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3 years ago

How much water should be added to 8 quarts of 90% antifreeze to get a 30% solution?

Mathematics

1 answer:

Serjik [45]3 years ago

7 0

Answer:

16 quarts

Step-by-step explanation:

Let's say x is the volume of water.

Initially, the amount of antifreeze is the concentration times the volume: 0.90 (8).

After the water is added, the amount of antifreeze is the new concentration times the new volume: 0.30 (8 + x).

Since the amount of antifreeze doesn't change, we can set these equal and solve for x.

0.90 (8) = 0.30 (8 + x)

7.2 = 2.4 + 0.30 x

4.8 = 0.30 x

x = 16

16 quarts of water should be added.

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What is the perimeter of rhombus WXYZ?

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<u>Answer</u>:

The perimeter of rhombus WXYZ is $4 \sqrt{13}$

<u>Step-by-step explanation:</u>

Step 1 :Finding length of XY

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$x_1$ = 5

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Step 2 :Finding length of YZ

Distance formula = $\sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}$

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$x_1$ = 3

$x_2$ =5

$y_1$ = 2

$y_2$ =5

YZ = $\sqrt{(5-3)^2 +(5-2)^2}$

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YZ = $\sqrt{(13)}$

Step 3 : :Finding length of ZW

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here

$x_1$ = 5

$x_2$ =7

$y_1$ = 5

$y_2$ =2

ZW = $\sqrt{(7-5)^2 +(5-2)^2}$

ZW = $\sqrt{(2)^2 +(3)^2}$

ZW = $\sqrt{4 +9}$

ZW = $\sqrt{(13)}$

Step 4 :Finding length of WX

Distance formula = $\sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}$

here

$x_1$ = 7

$x_2$ =5

$y_1$ = 2

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WX = $\sqrt{(7-5)^2 +((-1)-2)^2}$

WX = $\sqrt{(2)^2 +(-3)^2}$

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WX = $\sqrt{(13)}$

Step 5: finding the perimeter of the rhombus

Perimeter= 4 X side

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What is the answer to this question

umka21 [38]

First you bring the 9 over with the seven
$-2= \sqrt[3]{x+9}$
Now cube both sides
$-8=x+9$
Finally subtract 9 over and get
$x=-17$

Check
$7=9+ \sqrt[3]{-17+9}$
$7=9+ \sqrt[3]{-8}$
$7=9+(-2)$
7=7

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