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goldenfox [79]
3 years ago
8

A linear relationship has a constant rate of change. agree? or disagree? and why?

Mathematics
2 answers:
mestny [16]3 years ago
7 0
Yes, a linear relationship (mathematical) is a constant change happening to x, for example, if in the first step it adds 2 to the table chart or graph, it will always be adding 2 every step of the equason
slega [8]3 years ago
5 0
I agree based on the question
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49/10 as a mixed number
anastassius [24]
4 9/10 is your answer because you first have to divide the numerator by the denominator. this will give you 4, with a remainder of 9. The mix number can be created by using the 4 as the whole number, the 9 as the numerator and the 10 as a denominator. So, 49/10 = 4 9/10. Hope this works!
5 0
3 years ago
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Y= -0.25x + 4.7<br> Y=4.9x - 1.64
Andreas93 [3]

Answer:

The answer would be (1.23, 4.39)

Step-by-step explanation:

Because they are both equal to y, we can set the equations equal to each other and then solve.

4.9x - 1.64 = -0.25x + 4.7

5.15x - 1.64 = 4.7

5.15x = 6.34

x = 1.23

Now that we have the value for x, we can plug into either equation and find y.

y = -0.25x + 4.7

y = -0.25(1.23) + 4.7

y = -.31 + 4.7

y = 4.39

3 0
3 years ago
BRAINLIEST PLUS 22 POINTS
Vika [28.1K]
Hi there!

Angles that are complementary add up to 90. We know that in order to find the value of x, we'll need to create an equation. This equation would be (x + 15) + 48 = 90. This is because, together, the two angles must add up to 90.

ANSWER:
D - (x + 15) + 48 = 90

Hope this helps!! :)
If there's anything else that I can help you with, please let me know!
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3 years ago
Tim is an elementary school art teacher. His students are sculpting a replica of a shark out of clay. Tim has given them one blo
Darina [25.2K]
Radius = 1.25cm
81cm3 / 20 = 4.05 cm3 per cone
V cone = πr^2 (h/3)
So then 4.05 = π(1.25)^2 (h/3)
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3 years ago
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The area of a circle of radius 12 units is equal to the surface area of a sphere of radius 6 units. True or false
9966 [12]
The answer should be true
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3 years ago
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