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yKpoI14uk [10]
3 years ago
12

Is this right.need help

Mathematics
1 answer:
soldi70 [24.7K]3 years ago
5 0
Why yes it is right. good work
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I need help please..........
Mandarinka [93]
The formula for the slope of a line between two points is m=y2-y1/x2-x1. Substitute the given coordinates into this equation and solve:
m=(48-12)/(32-34)=36/-2= -18
The slope is -18
4 0
3 years ago
Select all that are like terms to 5a^5b^4.
tankabanditka [31]

Answer:

a^5b^4

-a^5b^4

9a^5b^4

Step-by-step explanation:

Like terms will all have the same bases and exponents.

a^5b^4

-a^5b^4

9a^5b^4

they all contain a^5 b^4

8 0
3 years ago
Is this equation correct? 63 ⋅ 73 = 423
alekssr [168]

Answer: Yes; 63 • 73 is equal to (6 • 7)3 or 423.

Step-by-step explanation:This is the answer because if you multiply the 6 and 7 you get 42 and you leave the exponent which in this case is 3 leaving you with 42^3

5 0
3 years ago
If it takes 1 second for sound to travel 1.5 km how much km will it travel in 3.7 second
kirza4 [7]
I think 5.55 will travel in 3.7 seconds
6 0
3 years ago
Two possible solutions of √11-2x=√x^2+4x+4 are –7 and 1. Which statement is true? A) Only x = –7 is an extraneous solution.
11111nata11111 [884]

Answer:

Option D)Neither solution is extraneous.

Step-by-step explanation:

we have

\sqrt{11-2x}=\sqrt{x^{2}+4x+4}

we know that

two possible solutions are x=-7 and x=1

<u><em>Verify each solution</em></u>

Substitute each value of x in the expression above and interpret the results

1) For x=-7

\sqrt{11-2(-7)}=\sqrt{-7^{2}+4(-7)+4}

\sqrt{25}=\sqrt{25}

5=5 ----> is true

therefore

x=-7 is not a an extraneous solution

2) For x=1

\sqrt{11-2(1)}=\sqrt{1^{2}+4(1)+4}

\sqrt{9}=\sqrt{9}

3=3 ----> is true

therefore

x=1 is not a an extraneous solution

therefore

Neither solution is extraneous

7 0
3 years ago
Read 2 more answers
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