Ask question

Ask question

All categories

Keith_Richards [23]

3 years ago

8

Jon’s weight loss for each week of the month is 4 lbs., 3.5 lbs. and 2 lbs. He gained 4.5 lbs. the last week. If Jon originally

weighted 153 lbs., how much does he weigh now?
***only enter the numerical value, do not put the units***

1 answer:

nekit [7.7K]3 years ago

4 0

Answer: 148 lbs

first you do 4+3.5+2=9.5 then you do 9.5-4.5=5 so then you do 153-5 to get your answer 148 lbs

i hope this helps :)

You might be interested in

Twice a number, decreased by 9 is 12

Serhud [2]

Answer:

10.5

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

In a Calculus class, there are 11 freshman and 15 sophomores; 5 of the sophomores are females,

KATRIN_1 [288]

Answer:

61.5%

Step-by-step explanation:

Let's find all the demographics first:

Sophomores in the class(total 15):

Female: 5

Male : 10

Freshmen in the class(total 11):

Female: 3

Male: 8

There are 11 freshmen(male and female) and 5 female sophomores. Thus, the probability of choosing one of these 16 people in a class of 26 is 16/26 or 61.5%.

4 0

3 years ago

Michael got a loan for the Playstation 5. His loan was $600. At a 3.75% interest rate he had to pay $7.50 in interest. How long

Semmy [17]

Answer:

4 mo.

Step-by-step explanation:

I = PRT P = 600 R = .0375 I = 7.50

7.50 = 600(.0375)T

7.5 = 22.5T

T = 7.5/22.5 = 1/3 yr = 4 mo.

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=Simplify%3A%20%5Cfrac%7B%205%C3%97%2825%29%5E%7Bn%2B1%7D%20-%2025%20%C3%97%20%285%29%5E%7B2n%7

Katen [24]

$\green{\large\underline{\sf{Solution-}}}$

<u>Given expression is </u>

$\rm :\longmapsto\:\dfrac{5 \times {25}^{n + 1} - 25 \times {5}^{2n} }{5 \times {5}^{2n + 3} - {25}^{n + 1} }$

can be rewritten as

$\rm \: = \: \dfrac{5 \times { {(5}^{2} )}^{n + 1} - {5}^{2} \times {5}^{2n} }{5 \times {5}^{2n + 3} - {( {5}^{2} )}^{n + 1} }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {( {x}^{m} )}^{n} \: = \: {x}^{mn}}}} \\$

And

$\purple{\rm :\longmapsto\:\boxed{\tt{ \: \: {x}^{m} \times {x}^{n} = {x}^{m + n} \: }}} \\$

So, using this identity, we

$\rm \: = \: \dfrac{5 \times {5}^{2n + 2} - {5}^{2n + 2} }{{5}^{2n + 3 + 1} - {5}^{2n + 2} }$

$\rm \: = \: \dfrac{{5}^{2n + 2 + 1} - {5}^{2n + 2} }{{5}^{2n + 4} - {5}^{2n + 2} }$

can be further rewritten as

$\rm \: = \: \dfrac{{5}^{2n + 2 + 1} - {5}^{2n + 2} }{{5}^{2n + 2 + 2} - {5}^{2n + 2} }$

$\rm \: = \: \dfrac{ {5}^{2n + 2} (5 - 1)}{ {5}^{2n + 2} ( {5}^{2} - 1)}$

$\rm \: = \: \dfrac{4}{25 - 1}$

$\rm \: = \: \dfrac{4}{24}$

$\rm \: = \: \dfrac{1}{6}$

<u>Hence, </u>

$\rm :\longmapsto\:\boxed{\tt{ \dfrac{5 \times {25}^{n + 1} - 25 \times {5}^{2n} }{5 \times {5}^{2n + 3} - {25}^{n + 1} } = \frac{1}{6} }}$

4 0

3 years ago

In order to pay off he car she bought lauri had to make 34 more payments of $145.98. How much does she still owe?

Talja [164]

4963.32 is the answer your looking for man.

8 0

3 years ago