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prisoha [69]
2 years ago
10

Pls help?!

Mathematics
1 answer:
pshichka [43]2 years ago
4 0

Answer:

See attached file.

Step-by-step explanation:

Firstly, plot the point (6,2).

Remember (x,y)

Next plot the slope point.

Remember rise/run or rise over run.

You would go up 1 unit from the point and to the right 3.

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mrs smith agrees to pay 5 neiborhood boys $20 to clean her yard. they do such a good job she gives them some extra money, for a
Artemon [7]
They all would get $4.70.
8 0
3 years ago
Help i need help with this 10x+16>6x+20
Mamont248 [21]

What does > mean?

10x + 16 ≥ 6x + 20

Subtract both sides by 16

10x ≥ 6x + 4

Subtract both sides by 6x

4x ≥ 4

Divide both sides by 4

x ≥ 1

I believe this is your answer good luck!!

<<<Emily>>>

4 0
3 years ago
What is the vertex of the absolute value function defined by ƒ(x) = |x + 5| + 7?
Temka [501]
To answer this, you need to know the general form of an absolute value function. the equation for this is f(x<span>) = </span>a|x<span> - </span>h<span>| + </span>k, and in this equation, the vertex is (h, k).

with that information, you can see that your vertex will be (-5, 7). you must take the negative for 5 because the general equation states that your h value is usually subtracted from x. to check your vertex, try plugging it into your general equation:
f(x) = a|x - (-5)| + 7
f(x) = a|x + 5| + 7 ... you see that this matches your given equation. this last part here was just to show why your 5 must be negative; your answer is bolded.
7 0
3 years ago
Which are the solutions of the quadratic equation? x2 = –5x – 3 –5, 0 StartFraction negative 5 minus StartRoot 13 EndRoot Over 2
boyakko [2]

<u>Given</u>:

The quadratic equation is x^{2}=-5 x-3

We need to determine the solutions of the quadratic equation.

<u>Solution</u>:

Let us solve the equation to determine the value of x.

Adding both sides of the equation by 5x and 3, we get;

x^{2}+5 x+3=0

The solution of the equation can be determined using quadratic formula.

Thus, we get;

x=\frac{-5 \pm \sqrt{5^{2}-4 \cdot 1 \cdot 3}}{2 \cdot 1}

x=\frac{-5 \pm \sqrt{25-12}}{2 }

x=\frac{-5 \pm \sqrt{13}}{2 }

Thus, the two roots of the equation are x=\frac{-5 + \sqrt{13}}{2 } and x=\frac{-5- \sqrt{13}}{2 }

Hence, the solutions of the equation are x=\frac{-5 + \sqrt{13}}{2 } and x=\frac{-5- \sqrt{13}}{2 }

6 0
3 years ago
Read 2 more answers
Jason is driving at 55 mph. He needs to keep his car within 5 miles per hour of his current speed. Write and solve an absolute v
djverab [1.8K]
Maximum speed: 55mph + 5mph = 60mph
minimum speed:  55mph - 5mph  = 50mph
Hope this helped?
5 0
3 years ago
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