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olya-2409 [2.1K]
3 years ago
14

What is 3000 divided by 25 *Brainliest Offered*

Mathematics
2 answers:
Ugo [173]3 years ago
7 0
 The answer is 120, All you have to do is find out how many times 3000 goes into 25.
Tema [17]3 years ago
3 0
3000 ÷ 25 = 120

Check: 
120(25) = 3000 <------------------True
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If a price decreases by 20% I can calculate 20% of the price and subtract it from the original price or I can calculate it in on
Irina-Kira [14]

Answer:

both statements are fine

Step-by-step explanation:

To find the price of a product after decreasing it, they can be done in the two ways that the statement tells us,

First, calculate that% that was decreased and then subtract it from the original price or simply multiply that original price by the percentage in which the new price would remain.

For example:

let "x" be the original price

in the first case it would be:

0.2 * x and then subtract from x, i.e .:

x - 0.2 * x

in the second case it is:

1 - 0.2 = 0.8

that is, the new price would be 0.8 * x

6 0
3 years ago
Silver's gym charges a $65 sign-up fee and $50 per month for their membership. Solar Fitness charges a $20 sign up fee and $55 p
alexandr1967 [171]
65+50x = 55x + 20
-50x -50
65=5x+20
- 20 - 20
55=5x
55/5= 5x/5
11=x
The solution represents that it 11 months will make both gyms the exact same price.
7 0
3 years ago
Find the missing terms of the geometric sequence.
Debora [2.8K]
Funny cause I just did something similar a few minutes ago

6 0
3 years ago
A. –30<br><br> B .–19<br><br> C. –22<br><br> D. –10
elena-s [515]

Answer:

-30

Step-by-step explanation:

substitute the values and get -30

5 0
3 years ago
Read 2 more answers
Please help with this I am completely stuck on it
vaieri [72.5K]

Answer:

f(x)=\sqrt[3]{x-4} , g(x)=6x^{2}\textrm{ or }f(x)=\sqrt[3]{x},g(x)=6x^{2} -4

Step-by-step explanation:

Given:

The function, H(x)=\sqrt[3]{6x^{2}-4}

Solution 1:

Let f(x)=\sqrt[3]{x}

If f(g(x))=H(x)=\sqrt[3]{6x^{2}-4}, then,

\sqrt[3]{g(x)} =\sqrt[3]{6x^{2}-4}\\g(x)=6x^{2}-4

Solution 2:

Let f(x)=\sqrt[3]{x-4}. Then,

f(g(x))=H(x)=\sqrt[3]{6x^{2}-4}\\\sqrt[3]{g(x)-4}=\sqrt[3]{6x^{2}-4} \\g(x)-4=6x^{2}-4\\g(x)=6x^{2}

Similarly, there can be many solutions.

7 0
3 years ago
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