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kati45 [8]
3 years ago
10

What is the product? −4 × ( −2 2/5)

Mathematics
1 answer:
elixir [45]3 years ago
5 0

Answer:

9.6

Step-by-step explanation:

Hope this help : ) . Bye.

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Write the following inequality. You are not solving it, just writing it.
igomit [66]

Answer:

1. 350 + 25x = 35x + 200

2. 35x - 350 = 25x - 200

3. 350 - 35x = 200 + 25x

4. 200x - 350 = 25 + 35x

Step-by-step explanation:

4 0
3 years ago
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What percent of 38 is 15
Nostrana [21]
Let's assume x=15. 38x=100×15=1,500. \frac{38x}{38}= \frac{1,500}{38}. x=39.472 which is rounded to 39%
8 0
3 years ago
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Please help! ASAP, no bots.
harkovskaia [24]

Answer:

perfect square (4x - 3)^2

Step-by-step explanation:

3 0
3 years ago
find the product with the exponent in the simplest form. Then, identify the values of x and y. 6^1/3 * 6^1/4 = 6^x/y
mr Goodwill [35]
<span>6^1/3 * 6^1/4 = 6^x/y

The key to solving this problem is understanding properties of exponents:
If you are multiplying two powers with the same base (in this case the base is 6), the result is the base raised to the power of the two exponents added together.

6^1/3+1/4 = 6^x/y
6^7/12 = 6^x/y
 
So the product would be 6^7/12
x = 7
y = 12</span>
6 0
3 years ago
How do you find the x intercepts of 3x^(5/3) - 4x^(7/3)
kaheart [24]
Set the whole expression = to 0 and solve for x.

3x^(5/3) - 4x^(7/3) = 0.  Factor out x^(5/3):     x^(5/3) [3 - 4x^(2/3)] = 0

Then either x^(5/3) = 0, or 3 - 4x^(2/3) = 0.

In the latter case, 4x^(2/3) = 3. 

To solve this:  mult. both sides by x^(-2/3).  Then we have

4x^(2/3)x^(-2/3) = 3x^(-2/3),            or 4 = 3x^(-2/3).  It'd be easier to work with this if we rewrote it as

4           3
---  = --------------------
1            x^(+2/3)

Then 

4
---  = x^(-2/3).  Then, x^(2/3) = (3/4), and x = (3/4)^(3/2).  According to my     3                      calculator, that comes out to x = 0.65 (approx.)

Check this result!  subst. 0.65 for x in the given equation.  Is the equation then true?

My method here was a bit roundabout, and longer than it should have been.  Can you think of a more elegant (and shorter) solution?
7 0
3 years ago
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