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1 year ago

HELP IVE BEEN STUCK ON THIS FOREVER

Mathematics

1 answer:

jeyben [28]1 year ago

5 0

Step-by-step explanation:

the figure I've circled is similar to yhe figure shown because when we compare the measurements of the length and breadth of both rectangles, we can see that both the rectangles have the see me measurements, 3ft and 7ft. The only difference however, is that the figures have been rotated differently, the figure shown is kept vertically, on the other hand, the figure I've circled is rotated horizontally.

hope that helps...

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1 year ago

Which is 14 divided by 0.50?

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1 year ago

Read 2 more answers

PLS TRY TO ANSWER ASAP.

Harrizon [31]

Answer:

0.075 m/min

Explanation:

You need to use derivatives which is an advanced concept used in calculus.

1. Write the equation for the volume of the cone:

$V=\dfrac{1}{3}\pi r^2h$

2. Find the relation between the radius and the height:

r = diameter/2 = 5m/2 = 2.5m

h = 5.2m

h/r =5.2 / 2.5 = 2.08

3. Filling the tank:

Call y the height of water and x the horizontal distance from the axis of symmetry of the cone to the wall for the surface of water, when the cone is being filled.

The ratio x/y is the same r/h

x/y=r/h

y = x . h / r

The volume of water inside the cone is:

$V=\dfrac{1}{3}\pi x^2y$

$V=\dfrac{1}{3}\pi x^2(2.08)\cdot x\\\\\\V=\dfrac{2.08}{3}\pi x^3$

4. Find the derivative of the volume of water with respect to time:

$\dfrac{dV}{dt}=2.08\pi x^2\dfrac{dx}{dt}$

5. Find x² when the volume of water is 8π m³:

$V=\dfrac{2.08}{3}\pi x^3\\\\\\8\pi=\dfrac{2.08}{3}\pi x^3\\\\\\ 11.53846=x^3\\ \\ \\ x=2.25969\\ \\ \\ x^2=5.1062$ m²

6. Solve for dx/dt:

$1.2m^3/min=2.08\pi(5.1062m^2)\dfrac{dx}{dt}$

$\dfrac{dx}{dt}=0.03596m/min$

7. Find dh/dt:

From y/x = h/r = 2.08:

$y=2.08x\\\\\\\dfrac{dy}{dx}=2.08\dfrac{dx}{dt}\\\\\\\dfrac{dy}{dt}=2.08(0.035964m/min)=0.0748m/min\approx0.075m/min$

That is the rate at which the water level is rising when there is 8π m³ of water.

4 0

2 years ago

PLEASE HELP -Given the graph of rectangle QUAD and its reflection.

V125BC [204]

Q is located at (-5,2)
Q" is located at (6,2)
they bot h have the same Y value (2) so we know it was only a reflection across the X axis
we need to find out where on the X axis it was reflected.
the distance between -5 and 6 = 11 units
11/2 = 5.5
the reflection line was 5.5 units to the right of the original Q
so the reflection line would be X = 0.5

7 0

2 years ago