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lukranit [14]
3 years ago
7

How do i find the perimeter of the triangle with vertices at three points (0,1),(2,1), and (0,3)?

Mathematics
2 answers:
Harlamova29_29 [7]3 years ago
7 0
Well I'm not just going to tell you the answer, you need to "experience" how you do it. Basically you use the distance formula, which I'll show you in a bit, to fine the length of the segments. Then you just add them up.
\sqrt{ (x_{2}- x_{1}) ^{2} +(y_{2}-y_{1})^{2}
suter [353]3 years ago
5 0
The perimeter is the sum of the lengths of the three sides.

To find the perimeter, you need to find the lengths of the three sides.

For each side, you are given the points at the two ends.You need to find the distance between those two points.

Once you have all thee distances, add them up.

That's how.
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Solve for b -233=6(3+7b)+1 a. -10 b. -3 c. -6 d. -13
Svet_ta [14]

Answer: b = -6

Step-by-step explanation:

-233 = 6(3+7b) +1

-233 = 18 + 42b +1

-233 - 18 - 1 = 42b

-252 = 42b

-252/42 = b

b = -6

4 0
3 years ago
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anyanavicka [17]
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Select the two values of x that are roots of this equation.<br> 2x2 + 11x+15= 0
USPshnik [31]

Answer:

The two values of x are -2.5 and -3

Step-by-step explanation:

we have

2x^{2}+11x+15=0

The formula to solve a quadratic equation of the form ax^{2} +bx+c=0 is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

a=2\\b=11\\c=15

substitute in the formula

x=\frac{-11(+/-)\sqrt{11^{2}-4(2)(15)}} {2(2)}

x=\frac{-11(+/-)\sqrt{1}} {4}

x=\frac{-11(+/-)1} {4}

x_1=\frac{-11(+)1} {4}=-2.5

x_2=\frac{-11(-)1} {4}=-3

therefore

The two values of x are -2.5 and -3

6 0
3 years ago
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Haley and mia's dad said if they combined the money they've saved. He would double it.
irina [24]

Answer:

I have no idea, I have been trying to figure it out

8 0
3 years ago
Which equation is equivalent to<br> y-1= 5(x+2) ?
Agata [3.3K]

Step-by-step explanation:

We can simply by opening the brackets ,

=> y - 1 = 5( x - 2 )

=> y - 1 = 5x - 10

=> y - 5x = -9

=> 5x - y + 9 = 0

8 0
3 years ago
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