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3 years ago

What is the sum of all possible multiples of 5 less than 500

Mathematics

1 answer:

Leona [35]3 years ago

8 0

Answer:

We need to find 5 + 10 + 15 + ... + 485 + 490 + 495. There are 99 numbers that are being added. To add this we can do 5 + 495, 10 + 490, 15 + 485 and so on. These pairs have a sum of 500 and since there are 98 / 2 = 49 pairs their sum is 500 * 49 = 24500. Since there is an odd number of numbers that are being added, there is one number that does not belong to a pair. That number is the 50th number in the list which is 50 * 5 = 250. Therefore the answer is 24500 + 250 = 24750.

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Factor the Trimonial y to the second +3y-54

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2 years ago

Help please :c!

LUCKY_DIMON [66]

Answer:

see the procedure

Step-by-step explanation:

we know that

m∠1+m∠2=180° -----> supplementary angles (form a linear pair)

we have

m∠1=27° ----> given problem

substitute the measure of m∠1 in the equation above and solve for m∠2

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4 0

3 years ago

It's been a while since I've taken a math class and I'm stressed please help me out. The question is circled.

brilliants [131]

-3/2x

The rational expression would not be defined only when the denominator is divided by 0

That is:

2x =0

x = 0/2

x =0

Hence it would not be defined when x = 0.

Hope this helps.

8 0

3 years ago

Read 2 more answers

What is the common denominator of (5/x^2-4) - (2/x+2) in the complex fraction (2/x-2) - (3/x^2-4)/(5/x^2-4) - (2/x+2)

PIT_PIT [208]

9514 1404 393

Answer:

common denominator: (x² -4)
simplified complex fraction: (2x +1)/(9 -2x)

Step-by-step explanation:

It is helpful to remember the factoring of the difference of squares:

a² -b² = (a -b)(a +b)

Your denominator of (x² -4) factors as (x -2)(x +2). You will note that one of these factors is the same as the denominator in the other fraction.

It looks like you want to simplify ...

$\dfrac{\left(\dfrac{2}{x-2}-\dfrac{3}{x^2-4}\right)}{\left(\dfrac{5}{x^2-4}-\dfrac{2}{x+2}\right)}=\dfrac{\left(\dfrac{2(x+2)}{(x-2)(x+2)}-\dfrac{3}{(x-2)(x+2)}\right)}{\left(\dfrac{5}{(x-2)(x+2)}-\dfrac{2(x-2)}{(x-2)(x+2)}\right)}\\\\=\dfrac{2(x+2)-3}{5-2(x-2)}=\boxed{\dfrac{2x+1}{9-2x}}$

3 0

3 years ago