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3 years ago

A rope is 50 meters long. It is cut into two pieces and one piece is 8 meters longer than the other. What are the two lengths?

Mathematics

1 answer:

Mashcka [7]3 years ago

6 0

Answer:

21 meters and 29 meters

Step-by-step explanation:

x + (x + 8) = 50

2x + 8 = 50

2xc = 42

x = 21

x + 8 = 29

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Area of a triangle with verticals of (-4,1),(-7,5),(0,1)

Lynna [10]

Answer:

$8 ft^2$

Step-by-step explanation:

I assume that the three points given are the vertices of the triangle. So the three points are:

A (-4,1)

B (-7,5)

C (0,1)

The area of a triangle is given by:

$A=\frac{1}{2}bh$

where b is the base and h is the height.

Here we can take as follows:

- We can take the base to be the side AC, so that its length is equal to the distance between the two points:

$AC=\sqrt{(0-(-4))^2+(1-1)^2}=4$

- The height of the triangle is therefore the perpendicular distance between AC and point B. This is equivalent the difference in the x-coordinate between any point on AC and B, so:

$h=(5-1)=4 ft$

Therefore, the area of the triangle is:

$A=\frac{1}{2}(4)(4)=8 ft^2$

7 0

2 years ago

The radius of the base of a cylinder is decreasing at a rate of 121212 kilometers per second. The height of the cylinder is fixe

WARRIOR [948]

Answer:

7536 $km^3/sec$

Step-by-step explanation:

Given that:

Rate of decreasing of radius = 12 km/sec

Height of cylinder is fixed at = 2.5 km

Radius of cylinder = 40 km

To find:

The rate of change of Volume of the cylinder?

Solution:

First of all, let us have a look at the formula for volume of a cylinder.

$Volume = \pi r^2h$

Where $r$ is the radius and

$h$ is the height of cylinder.

As per question statement:

$r$ = 40 km (variable)

$h$ = 2.5 (constant)

$\dfrac{dV}{dt} = \dfrac{d}{dt}\pi r^2h$

As $\pi, h$ are constant:

$\dfrac{dV}{dt} = \pi h\dfrac{d}{dt} r^2\\\Rightarrow \dfrac{dV}{dt} = \pi h\times 2 r\dfrac{dr}{dt} \\$Putting the values:$\\\Rigghtarrow\dfrac{dV}{dt} = 3.14 \times 2.5\times 2 \times 40\times 12 \\\Rigghtarrow\dfrac{dV}{dt} = 7536\ km^3/sec$