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7nadin3 [17]
3 years ago
12

Is y a function of x? 9y-3x-4=0

Mathematics
1 answer:
Elza [17]3 years ago
5 0
In other words, can 9y - 3x - 4 = 0 be solved for y, and, if so, is there always exactly one y value for every input (x) value?

9y = 3x + 4
Thus, y = (3/9)x + 4, or    y = (1/3)x + 4

No matter which real x you choose, y is always assigned exactly one value by this rule.  So, yes, this is a function.
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Turn 32/64 into a percent?
alexandr402 [8]


To turn 32/64 into a percent, you have to divide the numerator by the denominator and then multiply the answer by 100.

So, for this question, divide 32 by 64 which is 0.50. Multiply this decimal by 100 and you get 50.

So 50% is the percent for 32/64.

Hope it helps :)

5 0
3 years ago
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Answer this question I need help as soon as possible
torisob [31]

Answer

Because the function says 1.045 means 1+0.045

0.045=4.5%

So your answer is

4.50%

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3 years ago
4x - 2<br> What is BC ?<br> Enter your answer in the box.<br> 8x - 4<br> 5x + 11<br> units
arsen [322]
Is there a picture of the shape, and is 4x - 2 AC or AB?
7 0
3 years ago
What equals negative 4
galina1969 [7]
Your question is very vague, there is no multiple choice answers, your question lacks basically everything.

-2+(-2=4?
3 0
3 years ago
Triangle JKL has vertices J(2,5), K(1,1), and L(5,2). Triangle QNP has vertices Q(-4,4), N(-3,0), and P(-7,1). Is (triangle)JKL
Tems11 [23]

Answer:

Yes they are

Step-by-step explanation:

In the triangle JKL, the sides can be calculated as following:

  • J(2;5); K(1;1)

             => JK = \sqrt{(1-2)^{2} + (1-5)^{2}  } = \sqrt{(-1)^{2}+(-4)^{2}  } = \sqrt{1+16}=\sqrt{17}

  • J(2;5); L(5;2)

             => JL = \sqrt{(5-2)^{2} + (2-5)^{2}  } = \sqrt{3^{2}+(-3)^{2}  } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}

  • K(1;1); L(5;2)

             =>  KL = \sqrt{(5-1)^{2} + (2-1)^{2}  } = \sqrt{4^{2}+1^{2}  } = \sqrt{1+16}=\sqrt{17}

In the triangle QNP, the sides can be calculate as following:

  • Q(-4;4); N(-3;0)

             => QN = \sqrt{[-3-(-4)]^{2} + (0-4)^{2}  } = \sqrt{1^{2}+(-4)^{2}  } = \sqrt{1+16}=\sqrt{17}

  • Q (-4;4); P(-7;1)

   => QP = \sqrt{[-7-(-4)]^{2} + (1-4)^{2}  } = \sqrt{(-3)^{2}+(-3)^{2}  } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}

  • N(-3;0); P(-7;1)

             =>  NP = \sqrt{[-7-(-3)]^{2} + (1-0)^{2}  } = \sqrt{(-4)^{2}+1^{2}  } = \sqrt{16+1}=\sqrt{17}

It can be seen that QPN and JKL have: JK = QN; JL = QP; KL = NP

=> They are congruent triangles

7 0
3 years ago
Read 2 more answers
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