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nasty-shy [4]
2 years ago
5

What is 21n + 15 + n - 2 simplified ?

Mathematics
1 answer:
docker41 [41]2 years ago
4 0

Answer:

=22n+13

Let's simplify step-by-step:

21n+15+n−2

=21n+15+n+−2

<em>Combine Like Terms:</em>

=21n+15+n+−2

=(21n+n)+(15+−2)

=22n+13

Answer:

<u>=22n+13</u>

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AC=22,BC=x+14, and AB=x+10 find x
Elena-2011 [213]

Answer:

AB + BC = AC                 Segment Addition Postulate

(x + 10) + (x + 14) = 22     Substitution

2x + 24 = 22                   Simplify (added like terms)

2x = -2                            Subtraction Property of Equality

x = -1                               Divison Property of Equality

Answer: x = -1

Step-by-step explanation:

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7 0
2 years ago
The total number of pears and oranges in a basket is between 30 and 40 . There are 25% more apple than orange how many apples ar
Verdich [7]

The number of apples there are in the basket as described in the task content is; 20 apples.

<h3>What is the number of apples there are in the basket?</h3>

Inference drawn from the task content show that;

  • The ratio of apples to oranges in the basket is; 5:4. making the total partitions be 9.

On this note, we must identify a number divisible by 9 between 30 and 40.

The number is 36 and once divided by 9 is 4 units per partition.

On this note, the total number of apples in the basket is; 4 × 5 = 20 apples.

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7 0
2 years ago
Write the domain and range as an inequality.
BaLLatris [955]

Answer:

Step-by-step explanation:

domain is (-3,x) x- inter

range is (x,2) y-inter

4 0
2 years ago
Which equation is not equal to 6^3/6^6 ?: 1/6^2 6^-3 1/216 1/6^3
SCORPION-xisa [38]
The answer would be 60 because its multiplication all those numbers

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3 years ago
Determine whether the improper integral converges or diverges, and find the value of each that converges.
Ksju [112]

Answer:converge at I=\frac{1}{3}

Step-by-step explanation:

Given

Improper Integral I is given as

I=\int^{\infty}_{3}\frac{1}{x^2}dx

integration of \frac{1}{x^2}  is  -\frac{1}{x}

I=\left [ -\frac{1}{x}\right ]^{\infty}_3

substituting value

I=-\left [ \frac{1}{\infty }-\frac{1}{3}\right ]

I=-\left [ 0-\frac{1}{3}\right ]

I=\frac{1}{3}

so the value of integral converges at \frac{1}{3}

8 0
3 years ago
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