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2 years ago

\dfrac{-2}{3} \div 2\dfrac{1}{4} = 3 −2 ÷2 4 1

Mathematics

1 answer:

bekas [8.4K]2 years ago

7 0

Answer:

$\dfrac{-2}{3} \div 2\dfrac{1}{4} = \dfrac{-8}{27}$

Step-by-step explanation:

Given

$\dfrac{-2}{3} \div 2\dfrac{1}{4} =$

Required

Solve

Start by converting the mixed number to improper fraction

$\dfrac{-2}{3} \div \dfrac{9}{4} =$

Convert the division to muliplication

$\dfrac{-2}{3} * \dfrac{4}{9} =$

$\dfrac{-2*4}{3*9}=$

$\dfrac{-8}{27}$

Hence:

$\dfrac{-2}{3} \div 2\dfrac{1}{4} = \dfrac{-8}{27}$

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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

2 years ago

Read 2 more answers

14. Find the coordinates of the circumcenter for ∆DEF with coordinates D(1,1) E (7,1) and F(1,5). Show your work.

Bess [88]

Hello :
the coordinates circumcenter for ∆DEF :
x = (1+7+1)/3
y = (1+1+5//3
x=3
y= 7/3

6 0

3 years ago

What is the value of the expression 30+(-6)? A -180 B -5 C 5 D 24

Evgesh-ka [11]

Answer:

Step-by-step explanation:

from the question above,

30+(-6) =

when plus is multiplying minus it's minus

+ × - = -

30-6 = 24

4 0

1 year ago

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1. The point (–1, 13) is on the terminal side of an angle in standard position. Use this

Masja [62]

Uh i think it’s uhh.... i think it’s a carrot

4 0

2 years ago

What is f if the equation is -3/4f=5/4?

Lilit [14]

Answer:

Exact Form:

f=-5/3

Decimal Form:

f= -1.6

Mixed Number form:

f= -1 2/3

Step-by-step explanation:

Solve for f by simplifying both sides of the equation , then isolating the variable .

Exact Form:

f=-5/3

Decimal Form:

f= -1.6

Mixed Number form:

f= -1 2/3

8 0

2 years ago

\dfrac{-2}{3} \div 2\dfrac{1}{4} = 3 −2 ​ ÷2 4 1 ​

\dfrac{-2}{3} \div 2\dfrac{1}{4} = 3 −2 ÷2 4 1