All categories

3 years ago

Answer has to be in units

Mathematics

2 answers:

chubhunter [2.5K]3 years ago

7 0

It's a deal. Give me a question, and I'll give you an answer in units.

tatuchka [14]3 years ago

4 0

Yep I'll try to answer to if you would give us something to answer..

You might be interested in

Find the solution set of this inequality|10x+20| ≤10

Pavlova-9 [17]

Answer:

solution is

[-3,-1]

Step-by-step explanation:

we are given

$|10x+20|\leq 10$

Firstly, we will find critical values

so, let's assume it is equal

$|10x+20|= 10$

now, we can break absolute sign

For $|10x+20|= -(10x+20)$ :

$-(10x+20)= 10$

we can solve for x

$-10x-20= 10$

Add both sides by 20

$-10x-20+20= 10+20$

$-10x= 30$

Divide both sides by -10

and we get

$x=-3$

For $|10x+20|= (10x+20)$ :

$(10x+20)= 10$

we can solve for x

$10x+20= 10$

Subtract both sides by 20

$10x+20-20= 10-20$

$10x= -10$

Divide both sides by 10

and we get

$x=-1$

so, critical values are

$x=-3$

$x=-1$

now, we can draw a number line and locate these values

and then we can check inequality on each intervals

For $(-\infty,-3)$ :

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=-5

$|10\times -5+20|\leq 10$

$|-50+20|\leq 10$

$30\leq 10$

so, this is FALSE

For $[-3,-1]$ :

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=-2

$|10\times -2+20|\leq 10$

$|-20+20|\leq 10$

$0\leq 10$

so, this is TRUE

For $(-1,\infty)$ :

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=0

$|10\times 0+20|\leq 10$

$|0+20|\leq 10$

$20\leq 10$

so, this is FALSE

so, solution is

[-3,-1]

7 0

3 years ago

Each day Marvin exercises for 48 minutes how many days will it take him to exercise for 576 minutes?

Triss [41]

The answer is 12 because 576 divided 48 is 12 .

8 0

4 years ago

Read 2 more answers

2k/5+3/4=2 please help asap

charle [14.2K]

Answer:

k=25/8

Step-by-step explanation:

In the attached file

7 0

3 years ago

Can a vector have a component greater than the vector's magnitude ?<br>

Mars2501 [29]

Answer: No, it can never be greater. At most it can be equal though.

5 0

3 years ago

Yall please help me i need to turn this in asap

Monica [59]

Answer:

A)9

Step-by-step explanation:

so if we see the figure the angle are vertical opposite

and so are equal ,

3x+8=35

3x=35-8

3x=27

x=27/3

x=9

hope it helps

8 0

3 years ago