3 years ago

What is the value of x in the equation 4x+8y=40 when y=0.8

Mathematics

2 answers:

jolli1 [7]3 years ago

4 0

you plug in the y .8 to the 8 that is 6.4

<span>them you subtract 40 by 6.4 that is 33.6</span>

<span>then you divide that by 4 is 8.4</span>

GREYUIT [131]3 years ago

3 0

Substitute in 0.8 for y.

$4x+8y=40 \\ \\ 4x + 8(0.8) = 40 \\ \\ 4x + 6.5 = 40 \\ \\ 4x = 40 - 6.4 \\ \\ 4x = 33.6 \\ \\ x = 8.4 \\ \\$

The final result is: x = 8.4.

You might be interested in

Which of the following equations is equivalent to x - y = 6?

Agata [3.3K]

I believe the answer would be A. 2x-y=6 :)

8 0

3 years ago

Read 2 more answers

What is the product?

Sophie [7]

Answer:

Step-by-step explanation:

In exponent multiplication, if base are same just add the powers.

$(-6a^{3}b+2ab^{2})(5a^{2}-2ab^{2}-b)=(-6a^{3}b)*(5a^{2}-2ab^{2}-b)+2ab^{2}*(5a^{2}-2ab^{2}-b)\\\\$

$=(-6a^{3}b)*5a^{2}-(-6a^{3}b)*2ab^{2}-(-6a^{3}b)*b+ 2ab^{2}*5a^{2}-2ab^{2}*2ab^{2}-2ab^{2}*b\\\\=-30a^{3+2}b+12a^{3+1}b^{1+2}+6a^{3}b^{1+1}+10a^{1+2}b^{2}-4a^{1+1}b^{2+2}-2ab^{2+1}\\\\=-30a^{5}b+12a^{4}b^{3}+6a^{3}b^{2}+10a^{3}b^{2}-4a^{2}b^{4}-2ab^{3}\\$