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brilliants [131]
3 years ago
15

Graph the system of inequalities. Then use your graph to identify the point that represents a solution to the system.

Mathematics
1 answer:
natita [175]3 years ago
4 0
Hope this helps, the answer is A

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Solve using system of equations algebraically<br> y=2x-3 <br> x+y=18
Hitman42 [59]

Answer:

<em>(7, 11)</em>

Step-by-step explanation:

y = 2x - 3 .... (1)

x + y = 18 .... (2)

(1) ----> (2)

x + 2x - 3 = 18 ⇒ <u><em>x = 7</em></u>

y = 2(7) - 3 ⇒ <u><em>y = 11</em></u>

<em>(7, 11)</em>

6 0
3 years ago
Using the Factor Theorem, which of the polynomial functions has the zeros 2, radical 3 , and negative radical 3 ? f (x) = x3 – 2
butalik [34]

To solve this question, we use the factor theorem, and using it, the polynomial function is:

f(x) = x^3 - 2x^2 - 3x + 6

------------------------------

The factor theorem means that if k is a root of f(x), f(k) = 0.

Thus, applying the factor theorem for this question, we have to choose the function for which: f(2) = 0, f(\sqrt{3}) = 0, f(-\sqrt{3}) = 0

------------------------------

Function 1:

f(x) = x^3 - 2x^2 - 3x + 6

Testing the values:

f(2) = 2^3 - 2(2)^2 - 3(2) + 6 = 8 - 8 - 6 + 6 = 0

f(\sqrt{3}) = \sqrt{3^3} - 2(\sqrt{3})^2 - 3\sqrt{3} + 6 = \sqrt{3^2\times3} - 6 - 3\sqrt{3} + 6 = 3\sqrt{3} - 3\sqrt{3} = 0

f(-\sqrt{3}) = -\sqrt{3^3} - 2(-\sqrt{3})^2 - 3(-\sqrt{3}) + 6 = -\sqrt{3^2\times3} - 6 + 3\sqrt{3} + 6 = -3\sqrt{3} + 3\sqrt{3} = 0

Thus, since all three conditions are satisfied, f(x) = x^3 - 2x^2 - 3x + 6 is the polynomial function.

A similar question is given at brainly.com/question/11378552

5 0
2 years ago
a population of rabbits is growing naturally at 20% each month. if the colony began with 2 rabbits, how many will there be in 5
Brums [2.3K]
The answer is 112695

7 0
2 years ago
Justin is 2 years older
german
Answer:
5

Explanation:
1/4 of 12 = 3
2 + 3 = 5
5 0
2 years ago
Find the measure of angle A.<br> Can someone help
Zarrin [17]

80+x+45+2x+55=180

3x+100+80=180

3x=0

X=0

Measure of angle a=45

8 0
3 years ago
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