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Ivahew [28]
3 years ago
5

For which triangle is the length of the hypotenuse an INTEGER?

Mathematics
1 answer:
artcher [175]3 years ago
4 0

A right triangle :) to find the length of a missing side, you can use the pythagorean theorem (a^2 + b^2 = c^2)

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Candace runs a company that installs fences. She calculates the total cost C of installing a fence using the function rule C(x)
maksim [4K]

Answer:

g

Step-by-step explanation:

8 0
3 years ago
Which statement describes the inverse of m(x) = x2 – 17x?
stealth61 [152]

Answer:

The correct option is;

The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}

Step-by-step explanation:

The given information is that m(x) = x² - 17·x

The above equation can be written in the form;

y = x² - 17·x

Therefore;

0 = x² - 17·x - y

From the general solution of a quadratic equation, 0 = a·x² + b·x + c we have;

x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}

By comparison to the equation,0 = x² - 17·x - y, we have;

a = 1, b = -17, and c = -y

Substituting the values of a, b and c into the formula for the general solution of a quadratic equation, we have;

x = \dfrac{-(-17)\pm \sqrt{(-17)^{2}-4\times (1) \times (-y)}}{2\times (1)} = \dfrac{17\pm \sqrt{289+4\cdot y}}{2}

Which can be simplified as follows;

x =  \dfrac{17\pm \sqrt{289+4\cdot y}}{2}= \dfrac{17}{2} \pm \dfrac{1}{2}  \times \sqrt{289+4\cdot y}} = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +\dfrac{4\cdot y}{4} }}

And further simplified as follows;

x = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +y }} = \dfrac{17}{2} \pm \sqrt{y + \dfrac{289}{4} }}

Interchanging x and y in the function of the inverse, m⁻¹(x), we have;

m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}

We note that the maximum or minimum point of the function, m(x) = x² - 17·x found by differentiating the function and equating the result to zero, gives;

m'(x) = 2·x - 17 = 0

x = 17/2

Similarly, the second derivative is taken to determine if the given point is a maximum or minimum point as follows;

m''(x) = 2 > 0, therefore, the point is a minimum point on the graph

Therefore, as x increases past the minimum point of 17/2, m⁻¹(x) increases to give;

The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }} to increase m⁻¹(x) above the minimum.

8 0
3 years ago
For bills gym membership, he pays $25 per month plus a one dollar charge per class he takes. Bills total for last month was $32.
wlad13 [49]
7 classes because $32 -$25 = 7 (where $25 is for one month, and the 7 is how much is left over of the total and 1 class = $1 dollar)
5 0
3 years ago
Read 2 more answers
Solve for x in the following equation.
KonstantinChe [14]

Hello,

3 {}^{a}  = 3 {}^{b} \Leftrightarrow \: a = b

3 {}^{x}  =  {3}^{2} \Leftrightarrow \: x = 2

5 0
2 years ago
What is the volume of the right rectangular prism?
ddd [48]

Answer:

i think its 735

Step-by-step explanation:

since the base is a square it should be a whole square which is 7 and the height is 15 and v=whl and multiply all of it 7 times 7 is 49 times 15 is 735.

5 0
3 years ago
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