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3 years ago

Which of the following lists is ordered from least to greatest?

Mathematics

2 answers:

gladu [14]3 years ago

8 0

Answer:

Step-by-step explanation:

Inessa05 [86]3 years ago

5 0

Answer:

D! :)

Step-by-step explanation:

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Can someone help me out please??

olganol [36]

Answer:

the 2nd one is the answer.

x ≥ −8

Step-by-step explanation:

Hope that helps

Sorry if wrong

8 0

3 years ago

Function or not a Function

Vadim26 [7]

Answer:

Function; Not a function

Step-by-step explanation:

Write the ordered pair in the first set (x, y): (1, 11), (2, 7) , (3, 2)

Write the ordered pair in the second set (x, y): (1, 3), (1, 4), (2, 4) , (3, 2)

The first set in a function because no ordered pair has the same x value.

The second set in not a function because the x value is the same in two of the ordered pairs.

5 0

2 years ago

Help ASAP I stink at these i always give brainliest and thanks

denis23 [38]

Answer:

Step-by-step explanation:

Using the recursive formula with a₁ = 5 , then

a₂ = 2a₁ - 7 = 2(5) - 7 = 10 - 7 = 3

a₃ = 2a₂ - 7 = 2(3) - 7 = 6 - 7 = - 1

a₄ = 2a₃ - 7 = 2(- 1) - 7 = - 2 - 7 = - 9 → A

7 0

3 years ago

Can somebody check my answer

almond37 [142]

Answer:

(-4, -3)

Step-by-step explanation:

That's Correct!

x + 4 shifts the origin 4 left

y + 3 shifts the origin 3 down

(-4, -3)

4 0

3 years ago

Yea I can do tomorrow it would take me to get the money back to you

Tems11 [23]

Given, the equation that represents the height of an object:

$y(t)=100t-16t^2$

First, we will find the velocity of the object which is the first derivative of the height using the method of the limits

$\frac{dy}{dt}=\lim_{h\to a}\frac{y(3+h)-y(3)}{(3+h)-(3)}$

We will find the value of the function y(t) when t = 3, and when t = 3+h

$\begin{gathered} y(3+h)=100(3+h)-16(3+h)^2 \\ y(3+h)=300+300h-16(9+6h+h^2) \\ y(3+h)=300+300h-144-96h-16h^2 \\ y(3+h)=156+4h-16h^2 \\ y(3)=100(3)-16(3)^2=156 \end{gathered}$

Substitute y(3+h) and y(3) into the expression of the limit

$\begin{gathered} \frac{dy}{dt}|_{t=3}=\lim_{h\to a}\frac{156+4h-16h^2-156}{3+h-3}=\lim_{h\to a}\frac{4h-16h^2}{h} \\ \\ \frac{dy}{dt}|_{t=3}=\lim_{h\to a}(4-16h) \end{gathered}$

Where a = 0

d) compute the instantaneous velocity at t = 3

$\frac{dy}{dt}|_{t=3}=100-16*2*3=4$

So, the answer will be:

$\begin{gathered} \frac{dy}{dt}|_{t=3}=\lim_{h\to a}(4-16h) \\ a=0 \\ \\ \frac{dy}{dt}|_{t=3}=4\text{ ft/sec} \end{gathered}$

5 0

1 year ago