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vagabundo [1.1K]
3 years ago
13

What is { y= 4x+10 and y= 3x+9

Mathematics
2 answers:
Musya8 [376]3 years ago
5 0

Answer:

x=-1

Step-by-step explanation:

y=4x+10

y=3x+9

-----------

4x+10=3x+9

4x-3x=9-10

x=-1

Kaylis [27]3 years ago
4 0
Answer: (-1,6)

Step 1: Rewrite the equations

y=4x+10

y=3x+9

Step 2: Determine method of solving

When solving an equation, there are three ways you can do it:

1) Graphing
2) Substitution
3) Elimination

We will solve this using substitution

Step 3: Solve

When solving with substitution, you substitute the first equation into the second. So, we will get the value of y and substitute into where y is in the second equation.

y=4x+10 and y=3x+9 goes into:

4x+10=3x+9

Now let’s solve.

4x+10=3x+9

*subtract 3x on both sides*

4x+10=3x+9
-3x -3x
_________
x+10=9

*subtract 10 on both sides*

x+10=9
-10 -10
______
x=-1

Now we have the value of x. Let’s substitute this into the first equation and find y.

y=4x+10

y=4(-1)+10

y=-4+10

y=6

Now we have y! The ordered pair of this would be (-1,6). And that’s your answer.

Hope this helps comment below for more questions :)
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3 years ago
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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
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Answer:

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2. 3(2c+8)

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Answer:

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Answer:

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This also means B is not a choice.

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