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3 years ago

13

How many liters are in a pint?

1 answer:

Iteru [2.4K]3 years ago

3 0

0.473176473 litres are in a pint.

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This is fr due in 20 minutes. please help!

jeyben [28]

Answer:

i think 80/3

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

PLEASE SHOW ALL THE STEPS THAT YOU USE TO SOLVE THIS PROBLEM

Mademuasel [1]

Answer:

{x = 1 , y=1, z=0

Step-by-step explanation:

Solve the following system:

{-2 x + 2 y + 3 z = 0 | (equation 1)

{-2 x - y + z = -3 | (equation 2)

{2 x + 3 y + 3 z = 5 | (equation 3)

Subtract equation 1 from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x - 3 y - 2 z = -3 | (equation 2)

{2 x + 3 y + 3 z = 5 | (equation 3)

Multiply equation 2 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+3 y + 2 z = 3 | (equation 2)

{2 x + 3 y + 3 z = 5 | (equation 3)

Add equation 1 to equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+3 y + 2 z = 3 | (equation 2)

{0 x+5 y + 6 z = 5 | (equation 3)

Swap equation 2 with equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+3 y + 2 z = 3 | (equation 3)

Subtract 3/5 × (equation 2) from equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y - (8 z)/5 = 0 | (equation 3)

Multiply equation 3 by 5/8:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y - z = 0 | (equation 3)

Multiply equation 3 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Subtract 6 × (equation 3) from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y+0 z = 5 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Divide equation 2 by 5:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Subtract 2 × (equation 2) from equation 1:

{-(2 x) + 0 y+3 z = -2 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Subtract 3 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = -2 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Divide equation 1 by -2:

{x+0 y+0 z = 1 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Collect results:

Answer: {x = 1 , y=1, z=0

6 0

3 years ago

An online furniture store sells chairs for $150 each and tables for $400 each. Every day, the store can ship no more than 30 pie

ivolga24 [154]

Answer:

minimum of 13 chairs must be sold to reach a target of $6500

and a max of 20 chairs can be solved.

Step-by-step explanation:

Given that:

Price of chair = $150

Price of table = $400

Let the number of chairs be denoted by c and tables by t,

According to given condition:

t + c = 30 ----------- eq1

t(150) + c(400) = 6500 ------ eq2

Given that:

10 tables were sold so:

t = 10

Putting in eq1

c = 20 (max)

As the minimum target is $6500 so from eq2

10(150) + 400c = 6500

400c = 6500 - 1500

400c = 5000

c = 5000/400

c = 12.5

by rounding off

c = 13

So a minimum of 13 chairs must be sold to reach a target of $6500

i hope it will help you!

4 0

3 years ago

What are the first nine multiples of 5, and 3?

andrezito [222]

First 9 multiples of 5:

$\begin{gathered} 5\cdot1=5 \\ 5\cdot2=10 \\ 5\cdot3=15 \\ 5\cdot4=20 \\ 5\cdot5=25 \\ 5\cdot6=30 \\ 5\cdot7=35 \\ 5\cdot8=40 \\ 5\cdot9=45 \end{gathered}$

First 9 multiples of 3:

$\begin{gathered} 3\cdot1=3 \\ 3\cdot2=6 \\ 3\cdot3=9 \\ 3\cdot4=12 \\ 3\cdot5=15 \\ 3\cdot6=18 \\ 3\cdot7=21 \\ 3\cdot8=24 \\ 3\cdot9=27 \end{gathered}$

3 0

1 year ago

Helpppppppppppppppppp!

julia-pushkina [17]

Remember that the radicand (the area under the root sign) must be positive or zero for a radical with an even index (like the square root or fourth root, for example). This is because two numbers squared or to the fourth power, etc. cannot be negative, so there are no real solutions when the radicand is negative. We must restrict the domain of the square-root function.

If the domain has already been restricted to $x \geq -11$ , we can work backwards to add 11 to both sides. We see that $x+11$ must be under the radicand, so the answer is A.

7 0

3 years ago