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3 years ago

What is the sum of the first 100 positive multiples of 4?

Mathematics

2 answers:

Ann [662]3 years ago

7 0

This would be a series: 5 + 10 + 15 + 20 +...+ 500
there are n terms → n = 100
the nth term is 500

Masja [62]3 years ago

6 0

sum = (n/2)(a+b) in which n=number of multiples a = value of first multiple b = value of last multiple So : (100/2) X (4+400) 50 X 404 20200 The final answer is 20200. You got it?

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consider the equation a(4-x) =-3x + b What values of a and b make the solution of the equation x = -5

-BARSIC- [3]

Answer:

Step-by-step explanation:

hello :

put x = -5 in this equation : a(4-x) =-3x + b

a(4+5) =15 + b

9a = 15 +b

b = 9a - 15

the values of a and b is all order pair : ( a ; 9a-15) a real number

3 0

2 years ago

<img src="https://tex.z-dn.net/?f=-5%283.15-%5Cfrac%7B21%7D%7B20%7D%20%29%2B7" id="TexFormula1" title="-5(3.15-\frac{21}{20} )+7

Maksim231197 [3]

The answer is -7/2 or -3.5

8 0

3 years ago

3(x+1)+5(x+1)=64 What is x

Karo-lina-s [1.5K]

Answer:

x=7

Step-by-step explanation:

3(x+1)+5(x+1)=64

(3+5)(x+1)=64

8(x+1)=64

8x+8=64

8x=56

x=7

6 0

3 years ago

Read 2 more answers

A 32-ounce package of a snack mix is 24 percent pretzels. How many ounces are pretzels?

Igoryamba

Answer:

0.768 would be the amount of pretzels in ounces

Step-by-step explanation:

24% of 32 =0.768

how i got 24%?

24% of 100= 24

feel free to like and give brainliest

8 0

3 years ago

Read 2 more answers

Question : In the given figure , ∆ APB and ∆ AQC are equilateral triangles. Prove that PC = BQ.

lorasvet [3.4K]

Answer:

See Below.

Step-by-step explanation:

We are given that ΔAPB and ΔAQC are equilateral triangles.

And we want to prove that PC = BQ.

Since ΔAPB and ΔAQC are equilateral triangles, this means that:

$PA\cong AB\cong BP\text{ and } QA\cong AC\cong CQ$

Likewise:

$\angle P\cong \angle PAB\cong \angle ABP\cong Q\cong \angle QAC\cong\angle ACQ$

Since they all measure 60°.

Note that ∠PAC is the addition of the angles ∠PAB and ∠BAC. So:

$m\angle PAC=m\angle PAB+m\angle BAC$

Likewise:

$m\angle QAB=m\angle QAC+m\angle BAC$

Since ∠QAC ≅ ∠PAB:

$m\angle PAC=m\angle QAC+m\angle BAC$

And by substitution:

$m\angle PAC=m\angle QAB$

Thus:

$\angle PAC\cong \angle QAB$

Then by SAS Congruence:

$\Delta PAC\cong \Delta BAQ$

And by CPCTC:

$PC\cong BQ$

5 0

3 years ago

Read 2 more answers