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4 years ago

What set of coordinates does not describe a function?

Mathematics

2 answers:

masya89 [10]4 years ago

7 0

Answer:

its c

Step-by-step explanation:

Wittaler [7]4 years ago

7 0

D: as each input can only have one output and 2 has both 8 and 2 as outputs

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Paul has 16 fish pictures and 12 deer pictures. He wants to arrange the pictures in rows with the same numbers of fish and of de

VladimirAG [237]

Answer:

4 rows

Step-by-step explanation:

GCF of 12 and 16 is4

5 0

3 years ago

What expression is equivalent to <br> 2/3 X 5

Flura [38]

Answer:

1/3

Step-by-step explanation:

1/3 can be used as an expression that is exactly the same as 2/3x5.

When you multiply 2/3 x 5-

Look here if you need help solving 2/3 X 5: (2/3 x 5/1...is the simplest way to multiply the two, make sure to multiply 2 by 5 and 3 by 1 to get your answer.)(5x2=10...3x1=3)

-you get 10/3. This answer simplified is 1/3. So in the end, 1/3 is a great expression or answer to use as an equivalent to the equation 2/3 x 5.

Hoped this helped! :)

8 0

2 years ago

I can’t solve this. Please explain!

babunello [35]

Sorry I can’t help....

6 0

3 years ago

Read 2 more answers

Order the decimals from least to greatest 48.6, 1.6, 0.13, 153.7, 2.2, 49.2

postnew [5]

Answer:

0.13, 1.6, 2.2, 48.6, 49.2, 153.7

4 0

4 years ago

Read 2 more answers

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 43 ft/s. Its height

Lina20 [59]

Answer:

Instantaneous Velocity at t = 1 is 20 feet per second

Step-by-step explanation:

We are given he following information in the question:

$y(t)=43t-23t^{2}$

B) Instantaneous Velocity at t = 1

$y(1) = 43(1)-23(1)^2 = 20$

A) Formula:

Average velocity =

$\displaystyle\frac{\text{Displacement}}{\text{Time}}$

1) 0.01

$y(1 + 0.01)=y(1.01) = 43(1.01)-23(1.01)^{2} = 19.9677\\\\\text{Average Velocity} = \dfrac{y(1.01)-y(1)}{1.01-1} =\dfrac{19.9677-20}{1.01-1}= \dfrac{-0.0323}{0.01} = -3.230000 \text{feet per second}$

2) 0.005 s

$y(1 + 0.005)=y(1.005) = 43(1.005)-23(1.005)^{2} = 19.984425\\\\\text{Average Velocity} = \dfrac{y(1.005)-y(1)}{1.005-1} =\dfrac{19.984425-20}{1.005-1} = -3.1150000 \text{feet per second}$

3) 0.002 s

$y(1 + 0.002)=y(1.002) = 43(1.002)-23(1.002)^{2} = 19.993908\\\\\text{Average Velocity} = \dfrac{y(1.002)-y(1)}{1.002-1} =\dfrac{19.993908-20}{1.002-1} = -3.0460000 \text{feet per second}$

4) 0.001 s

$y(1 + 0.001)=y(1.001) = 43(1.001)-23(1.001)^{2} = 19.996977\\\\\text{Average Velocity} = \dfrac{y(1.001)-y(1)}{1.001-1} =\frac{19.996977-20}{1.001-1} = -3.0230000 \text{feet per second}$

3 0

3 years ago