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12345 [234]
3 years ago
15

I health club charges $35 a month for membership fees. Determine whether the cost of a membership is proportional to the number

of months. Explain your reasoning
Mathematics
2 answers:
mart [117]3 years ago
5 0
The porportion would be realating to the months because the longer you go on it adds up kind of like a slope of change or realating factors
Marina CMI [18]3 years ago
3 0
35 + 35 =70  i guess its saying to add up the months so i guess its saying whats 35 + 35 =70 thats what i got
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B ) 45°, 67.5°
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The first side of a triangle is 8 m shorter than the second side. The third side is 4 times as long as the first side. The perim
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Solve the following inequality: –1 6(–1 – 3x) > –39 – 2x
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If you would like to solve the inequality <span>–1 + 6 * (–1 – 3x) > –39 – 2x, you can do this using the following steps:

</span>–1 + 6 * (–1 – 3x) > –39 – 2x
-1 - 6 - 18x > -39 - 2x
-7 + 39 > -2x + 18x
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3 years ago
How many years in 9.7 scores?​
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Step-by-step explanation:

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7 0
3 years ago
Determine whether f(x) =3x^2+9x-2 has a maximum of a minimum value and find that value by hand.
xenn [34]

f(x) =3x^2+9x-2 has a minimum value. Minimum value of f(x) is \frac{-35}{4}

<u>Solution:</u>

Given, equation is f(x)=3 x^{2}+9 x-2

We have to find whether given equation has maximum or minimum for the given equation.

Now, we know that, f(x) is a quadratic equation and coefficient has x^2 is positive, then its graph is upward parabola. Which means that it will have minimum

The minimum value upward parabola will be its vertex.

So, let us convert f(x) into general form. That is,

f(x)=a(x-h)^{2}+k

where a is constant and (h, k) is vertex

\text { Now }_{,} f(x)=3 x^{2}+9 x-2

\text { Adding and subtracting } \frac{27}{4} \text { for easier calculations }

f(x)=3 x^{2}+9 x+\frac{27}{4}-\frac{27}{4}-2

Taking "3" as common from first three terms,

f(x)=3\left(x^{2}+3 x+\frac{9}{4}\right)-\frac{27}{4}-2

Now multiply and divide “2” with “3x” for easier calculations

f(x)=3\left(x^{2}+2 \times \frac{3}{2} \times x+\left(\frac{3}{2}\right)^{2}\right)-\frac{27+2 \times 4}{4}

\begin{array}{l}{\text { By using }(a+b)^{2}=a^{2}+2 a b+b^{2}, \text { we get }} \\\\ {\left(x^{2}+2 \times \frac{3}{2} \times x+\left(\frac{3}{2}\right)^{2}\right)=\left(x+\frac{3}{2}\right)^{2}}\end{array}

f(x)=3\left(x+\frac{3}{2}\right)^{2}-\frac{35}{4}

So, by comparison with general form we get,

\mathrm{h}=-\frac{3}{2} \text { and } \mathrm{k}=\frac{-35}{4}

\text { Here, }(\mathrm{h}, \mathrm{k})=(\mathrm{x}, \mathrm{f}(\mathrm{x}))=\left(\frac{-3}{2}, \frac{-35}{4}\right)

\text { So, minimum value of } f(x) \text { is } \frac{-35}{4}

5 0
3 years ago
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