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3 years ago

How many 0.4-liter glasses of water does it take to fill up a 3.4-liter pitcher. SHOW YOUR WORK!!!!!!!!!!

Mathematics

2 answers:

gulaghasi [49]3 years ago

7 0

3.4 divided by 0.4

Because a decimal can not be outside the box (the division house), you bring the decimal to the right and do the same for the number inside the box. It would turn 0.4 into 4 and 3.4 into 34. (That's a decimal point not a period). Now, align the decimals with the above quotient too. After doing the division steps, you get 8.5 glasses. So, it takes 8.5 0.4-liter glasses to fill up a 3.4-liter pitcher. Hope this helped! Follow me on ig: thats.nicolle

bulgar [2K]3 years ago

5 0

To get the result we just have to divide 3.4 liter by 0.4 liter.
3.4:0.4=8.5 - its our result.
To better undestand lets think that u have 4 piece of cake and 2 person. To know how much u can give each person u have to divide 4 piece by 2 person. Its similar situation here.

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What is the Equation of a line that has a slope<br> of -3 passes to the point (4,-5)

ololo11 [35]

Answer:

y= -3x + 7:)

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4 0

3 years ago

Read 2 more answers

Pre-Calc help:

stepladder [879]

Answer:

$8x^2+8y^2-79x-32y+95=0$

Step-by-step explanation:

Let the equation be : $x^2+y^2+2gx+2fy+c=0$

The point (1,1) must satisfy this equation.

$1^2+1^2+2g*1+2f*1+c=0\\2g+2f+c=-2---(1)$

Similarly, the point (1,3) must satisfy:

$1^2+3^2+2g*1+2f*3+c=0\\1+9+2g+6f+c=0\\2g+6f+c=-10---(2)$

Also for the point (9,2), we have:

$9^2+2^2+2g*9+2f*2+c=0\\81+4+18g+4f+c=0\\18g+4f+c=-85---(3)$

Solving the equations (1), (2) and (3) simultaneously, we get:

$g=-\frac{-79}{16},f=-2,c=\frac{95}{8}$

We Substitute this values to get:

$x^2+y^2+2(\frac{-79}{16})x+2(-2)y+\frac{95}{8}=0$

$x^2+y^2-(\frac{79}{8})x-4y+\frac{95}{8}=0$

Multiply through by 8 to get:

$8x^2+8y^2-79x-32y+95=0$

8 0

3 years ago

Hi guys please help meee

Lisa [10]

<h3>Answer: 10</h3>

=======================================================

Explanation:

We'll start off converting each mixed number into an improper fraction.

The formula to use is $A \frac{b}{c} = \frac{A*c+b}{c}$

So,

$A \frac{b}{c} = \frac{A*c+b}{c}\\\\2 \frac{1}{8} = \frac{2*8+1}{8}\\\\2 \frac{1}{8} = \frac{16+1}{8}\\\\2 \frac{1}{8} = \frac{17}{8}\\\\$

And,

$A \frac{b}{c} = \frac{A*c+b}{c}\\\\2 \frac{2}{3} = \frac{2*3+2}{3}\\\\2 \frac{2}{3} = \frac{6+2}{3}\\\\2 \frac{2}{3} = \frac{8}{3}\\\\$

So the task of computing $2 \frac{1}{8} \times 2 \frac{2}{3}$ is exactly the same as computing $\frac{17}{8} \times \frac{8}{3}$

Notice how we have an 8 up top and an 8 down below. Those 8's cancel out and we're left with $\frac{17}{3}$ . That fraction does not reduce any further.