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Alex Ar [27]
2 years ago
10

X - 2y = 3 5x + 3y = 2 The lines whose equations are shown intersect at which point?

Mathematics
1 answer:
Lady_Fox [76]2 years ago
8 0

The point of intersection of the two lines is at (1,-1)

<h3>System of equation</h3>

The given system of expression is shown below

x - 2y = 3

5x + 3y = 2

The solution to the system of equation is the point of intersection

From equation 1

x = 3 + 2y

Substitute into 2

5(3+2y) + 3y = 2

15 +10y + 3y = 2

13y = -13

y = -1

Substitute y = -1 into 3

x = 3 + 2y

x = 3+(-2)

x = 1

Hence the point of intersection of the two lines is at (1,-1)

Learn more on system of equation here: brainly.com/question/25976025

#SPJ1

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Answer:

1,00000,00,00

Step-by-step explanation:

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3 years ago
8. What is the domain of <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7Bx%5E%7B2%7D%20%2B20x%2B75%7D" id="TexFormula1" title
antiseptic1488 [7]

Answer:

8.  Domain: (-∞, -15) ∪ (-15, -5) ∪ (-5, ∞)

9.  Domain: [7/13, ∞)

    Range: [1, ∞)

Step-by-step explanation:

<u>Question 8</u>

Given rational function:

f(x)=\dfrac{x}{x^2+20x+75}

Factor the denominator of the given rational function:

\implies x^2+20x+75

\implies x^2+5x+15x+75

\implies x(x+5)+15(x+5)

\implies (x+15)(x+5)

Therefore:

f(x)=\dfrac{x}{(x+15)(x+5)}

<u>Asymptote</u>: a line that the curve gets infinitely close to, but never touches.

The function is <u>undefined</u> when the <u>denominator equals zero</u>:

x+15=0 \implies x=-15

x+5=0 \implies x=-5

Therefore, there are <u>vertical asymptotes</u> at x = -15 and x = -5.

<u>Domain</u>: set of all possible input values (x-values)

Therefore, the <u>domain of the given rational function</u> is:

(-∞, -15) ∪ (-15, -5) ∪ (-5, ∞)

---------------------------------------------------------------------------------

<u>Question 9</u>

Given function:

f(x)=\sqrt{13x-7}+1

<u>Domain</u>: set of all possible input values (x-values)

As the <u>square root of a negative number</u> is <u>undefined</u>:

\implies 13x-7\geq 0

\implies 13x\geq 7

\implies x\geq \dfrac{7}{13}

Therefore, the <u>domain of the given function</u> is:

\left[\dfrac{7}{13},\infty\right)

<u>Range</u>: set of all possible output values (y-values)

\textsf{As }\:\sqrt{13x-7}\geq 0

\implies \sqrt{13x-7}+1\geq 1

Therefore, the <u>range of the given function</u> is:

[1, ∞)

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In which year did the population grow the fastest ? Justify your answer using the average rate of change
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1957 avg rate of 500,000 per every person
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Two stands at a craft fair sell cone-shaped candles. They are made by pouring wax into a mold. Booth A sells a candle with a rad
frez [133]
In order to compare the amount of wax, you have to calculate the volume of both candles.

Candle of booth A = πr² h/3
A = π * 4² * 6/3
A = π * 16 * 2
A = 32π

Now, Candle of Booth B = π * 5² * 5/3
B = π * 25 * 5/3
B = 41.6π

In short, Booth B sells the candle with more wax.

Hope this helps!
6 0
3 years ago
Solve. 4x−y−2z=−8 −2x+4z=−4 x+2y=6 Enter your answer, in the form (x,y,z), in the boxes in simplest terms. x= y= z=
ladessa [460]

Answer:

(-2, 4, -2)

x=-2, y=4, z=-2.

Step-by-step explanation:

So we have the three equations:

4x-y-2z=-8\\-2x+4z=-4\\x+2y=6

And we want to find the value of each variable.

To solve this system, first look at it and consider what you should try to do.

So we can see that the second and third equations both have an x.

Therefore, we can isolate the variables for the second and third equation and then substitute them into the first equation to make the first equation all xs.

Therefore, let's first isolate the variable in the second and third equation.

Second Equation:

-2x+4z=-4

First, divide everything by -2 to simplify things:

x-2z=2

Subtract x from both sides. The xs on the left cancel:

(x-2z)-x=2-x\\-2z=2-x

Now, divide everything by -2 to isolate the z:

z=-\frac{2-x}{2}

So we've isolated the z variable. Now, do the same to the y variable in the third equation:

x+2y=6

Subtract x from both sides:

2y=6-x

Divide both sides by 2:

y=\frac{6-x}{2}

Now that we've isolated the y and z variables, plug them back into the first equation. Therefore:

4x-y-2z=-8\\4x-(\frac{6-x}{2})-2(-\frac{2-x}{2})=-8

Distribute the third term. The -2s cancel out:

4x-(\frac{6-x}{2})+(2-x)=-8

Since there is still a fraction, multiply everything by 2 to remove it:

2(4x-(\frac{6-x}{2})+(2-x))=2(-8)

Distribute:

8x-(6-x)+2(2-x)=-16\\8x-6+x+4-2x=-16

Combine like terms:

8x+x-2x-6+4=-16\\7x-2=-16

Add 2 to both sides:

7x=-14

Divide both sides by 7:

(7x)/7=(-14)/7\\x=-2

Therefore, x is -2.

Now, plug this back into the second and third simplified equations to get the other values.

Second equation:

z=-\frac{2-x}{2}\\ z=-\frac{2-(-2)}{2}\\z=-\frac{4}{2}\\z=-2

Third equation:

y=\frac{6-x}{2}\\y=\frac{6-(-2)}{2}\\y=\frac{8}{2}\\y=4

Therefore, the solution is (-2, 4, -2)

3 0
3 years ago
Read 2 more answers
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