All categories

3 years ago

A refrigerator cost 295 more than an oven costs. The oven cost 325 how much does the refrigerator cost

Mathematics

2 answers:

FinnZ [79.3K]3 years ago

7 0

Answer:620

Step-by-step explanation:

325+295=620

vampirchik [111]3 years ago

7 0

Answer: The refrigerator cost 620 dollars.

Step-by-step explanation: To solve you have to add 325 and 295.

You might be interested in

John can read 15 pages in 10 minutes. He had 1 hour and 10 minutes before football practice. How many pages can he read before p

krok68 [10]

Answer:

100

Step-by-step explanation:

15 times 6 plus 10 because there is 60 mintues in an hour

7 0

3 years ago

The product of two rational numbers is ______ rational.

Vika [28.1K]

The product of two rational numbers is ALWAYS rational

5 0

3 years ago

Read 2 more answers

Order these numbers from least to the greatest, 47/10,4.606,4+13/20,4.64

exis [7]

Answer:

From lowest to highest, the numbers would follow the following order:

1) 4.606, 2) 4.64, 3) 4 + 13/20, 4) 47/10.

Step-by-step explanation:

Given that the numbers 47/10, 4.606, 4 + 13/20 and 4.64 are presented, to order these from lowest to highest, the following calculations must be performed:

47/10 = 4.70

4,606

4 + 13/20 = 4 + 0.65 = 4.65

4.64

Therefore, from lowest to highest, the numbers would follow the following order:

1) 4.606, 2) 4.64, 3) 4 + 13/20, 4) 47/10.

6 0

3 years ago

Helpppppppppp plzzzzz

Ratling [72]

so do all the question like this types

thankyou for your question

4 0

3 years ago

Evaluate 1^3 + 2^3 +3^3 +.......+ n^3

Molodets [167]

Notice that

$(n+1)^4-n^4=4n^3+6n^2+4n+1$

so that

$\displaystyle\sum_{i=1}^n((n+1)^4-n^4)=\sum_{i=1}^n(4i^3+6i^2+4i+1)$

We have

$\displaystyle\sum_{i=1}^n((i+1)^4-i^4)=(2^4-1^4)+(3^4-2^4)+(4^4-3^4)+\cdots+((n+1)^4-n^4)$

$\implies\displaystyle\sum_{i=1}^n((i+1)^4-i^4)=(n+1)^4-1$

so that

$\displaystyle(n+1)^4-1=\sum_{i=1}^n(4i^3+6i^2+4i+1)$

You might already know that

$\displaystyle\sum_{i=1}^n1=n$

$\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2$

$\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6$

so from these formulas we get

$\displaystyle(n+1)^4-1=4\sum_{i=1}^ni^3+n(n+1)(2n+1)+2n(n+1)+n$

$\implies\displaystyle\sum_{i=1}^ni^3=\frac{(n+1)^4-1-n(n+1)(2n+1)-2n(n+1)-n}4$

$\implies\boxed{\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4}$

If you don't know the formulas mentioned above:

The first one should be obvious; if you add $n$ copies of 1 together, you end up with $n$ .
The second one is easily derived: If $S=1+2+3+\cdots+n$ , then $S=n+(n-1)+(n-2)+\cdots+1$ , so that $2S=n(n+1)$ or $S=\dfrac{n(n+1)}2$ .
The third can be derived using a similar strategy to the one used here. Consider the expression $(n+1)^3-n^3=3n^2+3n+1$ , and so on.

7 0

4 years ago