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Reika [66]
2 years ago
10

find the constant of variation for the relation and use it to write an equation for the statement. Then solve the equation. if y

varies directly as X and Z, and t=8/3 when x=1 and z=4, find y when x=6 and z=3

Mathematics
1 answer:
dlinn [17]2 years ago
7 0

Answer:

The Question isn't clear

Step-by-step explanation:

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Draw and upload a dot plot to represent the following data.
Nonamiya [84]

Answer:

I have the dot plot blow.

Step-by-step explanation:

Hope it's right and helps.

5 0
3 years ago
I need help on answering this question.​
-Dominant- [34]

Answer:

The value of m is 3 because (5^4)^3 = 244140625 and 5^12 = 244140625 also, this means they are both equal to each other. Another way of getting the correct answer is just seeing what times 4 will get you 12, as you can see

3 x 4 = 12.

So yea the answer is M = 3.

Step-by-step explanation:

5 0
3 years ago
Find the amount of time. I $2,064, P = $10,000, r = 6.88%
vagabundo [1.1K]

Answer:

Can u show an image

Step-by-step explanation:

5 0
3 years ago
The segments in each figure are tangent to the circle. Find each length.
zubka84 [21]
The tangents are equal length, so
.. 8x = x^2
We presume their length is not zero, so we can divide by x to get
.. 8 = 2x
.. 4 = x

Then the length of each tangent is 8*4 = 32 units.
3 0
3 years ago
One-stray Problem Solving
anygoal [31]
A) 5000 m²
b) A(x) = x(200 -2x)
c) 0 < x < 100
Step-by-step explanation:
b) The remaining fence, after the two sides of length x are fenced, is 200-2x. That is the length of the side parallel to the building. The product of the lengths parallel and perpendicular to the building is the area of the playground:
A(x) = x(200 -2x)
__
a) A(50) = 50(200 -2·50) = 50·100 = 5000 . . . . m²
__
c) The equation makes no sense if either length (x or 200-2x) is negative, so a reasonable domain is (0, 100). For x=0 or x=100, the playground area is zero, so we're not concerned with those cases, either. Those endpoints could be included in the domain if you like.
3 0
2 years ago
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