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mel-nik [20]
3 years ago
7

Determine which ordered pair is a solution of y=8x.

Mathematics
2 answers:
MAXImum [283]3 years ago
7 0
Y = 8x
y = 8(2)
y = 16
(x, y) = (2, 16)

The answer is D.
MrRissso [65]3 years ago
6 0

we know that

If the ordered pair is a solution of the equation, then the ordered pair must satisfied the equation

we will proceed to solve each case to determine the solution of the problem

we have

y=8x -------> equation 1

<u>case a)</u> (0,8)

x=0\\y=8

For x=0

substitute the value of x in the equation 1 and then compare the values of y

y=8*0=0

so

0\neq 8

The ordered pair case a) is not solution

<u>case b)</u> (-1,8)

x=-1\\y=8

For x=-1

substitute the value of x in the equation 1 and then compare the values of y

y=8*(-1)=-8

so

-8\neq 8

The ordered pair case b) is not solution

<u>case c)</u> (1.5,10)

x=1.5\\y=10

For x=1.5

substitute the value of x in the equation 1 and then compare the values of y

y=8*1.5=12

so

12\neq 10

The ordered pair case c) is not solution

<u>case d)</u> (2,16)

x=2\\y=16

For x=2

substitute the value of x in the equation 1 and then compare the values of y

y=8*2=16

so

16=16

The ordered pair case d) is a solution

therefore

<u>the answer is</u>

(2,16)

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4^3 = ?<br><br>A. 64<br>B. 7<br>C. 12<br>D. 1
Alexxandr [17]

By defining how exponents work, we conclude that 4^3 = 64, so the correct option is A.

<h3>How to find the cube of 4?</h3>

Let's remember how exponents work.

If we have a number a to the exponent n:

a^n

It means that we need to multiply the number a by itself n times.

Then, in this case:

4^3

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4*4*4

Now we can solve that product:

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Then we conclude that the correct option is A

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If you want to learn more about exponents:

brainly.com/question/847241

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8 0
2 years ago
(1/3)x-4(-1)=2 solve for x.
Elodia [21]

Answer: X = -6

Easy Peasy

3 0
3 years ago
5/8 + 3/4 -2/3 - 5\6​
Iteru [2.4K]

Answer:

5/8+3/4-2/3-5/6

= ( 5/8 + 3/4 ) - ( 2/3 + 3/4)

= ( 11/8) - (17 /12)

= 11/8 - 17/12

= 33-34 /24

= -1/24 (ans)

Hope it helps

5 0
3 years ago
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