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Sophie [7]
3 years ago
7

What expression is equivalent to 4/2 times 4/3

Mathematics
1 answer:
In-s [12.5K]3 years ago
6 0

Answer:

8/3

Step-by-step explanation:

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Graph the image of square TUVW after the following sequence of transformations:
GuDViN [60]
Move the shape down by one quadrant and flip it so where “U” was at, “W” is in its place and Vice versa
4 0
3 years ago
Which is a solution to the system of inequalities? y≤-1/2x+3 y≥x+1
Serggg [28]

Answer:

y≤-1/2(x=4)+3

Step-by-step explanation:

4 0
3 years ago
PLEASE HELP ASAP!!! I NEED CORRECT ANSWERS ONLY PLEASE!!!
stealth61 [152]

Answer:

BD = 8.8

Step-by-step explanation:

You have a right triangle.

The hypotenuse is BD

The side opposite the 27o angle is 4

Sin(27) = opposite / hypotenuse.

Sin(27) = 0.45399

opposite = 4

hypotenuse = ?

0.45399 = 4/hypotenuse

BD =  4/0.45399

BD = 8.811

The answer is 8.8

4 0
3 years ago
Read 2 more answers
What is the difference? StartFraction 2 x + 5 Over x squared minus 3 x EndFraction minus StartFraction 3 x + 5 Over x cubed minu
inn [45]

Answer:

\frac{(x+5)(x+2)}{x(x-3)(x+3)}

Step-by-step explanation:

The given expression is

\frac{2x+5}{x^{2} -3x} -\frac{3x+5}{x^{3} -9x} -\frac{x+1}{x^{2}-9}

First, we need to factor each denominator

\frac{2x+5}{x(x-3)} -\frac{3x+5}{x(x+3)(x-3)} -\frac{x+1}{(x-3)(x+3)}

So, the least common factor (LCF) is x(x-3)(x+3), because they are the factors that repeats.

Now, we diviide the LCF by each denominator, to then multiply it by each numerator.

\frac{(x+3)(2x+5)-3x-5-x(x+1)}{x(x-3)(x+3)} =\frac{2x^{2}+5x+6x+15-3x-5-x^{2}-x }{x(x-3)(x+3)}\\\frac{x^{2}+7x+10}{x(x-3)(x+3)}

Then, we factor the numerator, to do so, we need to find two numbers which product is 10 and which sum is 7.

\frac{(x+5)(x+2)}{x(x-3)(x+3)}

Therefore, the expression is equivalent to

\frac{(x+5)(x+2)}{x(x-3)(x+3)}

8 0
3 years ago
Read 2 more answers
Select all the correct answers. Which two inequalities can be used to find the solution to this absolute value inequality? 3|x +
ziro4ka [17]

The absolute value inequality can be decomposed into two simpler ones.

x < 0

x > -8

<h3></h3><h3>Which two inequalities can be used?</h3>

Here we start with the inequality:

3|x + 4| - 5 < 7

First we need to isolate the absolute value part:

3|x + 4| < 7 + 5

|x + 4| < (7 + 5)/3

|x + 4| < 12/3

|x + 4| < 4

The absolute value inequality can now be decomposed into two simpler ones:

x + 4 < 4

x + 4 > - 4

Solving both of these we get:

x < 4 - 4

x > -4 - 4

x < 0

x > -8

These are the two inequalities.

Learn more about inequalities:

brainly.com/question/24372553

#SPJ1

6 0
1 year ago
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