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IRINA_888 [86]
3 years ago
14

Help pleaseeeeeeeeeeeeeee

Mathematics
2 answers:
stiv31 [10]3 years ago
7 0

Answer: The answer would be 3.6

You first want to add 4 to both sides to get y on one side by itself. So it would look something like this: -2/5+4=y. Then you can add which gives you y=3.6

My name is Ann [436]3 years ago
6 0

y=3\frac{3}{5}

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2 years ago
What is 3/4(2x+5)=5(x+12)
lina2011 [118]

Answer:

--16 1/14, -225/14 or -16.07

Step-by-step explanation:

3/4(2x+5)=5(x+12)

3x/2+15/4=5x+60

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4 0
3 years ago
Solve the inqualities 3x-3<9
pentagon [3]

Answer: x < 4

Step-by-step explanation: When solving the inequality 3x - 3 < 9, just like an equation our first step is to isolate the <em>x</em> term which is 3x by adding 3 to both sides of the inequality.

On the left, -3 + 3 cancels and we have 3x.

Make sure to bring down the < sign.

On the right, 9 + 3 simplifies to 12.

So we have 3x < 12.

Now, divide both sides of the inequality by 3 to isolate <em>x</em>.

When we do that we get <em>x < 4</em>

So our answer is x < 4.

8 0
3 years ago
In 3 months John collected a debt of $70.00. What was the average amount of debt John collected per month?
Shalnov [3]

Answer:

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Step-by-step

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7 0
3 years ago
The graph of f(x) =(square root of "x") is reflected across the x-axis and then across the y-axis to create the graph of functio
vesna_86 [32]
Analysis:

1) The graph of function f(x) = √x is on the first quadrant, because the domain is x ≥ 0 and the range is y ≥ 0

2) The first transformation, i.e. the reflection of f(x) over the x axis, leaves the function on the fourth quadrant, because the new image is y = - √x.

3) The second transformation, i.e. the reflection of y = - √x over the y-axis, leaves the function on the third quadrant, because the final image is - √(-x). This is, g(x) = - √(-x).

From that you have, for g(x):

* Domain: negative x-axis ( -x ≥ 0 => x ≤ 0)

* Range: negative y-axis ( - √(-x) ≤ 0 or y ≤ 0).

Answers:

Now let's examine the statements:

<span>A)The functions have the same range:FALSE the range changed from y ≥ 0 to y ≤ 0

B)The functions have the same domains. FALSE the doman changed from x ≥ 0 to x ≤ 0

C)The only value that is in the domains of both functions is 0. TRUE: the intersection of x ≥ 0 with x ≤ 0 is 0.

D)There are no values that are in the ranges of both functions. FALSE: 0 is in the ranges of both functions.

E)The domain of g(x) is all values greater than or equal to 0. FALSE: it was proved that the domain of g(x) is all values less than or equal to 0.

F)The range of g(x) is all values less than or equal to 0. TRUE: it was proved above.</span>
4 0
4 years ago
Read 2 more answers
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