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3 years ago

5

Julian and Stacy needed 10 liters of water to fill a tank. Stacy filed the tank with 3 11/ 12 liters of water. Julian poured 1 2

/ 5 liters less than Stacy into the tank. how much more water is still needed to fill the tank?

1 answer:

Amiraneli [1.4K]3 years ago

8 0

7 29/60 liters of water

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<em>Trapezoid ABCD was reflected across the y-axis and then translated 7 units up.</em>

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3 years ago

What is the quotient of 35.86 divided by 2.2

Dima020 [189]

Answer:

16.3

Step-by-step explanation:

You simply divide the 35.86 by 2.2 using long division or any method you'd like!

8 0

3 years ago

In ΔUVW, w = 44 cm, u = 83 cm and ∠V=141°. Find the area of ΔUVW, to the nearest square centimeter.

Tema [17]

Answer:

$Area \approx 1149\ cm^2$

Step-by-step explanation:

<u>Given that:</u>

ΔUVW,

Side w = 44 cm, (It is the side opposite to $\angle W$ )

Side u = 83 cm (It is the side opposite to $\angle U$ )

and ∠V=141°

Please refer to the attached image with labeling of the triangle with the dimensions given.

Area of a triangle with two sides given and angle between the two sides can be formulated as:

$A = \dfrac{1}{2}\times a\times b\times sinC$

Where a and b are the two sides and

$\angle C$ is the angle between the sides a and b

Here we have a = w = 44cm

b = u = 44cm

and ∠C= ∠V=141

Putting the values to find the area:

$A = \dfrac{1}{2}\times 44\times 83\times sin141\\\Rightarrow A = \dfrac{1}{2} \times 3652 \times sin141\\\Rightarrow A =1826 \times 0.629\\\Rightarrow A \approx 1149\ cm^2$

So, the <em>area </em>of given triangle to the nearest square centimetre is:

$Area \approx 1149\ cm^2$

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3 years ago

a pet store keeps 4 small fish in every 10 gallons of water how many gallons of water is needed for 36 fish

faltersainse [42]

90 gallons of water

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3 years ago

Read 2 more answers

Find the perimeter of pentagon STUVW WITH VERTICES S(0,0) T(3,-2) U(2,-5) V(-2,-5) W(-3,-2)

andrew11 [14]

Use the distance formula.

$\sqrt{( x_{2} - x_{1} )^2 + (y_{2} - y_{1})^2}$

Points S and W.
$\sqrt{(3)^2 + (2)^2}$

$\sqrt{9+4}$

$\sqrt{13}$

~3.6

Points S and T
$\sqrt{(3 - 0)^2 + (-2 - 0)^2}$

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

$\sqrt{9+4}$

$\sqrt{13}$

~3.6

Points T and U
$\sqrt{(3 - 2)^2 + (-2 + 5)^2}$

$\sqrt{(1)^2 + (3)^2}$

$\sqrt{1+9}$

$\sqrt{10}$

~3.1

Points U and V
$\sqrt{(2+2)^2 + (-5 + 5)^2}$

$\sqrt{(4)^2 + (0)^2}$

$\sqrt{16}$

~4

Points V and W
$\sqrt{(-2+3)^2 + (-5 + 2)^2}$

$\sqrt{(1)^2 + (-3)^2}$

$\sqrt{2+9}$

$\sqrt{11}$

~3.3

Add all these together.

3.3 + 3.1 + 4 + 3.1 + 3.6
≈17

4 0

3 years ago