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Julli [10]
2 years ago
6

A weightlifter can increase the weight W(x) that she can lift according to W(x) = 315(1.05)", where x represents the number of t

raining cycles completed. How much will she
lift after 4 training cycles?
Mathematics
1 answer:
labwork [276]2 years ago
8 0

Answer:

She will lift 383 units.

Step-by-step explanation:

The amount of weight that she can lift after x cycles is given by:

W(x) = 315(1.05)^x

How much will she lift after 4 training cycles?

This is W when x = 4, that is, W(4). So

W(x) = 315(1.05)^4 = 383

She will lift 383 units.

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3 years ago
An ostrich run 6 mile in12 minut how far he could come in 40 minutes
bogdanovich [222]

The ostrich can run 20 miles in 40 minutes.

<u>Solution:</u>

Given that, An ostrich run 6 mile in 12 minutes

We have to find how far he could come in 40 minutes

Now, according to the given information  

Ostrich runs 6 miles ⇒ 12 minutes

Then, “n” miles ⇒ 40 minutes

Now, by criss cross multiplication we get,

\begin{array}{l}{\rightarrow 6 \times 40=n \times 12} \\\\ {\rightarrow 40=n \times 2} \\\\ {\rightarrow n=20}\end{array}

Hence, the ostrich can run 20 miles in 40 minutes

8 0
2 years ago
What is it called when writing a number by adding the value of each digit
Neko [114]

Answer:

Expanded form

Step-by-step explanation:

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5 0
2 years ago
What is the answer for $ 199 MP3 player 15% discount
Irina-Kira [14]
So discount means taken off
we find 15% of 199 and subtract from 199
% means parts out of 100 so 15%=15/100=1.5/10=.15/1=0.15
'of' means multiply
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8 0
3 years ago
Read 2 more answers
Miki has 104 nickels and 88 dimes. She wants to divide her coins into groups where each group has the same number of nickels and
pochemuha

Given :

Miki has 104 nickels and 88 dimes.

She wants to divide her coins into groups where each group has the same number of nickels and the same number of dimes.

To Find :

Largest number of groups she can have .

Solution :

In the given question we need to find the largest number of groups she can have i.e we have to find the LCM of 104 and 88 .

Now , factorizing both of them , we get :

104=1\times 2\times 2\times 2\times 13

88=1\times 2\times 2\times 2\times 11

Form above , we can say that common factors are :

HCF=2\times 2\times 2=8

Therefore , the largest number of groups she can have is 8 .

Hence , this is the required solution .

6 0
3 years ago
Read 2 more answers
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