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White raven [17]
2 years ago
7

please i need this asap, look at the graph to the right. What is the equation for the line (in y=mx+b form)?

Mathematics
1 answer:
vovikov84 [41]2 years ago
5 0

Answer:

The Answer is: y = - 2x + 8.

the y-intersept is 8, and the slope is -2x

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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
What is 16 percent of $36.48 ?
vlabodo [156]
16% of $36.48 is $5.84

 9s\frac{is}{36.48} =  \frac{16}{100}

Cross multiply 16 and 36.48 to 583.68, and then I divided by 100, to get the percent of 5.8368, which I rounded up to 5.84%.
4 0
3 years ago
Can I please get some help on this...I have tried to many times...
Bezzdna [24]

Answer: A

Step-by-step explanation:

First, the problem is g(f(x)). You would plug in f(x) wherever you see an x in g(x). To find the domain, you take the bottom function, and set it equal to 0.

\sqrt{x-2} =0

When you solve that, you get x=2. You know your domain is x≥2, but there is as asymptote at x=11. That means the graph never reaches x=11, but gets very close. You find that by setting the entire equation equal to 0 and solve from there.

5 0
3 years ago
Please help! [99 points]
anygoal [31]
This is a linear function.

A linear function is a straight line that has a slope, and follows the formula: <em>
y = mx + b</em>

In which: 
y = y
m = slope
x = x
b = y -intercept


hope this helps<em />
4 0
3 years ago
Read 2 more answers
- Which of the following is the least?<br> A).27<br> B) 1/4<br> C) 3/8<br> D) 2/11<br> E) 11%
mezya [45]

Step-by-step explanation:

A) 0.27

B) 1/4 = 0.25

C) 3/8 = 0.375

D) 2/ 11 = 0.1818

E) 11% = 0.11

E is the correct answer

4 0
3 years ago
Read 2 more answers
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