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3 years ago

14

Find the x- and y-intercepts of5x-3y = 12.x-intercept:y-intercept:

1 answer:

Lady bird [3.3K]3 years ago

8 0

Answer:

-5x -5x

-3y=-5x +12

/-5x /-5x

-5-3y=x+12

/-3 /-3

-5+y=x-4

+5 +5

y=x+1<---(y intercept(and it is its slop))

5x-3y = 12

/-3y /-3y

5x=-3y+12

/5 /5

x=-3y+2.4

/-3 /-3

-3+x=y+2.4

+3 +3

x=y+5.4<---(x intercept(not the slop because the y is not the first number))

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I need to know the answer please…

Triss [41]

Answer is a= 51/7 or 7 2/7

Step by step

-2/3 (a + 3) = 5/3a - 19

Distribute to resolve parentheses

-2/3a -2 = 5/3a - 19

Now to move variables to one side
Add 2/3a to both sides

-2/3a + 2/3a -2 = 5/3a + 2/3a -19

Combine like terms

-2 = 7/3a -19

Add 19 to both sides to isolate variable

-2 +19 = 7/3a -19 + 19

17 = 7/3a

Divide both sides by 7/3 to solve for a
(Remember when you divide a fraction, flip it and multiply.)

17 * 3/7 = a

51/7 = a

Picture attached shows steps in order, I think that is what you also needed?

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1 year ago

Element X decays radioactively with a half-life of eight minutes. If there are 800 g of element X, how long, to the nearest 10th

Harman [31]

Answer: 41.5 min

Step-by-step explanation:

This problem can be solved with the Radioactive Half Life Formula:

$A=A_{o}.2^{\frac{-t}{h}}$ (1)

Where:

$A=22 g$ is the final amount of the radioactive element

$A_{o}=800 g$ is the initial amount of the radioactive element

$t$ is the time elapsed

$h=8 min$ is the half life of the radioactive element

So, we need to substitute the given values and find $t$ from (1):

$22 g=(800 g) 2^{\frac{-t}{8 min}}$ (2)

$\frac{22 g}{800 g}=2^{\frac{-t}{8 min}}$ (3)

$\frac{11}{400}=2^{\frac{-t}{8 min}}$ (4)

Applying natural logarithm in both sides:

$ln(\frac{11}{400})=ln(2^{\frac{-t}{8 min}})$ (5)

$-3.593=-\frac{t}{8 min}ln(2)$ (6)

Clearing $t$ :

$t=41.46 min \approx 41.5 min$ This is the time elapsed

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Natali [406]

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Answer:

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Step-by-step explanation:

Please find the complete question in the attached file.

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That's why above given choice is correct.

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