Ask question

Ask question

All categories

2 years ago

12

Gordon recently started jogging. Last week he jogged for 15 minutes. Starting this week, he decided to add 2 minutes to his jog

each week. Which of the following equations represents how long his jog will be after x additional weeks?

1 answer:

MissTica2 years ago

5 0

This situation can be represented by this equation: y = 15 + 2x

You might be interested in

What is the solution of 5y + 1=31

gayaneshka [121]

Answer:

y=6

Step-by-step explanation:

edge2020

4 0

2 years ago

Read 2 more answers

What operation does the Product Rule use

kumpel [21]

Answer:

addition

Step-by-step explanation:

Example: 5^1 * 5^3= 5^4

4 0

2 years ago

how many pounds of nuts selling for $9 per lb and raisins selling for $3 per lb should a person combine to obtain 60 lb of a tra

marta [7]

Let n be the number of lbs of nuts.
Let r be the number of lbs of raisins.
9n + 3r = 60 * 5 = 300 .............(1)
n + r = 60
n = 60 - r ...............(2)
Plugging the expression for n in (2) into (1) we get:
9(60 - r) + 3r = 300
which expands to:
540 - 9r + 3r = 300 .............(3)
Simplifying and rearranging (3) we get:
-6r = -240
Therefore r, the weight of raisins = 40 lbs
and the weight of nuts is 60 - 40 = 20 lbs

8 0

2 years ago

[(1/(3+x))-(1/3)] / x<br> Limit 0

Effectus [21]

First find a common denominator and combine the fractions in the numerator:

$\displaystyle\lim_{x\to0}\frac{\dfrac1{3+x}-\dfrac13}x=\lim_{x\to0}\frac{\dfrac3{3(3+x)}-\dfrac{3+x}{3(3+x)}}x=\lim_{x\to0}\frac{3-(3+x)}{3x(3+x)}$

Now simplify and cancel out all the terms that you can:

$\displaystyle\lim_{x\to0}\frac{3-3-x}{3x(3+x)}=-\frac13\lim_{x\to0}\frac1{3+x}$

Since the remaining expression is continuous as a function of $x$ , you can directly substitute to end up with

$\displaystyle-\frac13\lim_{x\to0}\frac1{3+x}=-\frac13\times\frac1{3+0}=-\frac19$

8 0

2 years ago

Need help<br> 2x+10 <br> 2x<br> find x

Yuri [45]

Answer:5

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers