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3 years ago

Are all real numbers and imaginary numbers also complex numbers? Explain, using the parts of a complex number.

1 answer:

8 0

Numbrs are divided in real and imaginary, so they can't be together one thing
Complex number are imaginary numbers

in the other hand if you multiply am complex by i than you get a real number, but that is just a way of expressing, so not all numbers are complex.

(5i)²=-25----- a real number is a product of 2 imaginary but it doesn't mean that we can say all are complex

(the explaination is subjectiv, based just in my opinion and knowlige hope it helps)

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Solve the inequality. -1/3w ≤ -5

omeli [17]

Answer:

$\frac{ - 1}{3} w \leqslant - 5$

$w \leqslant - 5 \div \frac{ - 1}{3}$

$w \leqslant - 5 \times - 3$

$w \leqslant 15$

3 0

2 years ago

exis [7]

Answer:

Part 1) The length of the longest side of ∆ABC is 4 units

Part 2) The ratio of the area of ∆ABC to the area of ∆DEF is $\frac{1}{100}$

Step-by-step explanation:

Part 1) Find the length of the longest side of ∆ABC

we know that

If two figures are similar, then the ratio of its corresponding sides is proportional and this ratio is called the scale factor

The ratio of its perimeters is equal to the scale factor

Let

z ----> the scale factor

x ----> the length of the longest side of ∆ABC

y ----> the length of the longest side of ∆DEF

$z=\frac{x}{y}$

we have

$z=\frac{1}{10}$

$y=40\ units$

substitute

$\frac{1}{10}=\frac{x}{40}$

solve for x

$x=(40)\frac{1}{10}$

$x=4\ units$

therefore

The length of the longest side of ∆ABC is 4 units

Part 2) Find the ratio of the area of ∆ABC to the area of ∆DEF

we know that

If two figures are similar, then the ratio of its areas is equal to the scale factor squared

Let

z ----> the scale factor

x ----> the area of ∆ABC

y ----> the area of ∆DEF

$z^{2}=\frac{x}{y}$

we have

$z=\frac{1}{10}$

$z^2=(\frac{1}{10})^2$

$z^2=\frac{1}{100}$

therefore

The ratio of the area of ∆ABC to the area of ∆DEF is $\frac{1}{100}$

3 0

3 years ago

I will mark as brainliest

Nina [5.8K]

3/8. Divide 3/4 by two to get 3/8.

5 0

3 years ago

Read 2 more answers

Gianna is going to throw a ball from the top floor of her middle school. When she throws the hall from 48 feet above the ground,

vazorg [7]

Answer:

So, the times the ball will be 48 feet above the ground are t = 0 and t = 2.

Step-by-step explanation:

The height h of the ball is modeled by the following equation

$h(t)=-16t^2+32t+48$

The problem want you to find the times the ball will be 48 feet above the ground.

It is going to be when:

$h(t) = 48$

$h(t)=-16t^{2}+32t+48$

$48=-16t^{2}+32t+48$

$0=-16t^{2}+32t+48 - 48$

$16t^{2} - 32t = 0$

We can simplify by 16t. So

$16t(t-2)= 0$

It means that

16t = 0

t = 0

t - 2 = 0

t = 2

So, the times the ball will be 48 feet above the ground are t = 0 and t = 2.

6 0

3 years ago

What is the solution set for the inequality x < -2 or x > 6

bija089 [108]

he solution set is

{ x ∣ x > 1 } .

Explanation

For each of these inequalities, there will be a set of

-values that make them true. For example, it's pretty clear that large values of

(like 1,000) work for both, and negative values (like -1,000) will not work for either.

Since we're asked to solve a "this OR that" pair of inequalities, what we'd like to know are all the

-values that will work for at least one of them. To do this, we solve both inequalities for

, and then overlap the two solution set

8 0

3 years ago