Jack jogs at an average speed of 40 m/min. How far does he jog in 1.2 hours? m

1 answer:

4 0

So basically you would create an equation from y=mx+b, so you would put 40 in for m and to find x you convert 1.2 hours into minutes which is 72 minutes. so multiply 40 times 72 and that’s how far Jack will be i believe. Jack jogged 2,880m/min. Hope it’s correct.

You might be interested in

AB = x + 3

torisob [31]

Answer:

Step-by-step explanation:

IF it is a SYMMETRIC kite

CB = CD

x + 4 = 2x - 1

x - x + 4 + 1 = 2x - x - 1 + 1

5 = x

AB = AD

x + 3 = 3x - y

5 + 3 = 3(5) - y

8 = 15 - y

y = 7

4 0

2 years ago

I need a little bit of help on this. Thought I knew how to do it

Mariana [72]

Answer:

U should do like this...

6 0

3 years ago

What is one of the solutions to the following system of equations? y2 x2 = 53 y − x = 5 (−10, −5) (−7, −2) (7, 2) (5, 10).

Aliun [14]

The solution of the system of equation is (-7, -2).

<h2>Given</h2>

The system of the equations;

$\rm y^2 + x^2 = 53 \\\\y -x = 5$

From equation 1

$\rm y -x=5\\\\y = 5+x$

Substitute the value of y in equation 1

$\rm y^2+x^2=53\\\\(5+x)^2+x^2=53\\\\25+x^2+10x+x^2=53\\\\2x^2+10x+25-53=0\\\\2 x^2 + 10x - 28 = 0 \\\\Divide \ by \ 2 \ both \ side \ the \ equation\\\\x^2 + 5x - 14 = 0\\\\x^2-7x+2x-14=0\\\\x(x-7) + 2(x-7) =0\\\\(x-7) \ (x+2)=0\\\\x-7=0 \ \ x = 7\\\\x +2 =0 \ \ x=-2$