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3 years ago

I need help with this problem

Mathematics

1 answer:

erik [133]3 years ago

8 0

I think the answer is 400 square centimetres

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The total cost to attend the university Daniel wants to attend is going to be $12,000 for the first year.

Zielflug [23.3K]

Answer:

$395.83

Step-by-step explanation:

to solve, we first need to subtract what we already have, first the scholarship, wich is a set amount being taken from the original amount

12'000 - 2'500 = 9'500

now we have the second subtracting factor, but this one isn't set in stone and defined, it is half of the amount his parents will pay, so what we can do is divide what we have by two, wich will give us 2 halves

9'500 / 2 = 4'750

now all we have to do is divide again but this time for each month that Daniel needs to save up in, in this case 12

4'750 / 12 = 395.83 (note1)

and there we have it, that is the minimum amount Daniel would save each month

note1: (3 goes on for infinity, usualy this is represented by a line above the repeating number, this is the case of a repeating decimal)

5 0

3 years ago

Plz help

Sergio039 [100]

2(5*4)+2(4*3)+2(3*5)=94 sq.m (Orginial SA)
2(4*5)+2(5*6)+2(6*4)=148 sq.m (Total SA)

148-94=54 (New SA)

4 0

3 years ago

Read 2 more answers

Graph each inequality and graph its solution.<br> 3) n - 11 > -21

Vera_Pavlovna [14]

The correct answer is
Inequality: n>-10
Interval Notation: (-10,infinity)

7 0

3 years ago

On a 40-question test, Peggy answers 34 questions correctly. What percent of the questions does she answer correctly?

marta [7]

Answer:

A.) 85%

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

*Find the formula of the series 1+2²+3²+4²+....+n²*<br>

Sophie [7]

The formula is $\frac{n(n+1)(2n+1)}{6}$

What are series?

In mathematics, we can describe a series as adding infinitely many numbers or quantities to a given starting number or amount.

We will find the formula as shown as below:

Let $S=1+2^2+3^2+4^2+................+n^2$

We know $(n+1)^3=n^3+3n^2+3n+1$

$(1+1)^3=1^3+3(1)^2+3(1)+1$

$(2+1)^3=2^3+3(2)^2+3(2)+1$

$(3+1)^3=3^3+3(3)^2+3(3)+1$

$(n+1)^3=n^3+3(n)^2+3(n)+1$

On adding

$2^3+3^3+4^3......(n+1)^3=(1^3+2^3+3^3+.....+n^3)+3(1^2+2^2+.....+n^2)+3(1+2+3....n)+(1+1+1+....+1)$

$2^3+3^3+4^3......(n+1)^3-(1^3+2^3+3^3+.....+n^3)=3S+\frac{3n(n+1)}{2}+(1+1+1+....+1)$

$(n+1)^3-1^3=3S+\frac{3n(n+1)}{2} +n$

$n^3+3(n)^2+3(n)+1-1=3S+\frac{3n(n+1)}{2} +n$

$2n^3+6n^2+6n=6S+3n^2+3n+2n$

$6S=2n^3+3n^2+n$

$6S=2n^2(n+1)+n(n+1)$

$6S=(n+1)(2n^2+n)$

$6S=n(n+1)(2n+1)$

$S=\frac{n(n+1)(2n+1)}{6}$

Hence, the formula is $\frac{n(n+1)(2n+1)}{6}$

Learn more about Series here:

brainly.com/question/24643676

#SPJ1

3 0

2 years ago