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3 years ago

Define a variable and write an expression for the phrase.

Mathematics

1 answer:

kondaur [170]3 years ago

5 0

n / 82 is your answer

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1. The area of a circle is 60 square units. What is the area of ...

AVprozaik [17]

Multiply the area of the circle by the percentage:

60 square units x 50% = 60 x 0.50 = 30

50% 0f the circle id 30 square units.

3 0

2 years ago

(8x^3 - x) - (10x - 5x^2 - 3x^3) Show your work.

sergeinik [125]

Let's simplify step-by-step.
8
x
3
−
x
−
(
10
x
−
5
x
2
−
3
x
3
)
Distribute the Negative Sign:
=
8
x
3
−
x
+
−
1
(
10
x
−
5
x
2
−
3
x
3
)
=
8
x
3
+
−
x
+
−
1
(
10
x
)
+
−
1
(
−
5
x
2
)
+
−
1
(
−
3
x
3
)
=
8
x
3
+
−
x
+
−
10
x
+
5
x
2
+
3
x
3
Combine Like Terms:
=
8
x
3
+
−
x
+
−
10
x
+
5
x
2
+
3
x
3
=
(
8
x
3
+
3
x
3
)
+
(
5
x
2
)
+
(
−
x
+
−
10
x
)
=
11
x
3
+
5
x
2
+
−
11
x
Answer:
=
11
x
3
+
5
x
2
−
11
x

Brainliest Please:)

6 0

3 years ago

Read 2 more answers

Help #23 pleaseeeeeeeee

GuDViN [60]

A bus holds 40 people and a van holds 10

7 0

2 years ago

12x-4x+3+11 help me solve this problem

daser333 [38]

Answer:

8x + 14

Step-by-step explanation:

12x - 4x + 3 + 11 (Subtract 12x - 4x)

8x + 3 + 11 (Add 3 + 11)

8x + 14

Tip: the like-terms 12x and 4x can be subtracted like regular numbers because they have the same variable and exponent.

6 0

3 years ago

Let n be a natural number. Show that 3 | n3 −n

Vladimir [108]

Let $n=1$ . Then $n^3-n=1^3-1=0$ . By convention, every non-zero integer $n$ divides 0, so $3\vert n^3-n$ .

Suppose this relation holds for $n=k$ , i.e. $3\vert k^3-k$ . We then hope to show it must also hold for $n=k+1$ .

You have

$(k+1)^3-(k+1)=(k^3+3k^2+3k+1)-(k+1)=(k^3-k)+3(k^2+k)$

We assumed that $3\vert k^3-k$ , and it's clear that $3\vert 3(k^2+k)$ because $3(k^2+k)$ is a multiple of 3. This means the remainder upon divides $(k+1)^3-(k+1)$ must be 0, and therefore the relation holds for $n=k+1$ . This proves the statement.

4 0

2 years ago