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3 years ago

15

I have 3 equations 9x+6, 2(5x-3) and 5y+14 how do i solve for x?

1 answer:

snow_tiger [21]3 years ago

8 0

Set the equation equal to zero.

9x+6=0
Subtract 6 from both sides
9x = -6
Divide both sides by 9
x=-6/9
or -2/3

2(5x-3)=0
Distribute
10x-6=0
add 6 to both sides
10x=6
Divide by 10 on both sides
x=6/10 or 4/5

5y+14=0
-14 from both sides
5y=-14
divide both sides by 5
y=-14/5
or -2 and 4/5

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Answer:

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Step-by-step explanation:

Given the endpoints of the diameter of a circle: (-2,-3) and (4,-1)

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$(x-h)^{2}$ + $(y-k)^{2}$ = $r^{2}$

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3 years ago

Read 2 more answers

Why did I get this question wrong?

lord [1]

Step-by-step explanation:

∫₀³⁰ (r/V C₀ e^(-rt/V)) dt

If u = -rt/V, then du = -r/V dt.

∫ -C₀ e^u du

-C₀ ∫ e^u du

-C₀ e^u + C

-C₀ e^(-rt/V) + C

Evaluate between t=0 and t=30.

-C₀ e^(-30r/V) − -C₀ e^(-0r/V)

-C₀ e^(-30r/V) + C₀

C₀ (-e^(-30r/V) + 1)

I got the same answer. Try changing the lowercase v to an uppercase V.

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3 years ago

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