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3 years ago

How do i solve 3 (2y-4)=8y+9-9y

2 answers:

7 0

The first step for solving this equation is to distribute 3 through the parenthesis.
6y - 12 = 8y + 9 - 9y
Collect the like terms with y.
6y - 12 = -y + 9
Move the variable to the left side and then change its sign.
6y + y - 12 = 9
Now move the constant to the right side and change its sign.
6y + y = 9 + 12
Add the terms on the left side of the equation.
7y = 9 + 12
Add the numbers on the right side.
7y = 21
Lastly,, divide both sides of the equation by 7 to find your final answer.
y = 3
This means that y = 3 is the correct answer to your question.
Let me know if you have any further questions.
:)

bulgar [2K]3 years ago

4 0

<span>3(2y - 4) = 8y + 9 - 9y

Distribute 3:
6y - 12 = 8y + 9 - 9y

Combine like terms:
6y - 12 = -y + 9

Add y to both sides:
7y - 12 = 9

Add 12 to both sides:
7y = 21

Divide both sides by 3:
y = 3</span>

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3 years ago

A truck is 250 inches long. A car is 35% shorter than the truck. How long is the car

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3 years ago

Read 2 more answers

1) Use the Euclidean Algorithm to nd the greatest common divisor of 71407 and 2020. (Write down all of the steps, so that you ar

Olenka [21]

Answer:

Step-by-step explanation:

We have

$71407 = 2000(35)+1407\\2000 = 1407+1(593)\\1407= 2(593) + 221\\593 = 2(221) = 151\\221=151+70\\151=2(70)+11\\70 = 6(11)+4\\11=2(4)+3\\4 = 3+1\\3 = 3(1)+0\\$

So we find GCD = 1

2) By back substitution we get

71407 (543) - 19387(2000) =1

So x = 543 and y = -19387

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3 years ago

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3 years ago

How would you use the Fundamental Theorem of Calculus to determine the value(s) of b if the area under the graph g(x)=4x between

-BARSIC- [3]

The Fundamental Theorem of Calculus regarding geometry states that
$\int\limits^b_a g{(x)} \, dx = F(b)-F(a)$

Where F is the indefinite integral of $g(x)$

The first step is to integrate $g(x)$
$\int\ {4x} \, dx = \frac{4x^{1+1} }{1+1} = \frac{4x^{2} }{2} =2 x^{2}$

Then substitute the value of $b$ and $a=1$ into $2x^{2}$

$[2 (b)^{2}]-[2 (1)^{2}] = 240$
$2b^{2} -2=240$
$2b^{2}=240+2$
$2b^{2}=242$
$b^{2}= \frac{242}{2}$
$b^{2}=121$
$b=11$

Hence the limit of the area under $g(x)$ is between $a=1$ and $b=11$

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3 years ago