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3 years ago

12

You are ima room with other students and are asked to get in groups of 3. When finished, there are 21 groups of 3. How manny stu

dents are in the room?

2 answers:

kap26 [50]3 years ago

7 0

Answer:

63

Step-by-step explanation:

21 goups of 3 students each.

21 * 3 = 69

Answer: There are 63 students.

ivolga24 [154]3 years ago

6 0

There are 63 students in the room, this is because 21 time 3 gives you 63.

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One canned juice drink is 25% orange juice another is 10% orangw juice. How many liters of each should be mixed together in orde

Vsevolod [243]

Answer:

There should be 10 l of 25% orange juice and 5 l of 10% orange juice.

Step-by-step explanation:

Let the amount of the 25% orange juice be "x", while the amount of the 10% one be "y". The sum of these must be equal to 15 l, therefore:

$x + y = 15$

The sum of the concentration of juice on each can must be equal to the final product, therefore:

$25\%*x + 10\%*y = 15*20\%\\0.25*x + 0.1*y = 3$

We can now solve the system of equations as shown below:

$\left \{ ({{x + y=15})*(-0.1) \atop {0.25*x + 0.1*y=3}} \right. \\\left \{ {{-0.1*x - 0.1*y=-1.5} \atop {0.25*x + 0.1*y=3}} \right. \\-0.1*x + 0.25*x = 3 - 1.5\\0.15*x = 1.5\\x = \frac{1.5}{0.15} = 10\\y = 15 - x = 15 - 10 = 5$

There should be 10 l of 25% orange juice and 5 l of 10% orange juice.

4 0

3 years ago

The largest set of x values satisfying

velikii [3]

Answer:

If $p\ge -\dfrac{1}{3},$ then $x>\dfrac{2}{3}+p$ and $mn=\dfrac{2}{3}+p=\left(\dfrac{2}{3}+p\right)\cdot 1,\ \ m+n=\dfrac{5}{3}+p$

Step-by-step explanation:

Solve two inequalities for x.

1. $2,018x-p$

Separate terms with x and without x into two sides:

$2,018x-2,020x$

Multiply by -1:

$2x>-2p\\ \\x>-p$

2. $7x+3p$

Separate terms with x and without x into two sides:

$7x-10x$

Multiply by -1:

$3x>2+3p\\ \\x>\dfrac{2}{3}+p$

Find the largest set of x values satisfying both inequalities:

$x>-p\\ \\x>\dfrac{2}{3}+p$

If

$-p>\dfrac{2}{3}+p\\ \\-2p>\dfrac{2}{3}\\ \\2p$

then $x>-p$ and $mn=-p=p\cdot (-1),\ m+n=p-1.$ In this case both m and n are negative.

If $p>-\dfrac{1}{3},$ then $x>\dfrac{2}{3}+p$ and $mn=\dfrac{2}{3}+p=\left(\dfrac{2}{3}+p\right)\cdot 1,\ \ m+n=\dfrac{5}{3}+p$

If $p=-\dfrac{1}{3},$ then $x>\dfrac{1}{3}$ and $mn=\dfrac{1}{3}=\dfrac{1}{3}\cdot 1,\ \ m+n=\dfrac{4}{3}$

7 0

3 years ago

833 ÷ 64 showing remainders

shtirl [24]

Answer:

13.015625

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

If <img src="https://tex.z-dn.net/?f=%5Cint%5Climits%5E6_1" id="TexFormula1" title="\int\limits^6_1" alt="\int\limits^6_1" align

navik [9.2K]

Answer:

32

Step-by-step explanation:

[1 , 6] = [1 , 3] + [3 , 6]

[6 , 3] = 10

[3 , 6] = -10

22 = [1 , 3] + (-10)

[1 , 3] = 22 + 10 = 32

7 0

3 years ago

What is the answer??

Volgvan

Answer:

x=3

z=25

Step-by-step explanation:

4 0

3 years ago