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2 years ago

help! I only have a certain amount of time.... easy points! The sum of 3 times a number and 4 is 13. Find the number.

Mathematics

2 answers:

Ber [7]2 years ago

6 0

Answer:

The number is 3

Step-by-step explanation:

Ok, first thing's first: we need to make the words into an actual equation.

since it just says a number I'll use x

3 times x + 4 = 13

3x +4 =13

Now that we have our equation, we can solve for x

3x + 4 = 13

-4 -4

3x = 9

I subtracted 4 on both sides to get rid of the 4 and move it to the other side of the equation, so we can try to get x by itself.

3x =9

/3 /3

x = 3

I divided both sides by 3 to get rid of the 3 that's being multiplied by x

Now we have our number : x = 3

I hope this helps :)

Anni [7]2 years ago

3 0

3x+4=13
3x=13-4
3x=9
x=9/3
x=3

The number is 3 :)

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3 years ago

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3 years ago

Read 2 more answers

Hi, Could anyone help me with my homework

tatiyna

Answer:

$\hookrightarrow \sf x^6+24x^5+240x^4+1280x^3+3840x^2+6144x+4096$

solving steps:

$\rightarrow \sf (x + 4)^6$

$\bold{rewrite \ the \ following}$

$\rightarrow \sf (x + 4)^2 (x + 4)^2 (x + 4)^2$

$\bold {formula \ used : \sf (x+a)^2 = (x^2 + 2xa + a^2)}$

$\rightarrow \sf (x^2 + 8x+16) (x^2 + 8x+16) (x^2 + 8x+16)$

$\bold{simplify \ by \ removing \ parenthesis}$

$\rightarrow \sf (x^4 +8x^3 + 16x^2 + 8x^3 +64x^2 + 128x+16x^2+128x+256 ) (x^2 + 8x+16)$

$\bold{basic \ addition \ of \ integers }$

$\rightarrow \sf (x^4+16x^3+96x^2+256x+256) (x^2 + 8x+16)$

$\bold{remove \ parenthesis}$

$\rightarrow \sf (x^6 + 16x^5 + 96x^4 + 256x^3 + 256x^2 + 8x^6 + 128x^4 + 768x^3 + 2048x^2 + 2048 + 16x^4 + 256x^3 + 1536x^2 + 4096x + 4096)$

$\bold {final \ answer:}$

$\rightarrow \sf x^6+24x^5+240x^4+1280x^3+3840x^2+6144x+4096$

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2 years ago