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sp2606 [1]
3 years ago
5

A cubical tank of edge 12 cm is half filled with water. The water is poured into an empty rectangular tank measuring 10 cm by 8

cm by 7 cm until it is full. How much water is left in the cubical tank? Give your answer in millilitres.
Mathematics
1 answer:
qaws [65]3 years ago
3 0

Answer:

3040000mm³

Step-by-step explanation:

Volume of a cube = a³ ; a = size of edge

Volume of cubical tank = 12³ = 12 * 12 * 12 = 1728 cm³

Half filled cube = 1728 / 2 = 864 cm³

Volume is f rectangular tank :

V = length * width * height

V = 10 * 8 * 7 = 560 cm³

Volume of water left in cubical tank :

(864 - 560) = 304cm³

1cm³ = 1000mm³

304cm³ = 304 * 1000 = 3040000mm³

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Solve the system by elimination<br><br> 3x-2y=3<br> 15x-4y=-3
stiv31 [10]

Answer:

x=-1

y=-3

Step-by-step explanation:

3x-2y=3(1)

5.(1) <=> 15x-10y=15 (2)

15x-4y=-3 (3)

(2)-(3) <=> -6y=18

4 0
2 years ago
If 0 lies in the quadrant IV. What can be the value of cos 0?
Andru [333]

Answer:

Positive one

Step-by-step explanation:

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4 0
3 years ago
Your answer should be a polynomial in standard form. (x + 5)(x +3)=(x+5)(x+3)
sattari [20]

Answer:

∴(x+5)(x+3) =x²+8x+15

Step-by-step explanation:

Standard form of a polynomial:

Standard of polynomial means that the terms ordered from biggest exponent of variable to lowest exponent of variable.

Example:

a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+......

Given,

(x+5)(x+3)

=x(x+3)+5(x+3)

=x²+3x+5x+15

=x²+8x+15

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7 0
2 years ago
Hello, I'm new! I need help
melamori03 [73]

The interest in the first month is given as $ 97.1. The principal balance in the second question is $15,030.02

<h3>How to solve for the interest in the first month</h3>

1. We have to solve for the cost of the car

This would be = 19,725*(1.0475)

= 20,661.9375

There is a Down payment = 2,175

balance would be 20661.9375-2,175 = 18,486.94

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principal balance = 16700 - 1670

= $15,030.02

Read more on interest rate here

brainly.com/question/25793394

7 0
1 year ago
The observation deck is about 20 m above sea level. From the observation deck, the angle of depression of the boat is 6 degrees.
Leno4ka [110]
 θ\ 6 degrees
     \
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