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3 years ago

How do you simplify 21+(n-5)

Mathematics

2 answers:

dem82 [27]3 years ago

4 0

Answer:

n+16

Step-by-step explanation:

21+(n-5)=21+n-5=n+21-5=n+16

tatyana61 [14]3 years ago

3 0

Answer: 16+n

Let’s first rewrite the equation

21+(n-5)

The first thing we need to do is distribute. The + acts as a +1. Let’s multiply n • 1 and -5•1.

21+(n-5)

21+n-5

Now we need to combine like terms. Like terms are numbers that share the same variable, or lack of. In this equation, these numbers are 21 and -5.

21+n-5

16+n

This is your answer! Hope this helps comment below for more questions :)

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3 years ago

Solve 9 is subtracted from 2 times the sum of 7 and 4

Goshia [24]

2×11+9

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3 years ago

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What is -(1) Please help me.

KatRina [158]

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3 years ago

Marshall drove to work in rush hour traffic, going just 27 miles in 1 hour and 20 minutes. Later that day, he left work early to

iragen [17]

Answer:

The average speed of Marshall is 29.7 mph

Step-by-step explanation:

Given;

distance traveled to work, d₁ = 27 miles

distance traveled back home, d₂ = 27 miles

initial time taken, t₁ = 1 hour and 20 minutes

final time taken, t₂ = 29 minutes

Total distance traveled, d = d₁ + d₂ = 27 miles + 27 miles = 54 miles

Total time taken for the journey, t = t₁ + t₂ = 1 hour + ( 20 mins + 29 mins)

t = 1 hour + 49 mins = 1 hour + 0.817 hour = 1.817 hour

The average speed is given by;

v = total distance / total time

v = 54 miles / 1.817 hour

v = 29.7 mph

Therefore, the average speed of Marshall is 29.7 mph

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3 years ago

Use any method to multiply (3a + 2b - c)(a - b + 2c).

Alla [95]

Answer: $3a^{2}-ab + 5ac- 2b^{2} + 5bc-2c^{2}$

Step-by-step explanation:

(3a + 2b − c) (a − b + 2c)

Expand (3a + 2b − c) (a − b + 2c) by multiplying each term in the first expression by each term in the second expression.

3a ⋅ a + 3a (−b) + 3a (2c) + 2ba + 2b (−b) + 2b (2c) − ca − c (−b) − c (2c)

3 (a ⋅ a) + 3a (−b) + 3a (2c) + 2ba + 2b (−b) + 2b (2c) − ca − c (−b) − c (2c)

3a^2 + 3a (−b) + 3a (2c) + 2ba + 2b (−b) + 2b (2c) − ca − c (−b) − c (2c)