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3 years ago

Jordy walked 1/3 mile in 15 minutes walking at this same pace how far would he get in an hour

2 answers:

5 0

You use ratio here: in 15 minutes he went 1/3 miles so it could be written like this, 15 min: 1/3 miles. And then write 60 under 15 because 1 hour has 60 minutes. Now multiply 1/3 by 60 and divide the answer by 15

Digiron [165]3 years ago

3 0

Answer:

1.3 miles is your answer

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Yes, SSS

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3 years ago

13x -2y -5x +8

Sati [7]

Step-by-step explanation:

1. How many terms are in the expression?

The terms are sorted by looking at the sing/indicator in front of them. For this expression there are 4 terms

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2. Which terms are “like terms”?

13x and -5x are like terms.

3. Are there any constants? If so, what are they?

Yes there is a constant terms and it's 8

4. What are the variables in the expression?

The variables in this expression are x, and y

5. What are the coefficients in the expression?

Coefficients is the number in front of variables in this expression the coefficients are 13, -2, and -5

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3 years ago

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You leave a 10% tip on a $30 dinner bill. Find he amount of the tip and add it to

Korolek [52]

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3 years ago

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What is the distance between these two points? Write in simplest radical form.

kaheart [24]

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Step-by-step explanation:

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2 years ago

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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:

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

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1 year ago