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3 years ago

What is the value of K?

Mathematics

1 answer:

ANEK [815]3 years ago

4 0

Answer:

k is 2, l=2√5, m=8√5

Step-by-step explanation:

Triangle MNO is half size of the triangle LON

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kenny6666 [7]

The cost to equip all the stations in the chemistry lab is calculated as: $393.75.

<h3>How to Calculate Total Cost?</h3>

In this scenario, we are given the following:

Total number of stations = 21 stations

Length of rubber tubing each of the stations in the chemistry lab needs = 5 feet

Total length of rubber tubing needed for all stations in the chemistry lab = 21 × 5 = 105 feet

Cost of 1 rubber tubing = $6.25 per yard

Convert 5 feet to yard:

1 yard = 3 feet

x yard = 5 feet

x = (5 × 1)/3

x = 5/3 feet.

So, the cost of 1 rubber tubing = $6.25 per 5/3

Cost of total length of tubbing needed = (105 × 6.25)/5/3 = (105 × 6.25) × 3/5

Cost of total length of tubbing needed = $393.75

Therefore, the cost to equip all the stations in the chemistry lab is calculated as: $393.75.

Learn more about the How to Calculate Total Cost on:

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6 0

2 years ago

An inner city revitalization zone is a rectangle that is twice as long as it is wide. The width of the region is growing at a ra

Ket [755]

Answer:

The area is changing at the point of $\mathbf{61200 m^2/year}$

Step-by-step explanation:

From the given information:

Let's recall from our previous knowledge that the formula for finding the area of a rectangle = L × w

where;

L = length and w = width of the rectangle

Suppose the Length L is twice the width w

Then L = 2w --- (1)

From The area of a rectangle

A = L × w

A = 2w × w

A = 2w²

Taking the above differentiating with respect to time

$\dfrac{dA}{dt }= 4w \times \dfrac{dw}{dt} --- (2)$

At the time t

$\dfrac{dw}{dt}= 34 m \ per \ year ; w = 450 \ m$

Replacing the values back into equation 2, we get:

$\dfrac{dA}{dt }= 4 \times 450 \times 34$

$\mathbf{\dfrac{dA}{dt }= 61200 m^2/year}$

7 0

3 years ago

The radius of the large sphere is double the radius of the small sphere. How many times does the volume of the large sphere than

pshichka [43]

Answer:

8 times larger.

Step-by-step explanation:

The radius of the large sphere is double the radius of the small sphere.

Question asked:

How many times does the volume of the large sphere than the small sphere

Solution:

Let radius of the small sphere =  $x$ 

As the radius of the large sphere is double the radius of the small sphere:

Then, radius of the large sphere = $2x$

To find that how many times is the volume of the large sphere than the small sphere, we will divide the volume of large sphere by volume of small sphere:-

For smaller sphere: $Radius = x$

$Volume \ of \ sphere = \frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \pi x^{3}$

For larger sphere: $Radius = 2x$

$Volume \ of \ sphere = \frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \pi (2x)^{3}$

Now, we will divide volume of the larger by the smaller one:

$=\frac{4}{3} \pi (2x)^{3}\div\frac{4}{3} \times\frac{\pi }{1 } \times x^{3}\\ \\ =\frac{4}{3} \pi\times8x^{3} \times\frac{3}{4\pi }\ \times\frac{1}{x^{3} }$