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lisabon 2012 [21]
3 years ago
14

A cylindrical tank that holds 375 L of water is completely full. A pump removes water at a rate of 0.6 L/s. For how many minutes

must the pump work until 240 L of water remain in the tank?
Mathematics
1 answer:
Misha Larkins [42]3 years ago
6 0
375L-240L=135L
135÷0.6=81
The pump must work for 81 min or 1hr and 21 min until the tank reaches 240L
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I don’t know how to do it
mote1985 [20]
Basically, we need to find the diagonal of the square. Once we have the diagonal, we need to find if it's more or less than 11 cm, which is diameter of the circle.
This is the formula we'll use:
a²+b²=c²
Since it's a square, a and b are the same.
7²+7²=c²
49+49=c²
98=c²
√98=c
c=9.8995
So, the diagonal is 9.8995
9.8995<11

So, since the diagonal of the square is 9.8995 cm, it is less than the circumference of the circle meaning that it will fit in to the circle without touch the circle's circumference.
6 0
3 years ago
How to solve? Would you divide by 4? 4 representing the four corners
charle [14.2K]

Answer:

paper left = 168.56 units²

Step-by-step explanation:

first, figure the area of the square:

area of square = 28x28 = 784 units²

second, figure the area of the circle:

radius = 1/2of 28 = 14

area of circle = Pi x Radius² = (3.14)(14)² = 615.44 units²

subtract area of circle from area of square:

784 - 615.44 = 168.56 units²

3 0
3 years ago
What is the answer to this please??
lianna [129]
The formula for the circumference of a circle is C = 2pi*r, where r is the radius.  Here, 7850 units = C = 2pi*r.

Dividing 7850 units by 2pi, we get
        7850 units
r  = ----------------- = 1249 units (to the nearest unit)
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3 0
3 years ago
4) Point (1, 4) is reflected over the y-axis. What are the new coordinates?​
Viefleur [7K]

Answer:

(-1,4)

Step-by-step explanation:

Reflection Rule for over the y axis: A(x,y) A'(-x,y)

Hope this helps!

3 0
3 years ago
Read 2 more answers
Help me plss, number 2.​
Gnesinka [82]

Answer:

At (-2,0) gradient is -4 ; At (2,0) gradient is 4

Step-by-step explanation:

For this problem, we simply need to take the derivative of the function and evaluate when y = 0 (when crossing the x-axis).

y = x^2 - 4

y' = 2x

The function y = x^2 - 4 cross the x-axis when:

y = x^2 - 4

0 = x^2 - 4

4 = x^2

2 +/- = x

Hence, this curve crosses the x-axis twice, once at (-2,0) and again at (2,0).

The gradient at these points are as follows:

y' = 2(-2) = -4

y' = 2(2) = 4

Cheers.

3 0
4 years ago
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