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goldenfox [79]
3 years ago
13

Solve for x.

Mathematics
1 answer:
alexandr402 [8]3 years ago
7 0

this is how i calculated for x!

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The least common multiple of two numbers is 40. The difference of the two numbers is 2.
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4 0
4 years ago
Read 2 more answers
Each big square below represents one whole.
tatyana61 [14]

Answer:

45 percent /100 percent

4 0
3 years ago
Please help me with this
Evgesh-ka [11]

B. The graph that best represents the equation y =  |x| - 1 is the option B.

To solve this problem we have to try with some values, the symbol |x| is the absolute value which means any number either positive or negative always is positive |-5| = 5 and |5| = 5.

Let's take x = -3, -2, -1, 0, 1, 2, 3.

For x = -3

y = |-3| - 1 = 3 - 1

y = 2

For x = -2

y =  |−2| - 1 = 2 - 1

y = 1

For x = -1

y =  |−1| - 1 = 1 - 1

y = 0

For x = 0

y =  |0| - 1 = 0 - 1

y = -1

For x = 1

y =  |1| - 1 = 1 - 1

y = 0

For x = 2

y =  |2| - 1 = 2 - 1

y = 1

For x = 3

y =  |3| - 1 = 3 - 1

y = 2

y  ║  x

2      -3

1       -2

0       -1

-1       0

0       1

1        2

2       3

If we graph the points obtain in the table above, the result is a graph with the characteristics of the option B.

6 0
3 years ago
Simplify the sum. (8u^3+8u^2+6)+(4u^3-6u+3)​
rewona [7]

Answer:

\boxed{12u^3+8u^2-6u+9}

Step-by-step explanation:

(8u^3+8u^2+6)+(4u^3-6u+3) = 8u^3+8u^2+6+4u^3-6u+3

Once

8x+4x = 12x, \text{ as } x = u^3

8u^3+8u^2+6+4u^3-6u+3 = 12u^3+8u^2+6-6u+3 = \boxed{12u^3+8u^2-6u+9}

You don't need to factor.

7 0
3 years ago
Answer must be in simplest form and an exact value.
DIA [1.3K]

Answer: 0.3846153846

Step-by-step explanation:

3 0
3 years ago
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