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Gnesinka [82]
3 years ago
7

What's the correct choice please help

Mathematics
1 answer:
nadezda [96]3 years ago
6 0

Answer:

$77,886

Step-by-step explanation:

We solve this question by proportions, using a rule of three.

The table states that:

$201.6 in 2006 is worth $232.957 in 2013.

We want to find:

The value of $90,000 in 2013 in 2006:

$201.6 - $232.957

x - $90000

Applying cross multiplication

232957x = 201.6*90000

x = \frac{201.6*90000}{232957}

x = 77886

$77,886 is the answer

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A= lxw = 34 Times 7 meaning the answer is 238
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Answer: Number one would be 3x+2y+-14 and x+y=-4, second one is x=-3, y=7

Step-by-step explanation:

The first one is solved by inputting the x and y in each one and finding which one comes out true, the second on is solved by substitution. to find x you would subtract x in the first equation and make it y=4-x then input that equation in the y in the second equation.

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4 years ago
If y = x² - 6x + 11 is written in the form y = a ( x - h )² + k, find the value of ( a + h + k).
Crazy boy [7]

Answer:

6

Step-by-step explanation:

y = x² - 6x + 11

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y = x² - 6x + 9 + 2

y = (x - 3)² + 2

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3 years ago
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Solve the following equation by factoring:9x^2-3x-2=0
olya-2409 [2.1K]

Answer:

The two roots of the quadratic equation are

x_1= - \frac{1}{3} \text{ and } x_2= \frac{2}{3}

Step-by-step explanation:

Original quadratic equation is 9x^{2}-3x-2=0

Divide both sides by 9:

x^{2} - \frac{x}{3} - \frac{2}{9}=0

Add \frac{2}{9} to both sides to get rid of the constant on the LHS

x^{2} - \frac{x}{3} - \frac{2}{9}+\frac{2}{9}=\frac{2}{9}  ==> x^{2} - \frac{x}{3}=\frac{2}{9}

Add \frac{1}{36}  to both sides

x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{2}{9} +\frac{1}{36}

This simplifies to

x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{1}{4}

Noting that (a + b)² = a² + 2ab + b²

If we set a = x and b = \frac{1}{6}\right) we can see that

\left(x - \frac{1}{6}\right)^2 = x^2 - 2.x. (-\frac{1}{6}) + \frac{1}{36} = x^{2} - \frac{x}{3}+\frac{1}{36}

So

\left(x - \frac{1}{6}\right)^2=\frac{1}{4}

Taking square roots on both sides

\left(x - \frac{1}{6}\right)^2= \pm\frac{1}{4}

So the two roots or solutions of the equation are

x - \frac{1}{6}=-\sqrt{\frac{1}{4}}  and x - \frac{1}{6}=\sqrt{\frac{1}{4}}

\sqrt{\frac{1}{4}} = \frac{1}{2}

So the two roots are

x_1=\frac{1}{6} - \frac{1}{2} = -\frac{1}{3}

and

x_2=\frac{1}{6} + \frac{1}{2} = \frac{2}{3}

7 0
2 years ago
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