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3 years ago

Which is bigger, 80% of 150 or 3/4 of 200

1 answer:

7 0

80% of 150
step 1: 80% is a percentage, so you convert that into a decimal(just move it over two places)

step 2: Multiply .80(150)= 120

3/4 of 200
step 1: 3/4 is also 75%
step 2: 75% is a percent, so you change that into .75
step 3: Multiply .75(200)= 150

So 3/4 of 200 is bigger

You might be interested in

If the temperature degrees above zero are denoted by positive numbers, what does - 10 degrees denoted?

kogti [31]

Answer:

-10 degrees denotes 10 degrees below zero.

Step-by-step explanation:

3 0

3 years ago

A line goes through the point (2, 4) and has a slope of 3/2. Which of the following points is also on the line? CHOOSE ALL THAT

Korolek [52]

Hmmmm not sure

Explanation: I took this test and got it correct

8 0

3 years ago

Use complete sentences to describe why √-1 ≠ -√1

tekilochka [14]

Well let's say that to compare these two numbers, we have to start with the definition first.

Definition

$\displaystyle \large{ {y}^{2} = x} \\ \displaystyle \large{ y = \pm \sqrt{x} }$

Looks like we can use any x-values right? Nope.

The value of x only applies to any positive real numbers for one reason.

As we know, any numbers time itself will result in positive. No matter the negative or positive.

Definition II

$\displaystyle \large{ {a}^{2} = a \times a = |b| }$

Where b is the result from a×a. Let's see an example.

Examples

$\displaystyle \large{ {2}^{2} = 2 \times 2 = 4} \\ \displaystyle \large{ {( - 2)}^{2} = ( - 2) \times ( - 2) = | - 4| = 4}$

So basically, their counterpart or opposite still gives same value.

Then you may have a question, where does √-1 come from?

It comes from this equation:

$\displaystyle \large{ {y}^{2} = - 1}$

When we solve the quadratic equation in this like form, we square both sides to get rid of the square.

$\displaystyle \large{ \sqrt{ {y}^{2} } = \sqrt{ - 1} }$

Then where does plus-minus come from? It comes from one of Absolute Value propety.

Absolute Value Property I

$\displaystyle \large{ \sqrt{ {x}^{2} } = |x| }$

Solving absolute value always gives the plus-minus. Therefore...

$\displaystyle \large{ y = \pm \sqrt{ - 1} }$

Then we have the square root of -1 in negative and positive. But something is not right.

As I said, any numbers time itself of numbers squared will only result in positive. So how does the equation of y^2 = -1 make sense? Simple, it doesn't.

Because why would any numbers squared result in negative? Therefore, √-1 does not exist in a real number system.

Then we have another number which is -√1. This one is simple.

It is one of the solution from the equation y^2 = 1.

$\displaystyle \large{ {y}^{2} = 1} \\ \displaystyle \large{ \sqrt{ {y}^{2} } = \sqrt{1} } \\ \displaystyle \large{ y = \pm \sqrt{1} }$

We ignore the +√1 but focus on -√1 instead. Of course, we know that numbers squared itself will result in positive. Since 1 is positive then we can say that these solutions exist in real number.

Conclusion

So what is the different? The different between two numbers is that √-1 does not exist in a real number system since any squared numbers only result in positive while -√1 is one of the solution from y^2 = 1 and exists in a real number system.

5 0

3 years ago

Read 2 more answers

How do you figure out the reciprocal of a numer? E.g 5, 2/7 and 1&2/5 ??

eimsori [14]

You need to multiply 1 and 5 plus 2.
and for 5 2/7 it would be 5 times 7 plus 2.

So...
5 2/7= 37/7

1 2/5= 7/5

6 0

4 years ago

What is 13 - 4x = 1 - x

JulsSmile [24]

13-4x=1-x
13=4x+x=1-x+x
13-3x=1
13-1-3x=1-1
12-3x=0
12=3x
12/3=3x/3
4=x
Therefore, x=4

5 0

3 years ago