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2 years ago

Simplify (3 × 22) ÷ 6 + [28 – (4^2] =

2 answers:

8 0

Use PEMDAS:
Parentheses
Exponents
Multiplication
Division
Addition
Subtraction
(3x22) /6+(28-(4^2))
66/6+(28-16)
66/6+12
11+12
23

Kitty [74]2 years ago

6 0

The Final result is 23

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Pienso en un número lo multiplicó por 2 y el resultado es lo mismo cuando le sumó 1

MrRa [10]

Translate that to english please?

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2 years ago

Michael has a set of 24 math problems to do for homework. Michael completed 1/4 of his homework in 3/4 of an hour. If he continu

gtnhenbr [62]

Given:

Total number of problems = 24

Michael completed $\dfrac{1}{4}$ of his homework in $\dfrac{3}{4}$ of an hour.

To find:

What fraction of his homework would he complete in one hour.

Solution:

We have,

Part of work done in $\dfrac{3}{4}$ of an hour $=\dfrac{1}{4}$

Part of work done in an hour $=\dfrac{4}{3}\times \dfrac{1}{4}$

$=\dfrac{1}{3}$

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2 years ago

Kevin is $25 in his checking account with Kevin has 2.20 less than 4 times of the amount that Molly has in her account what is t

Nikolay [14]

Answer:

$6.8

Step-by-step explanation:

suppose Molly's account balance = X

25 + 2.2 = X * 4

27.2 = X*4

27.2/4 = X = 6.8

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2 years ago

Read 2 more answers

Find the answers to this activity

diamong [38]

Answer:

(224)10 = (E0)16

I dont know why it wouldnt be right this is the answer...Im acually confused now

Step-by-step explanation:

(224)10 = (E0)16

Step by step solution

Step 1: Divide (224)10 successively by 16 until the quotient is 0:

224/16 = 14, remainder is 0

14/16 = 0, remainder is 14

Step 2: Read from the bottom (MSB) to top (LSB) as E0. This is the hexadecimal equivalent of decimal number 224

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2 years ago

There are two unknowns you are trying to find, and you cannot express one in terms of the other. Describe how you could use alge

serious [3.7K]

Answer:

When you solve systems with two variables and therefore two equations, the ... of any variable is 1, which means you can easily solve for it in terms of the other ... In the substitution method, you use one equation to solve for one variable and ... Look for a variable with a coefficient of 1 … that's how you'll know where to begin.

Step-by-step explanation:

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2 years ago