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2 years ago

Solve for x. 4x − 33=7

Mathematics

2 answers:

tatuchka [14]2 years ago

5 0

Note the equal sign. What you do to one side, you do to the other.

Do the opposite of PEMDAS. First, add 33 to both sides

4x - 33 (+33) = 7 (+33)

4x = 40

Next, divide by 4

4x/4 = 40/4

x = 40/4

x = 10

10 is your answer

hope this helps

muminat2 years ago

5 0

Answer:

x = 6

Step-by-step explanation:

I believe the question is really this:

Solve for x.

$\frac{4x - 3}{3} = 7$

Multiply both sides by 3

$\frac{3(4x-3)}{3} = 7 * 3$

Simplify

4x - 3 = 21

Add 3 to both sides

4x - 3 + 3 = 21 + 3

Simplify

4x = 24

Divide both sides by 4

$\frac{4x}{4} = \frac{24}{4}$

Simplify

x = 6

top problems:

1-) x²-2x-8

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2-) 4s²-4s+1

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3-) 6x²-3-7x

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4-) 4u²+8u

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5-) t²-15

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6-) 3x²-x-4

bottom problems:

1-) $\frac{1}{4},-1$

y=x²+ $\frac{3x}{4}-\frac{1}{4}$

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2-) $\sqrt{2},\frac{1}{3}$

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3-) 0, √5

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4-) 1,1

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5-) $-\frac{1}{4}, \frac{1}{4}$

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6-) 4,1

7 0

2 years ago

Can someone explain step by step to me

skad [1K]

- 2 /15 - ( - 9/10 )
-2/15 + 9/10 multiply fraction #1 by 2 an fraction #2 by 3 to get
-4 /30 +27/ 30 = 23/30

7 0

2 years ago

Over What Interval is The Function In This Graph Increasing?

irina1246 [14]

The correct answer is option D.which is -4 ≤ x ≤ 1.

<h3>What is a graph?</h3>

A graph is the representation of the data on the vertical and horizontal coordinates so we can see the trend of the data.

in the graph, we can see that the value of x is increasing from -4 coordinate of x to the 1 coordinate of x so the interval will be given as -4 ≤ x ≤ 1.

Therefore the correct answer is option D.which is -4 ≤ x ≤ 1.

To know more about graphs follow

brainly.com/question/25020119

#SPJ1

8 0

1 year ago

Inverse laplace of [(1/s^2)-(48/s^5)]

Katen [24]

**Refresh page if you see [ tex ]**

I am not familiar with Laplace transforms, so my explanation probably won't help, but given that for two Laplace transform $F(s)$ and $G(s)$ , then $\mathcal{L}^{-1}\{aF(s)+bG(s)\} = a\mathcal{L}^{-1}\{F(s)\}+b\mathcal{L}^{-1}\{G(s)\}$

Given that $\dfrac{1}{s^2} = \dfrac{1!}{s^2}$ and $-\dfrac{48}{s^5} = -2\cdot\dfrac{4!}{s^5}$

So you have $\mathcal{L}^{-1}\left\{\dfrac{1}{s^2} - 2\cdot\dfrac{4!}{s^5}\right\} = \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\}$

From Table of Laplace Transform, you have $\mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}}$ and hence $\mathcal{L}^{-1}\left\{\dfrac{n!}{s^{n+1}}\right\} = t^n$

So you have $\mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\} = \boxed{t-2t^4}$ .

Hope this helps...

7 0

3 years ago

PLS HELP ILL GIVE BRAINLIEST

cluponka [151]

So you start of by finding the median. The median in this case would be 43. Then you find the median of both the lower (36) and the upper (53) half of the data. Then the find the difference between those two number. So you would do 53-36 which would make the answer 17

8 0

3 years ago