How much water can be contained by a cylindrical tin can with a radius of 15m and a height of 268m if its 5/6 full?

1 answer:

5 0

First you find the area of the circle using the radius then multiplying that to the height. That gets the full volume. Using that measurement multiply it by 5/6 to figure out what 5/6 of the volume is. Good luck!

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Is 0.75 greater than or less than 7/8

Marina86 [1]

0.75 is greater then 7/8

8 0

2 years ago

7x^2=9+x what are the values of x Will give medal and points

DENIUS [597]

7x² = 9 + x Subtract x from both sides
7x² - x = 9 Subtract 9 from both sides
7x² - x - 9 = 0 Use the Quadratic Formula

a = 7 , b = -1 , c = -9

x = $\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$ Plug in the a, b, and c values
x = $\frac{- (-1) \pm \sqrt{(-1)^2 - 4(7)(-9)} }{2(7)}$ Cancel out the double negative
x = $\frac{1 \pm \sqrt{(-1)^2 - 4(7)(-9)} }{2(7)}$ Square -1
x = $\frac{1 \pm \sqrt{1 - 4(7)(-9)} }{2(7)}$ Multiply 7 and -9
x = $\frac{1 \pm \sqrt{1 - 4(-63} }{2(7)}$ Multiply -4 and -63
x = $\frac{1 \pm \sqrt{1 + 252} }{2(7)}$ Multiply 2 and 7
x = $\frac{1 \pm \sqrt{1 + 252} }{14}$ Add 1 and 252
x = $\frac{1 \pm \sqrt{253} }{14}$ Split up the $\pm$
x = $\left \{ {{ \frac{1 + \sqrt{253} }{14} } \atop { \frac{1 - \sqrt{253} }{14} }} \right.$
The approximate square root of 253 is 15.905973.
x ≈ $\left \{ { \frac{1 + 15.905973}{14} } \atop { \frac{1 - 15.905973}{14} }} \right$ Add and subtract
x ≈ $\left \{ {{ \frac{16.905973}{14} } \atop { \frac{14.905973}{14} }} \right.$ Divide
x ≈ $\left \{ {{1.2075} \atop {1.0647}} \right.$ Round to the nearest hundredth
x ≈ $\left \{ {{1.21} \atop {1.06}} \right.$

7 0

3 years ago

6 [ 13 - 2 ( 4 + 1 ) ] simply step by step

sp2606 [1]

Answer:

Step-by-step explanation:

$6(13 - 2(4 + 1)) \\ 6(13 - 2 \times 5) \\ 6(13 - 10) \\ 6 \times 3 \\ = 18$