 \times 1}{100}\\\\S.I = 0.05y

<em><u>Since, Combined S.I = 152.75</u></em>

Therefore,

0.42y + 0.05y = 152.75

0.47y = 152.75

Divide both sides by 0.47

y = 325

Therefore,

x = 3y

x = 3(325)

x = 975

<em><u>how much did he invest altogether?</u></em>

Amount invested together = x + y = 975 + 325 = 1300

Thus he invested $ 1300 altogether

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3 years ago
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