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kirill115 [55]
3 years ago
6

Write the equation of the line in point-slope form. PLEASE HELP!!

Mathematics
1 answer:
Andru [333]3 years ago
3 0

The answer would be option D.

You can use desmos.com/calculator for graphing these equations.

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Which of the following sets could be the sides of a right triangle?
coldgirl [10]

Answer: Choice B) {3, 5, sqrt(34)}

=====================================

Explanation:

We can only have a right triangle if and only if a^2+b^2 = c^2 is a true equation. The 'c' is the longest side, aka hypotenuse. The legs 'a' and 'b' can be in any order you want.

-----------

For choice A,

a = 2

b = 3

c = sqrt(10)

So,

a^2+b^2 = 2^2+3^2 = 4+9 = 13

but

c^2 = (sqrt(10))^2 = 10

which is not equal to 13 from above. Cross choice A off the list.

-----------

Checking choice B

a = 3

b = 5

c = sqrt(34)

Square each equation

a^2 = 3^2 = 9

b^2 = 5^2 = 25

c^2 = (sqrt(34))^2 = 34

We can see that

a^2+b^2 = 9+25 = 34

which is exactly equal to c^2 above. This confirms the answer.

-----------

Let's check choice C

a = 5, b = 8, c = 12

a^2 = 25, b^2 = 64, c^2 = 144

So,

a^2+b^2 = c^2

25+64 = 144

89 = 144

which is a false equation allowing us to cross choice C off the list.

7 0
3 years ago
Write two integers with different signs that have a sum of -25.
Scilla [17]

Answer:

1.

-25+0

-30+5

2.

-15-10

-20-5

3 0
3 years ago
A line passes through (−1, 7) and (2, 10).
Murljashka [212]
The formula is y=mx+b
To get the slope or m, use this formula
(the second y minus the first y)/(the second x minus the first x)
Now set it up.
(10-7)/(2--1)
3/3=slope is 1.
y=1x+b

Insert one of the points for x and y.
i will do (-1,7)
7=1(-1)+b
7=-1+b
8=b
Insert this into the final equation:
y=1x+8

Try it out. If you're not sure, try both points. If it works, then you set it up correctly.
6 0
3 years ago
Read 2 more answers
What is the value of 6x-3x +7 if x=9?
Lynna [10]

6x - 3x + 7 \\ 3x + 7 \\ 3(9) + 7 \\ 27 + 7 \\  = 34
Combine like terms and "plug in" x into the equation to eventually get 34 after some addition
8 0
3 years ago
The radius of one circle is three the time radius of another circle. The sum of their areas is 12pi. Find the radius of each cir
zlopas [31]
   
\displaystyle\\
\begin{cases}
R_1 = 3R_2\\
\pi R_1^2 +\pi R_2^2=12\pi ~~~\Big|~~:\pi 
\end{cases} \\  \\ 
\begin{cases} 
R_1 = 3R_2\\
R_1^2 + R_2^2=12 
\end{cases} \\  \\ 
(3R_2)^2+ R_2^2=12 \\  \\ 
9R_2^2 + R_2^2=12 \\  \\ 
10R_2^2=12\\\\
R_2^2 = \frac{12}{10} = \frac{6}{5}  \\  \\ 
R_2 =  \boxed{\sqrt{\frac{6}{5}} } \\  \\ 
R_1 = \boxed{3\sqrt{\frac{6}{5}} }



7 0
3 years ago
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