Step-by-step explanation:because that's how we've defined it, in the same way that the word "cat" refers to small, furry, feline creatures because that's how we've defined it.

6 0

2 years ago

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PLEASE HELP ME.Math work: The Head Chef has asked you to create a delicious baked dessert that will have the same volume if it i

nikklg [1K]

Answer:

Large baking dish of cylinder, cone and sphere.

Step-by-step explanation:

When we compute the volume of each of the small and large baking dish taking $\pi$ =3.14, we get

Volume of cylinder (small dish)= $\pi$ (r^2)(h)

=3.14( $\frac{3}{2}$ )^2*3

= $\frac{3.14*9*3}{4}$

= 21.195

Volume of cylinder (large dish)= $\pi$ (r)^2(h)= 3.14*(3)^2(3)

=84.78

Similarly, volume of cone (smaller dish)= $\frac{1}{3} \pi r^{2} h$

= 7.065

Volume of cone (larger dish)= 84.78

Volume of sphere(smaller dish)= $\frac{4}{3} \pi r^{3}$

= 14.13

Volume of sphere (larger dish) = 85.17 (approx)

Hence, we get that larger baking dish have approximately same volume.

Therefore, large baking dish of cylinder, cone, sphere will be chosen.

4 0

3 years ago

ACTIVITY 2 (19) Mr Duma recently inherited a rectangular plot, part of the estate left by his late father. The plot with the fol

seraphim [82]

The sum of the lengths of three sides of the rectangle, gives the length

of the fencing, while one third of the rectangle area is for the pavement.

Responses:

Project A: The formula for the length of the fencing is, L = 4·x - 1

Project B: Length of the fancy wall = 2·x + 1

$Project \ C:Area \ of \ the \ paving = \underline{ \dfrac{2\cdot x^2 }{3} - \dfrac{x }{3} - \dfrac{1}{3}}$

<h3>Which method can be used to find the length and area of paving from the given equations?</h3>

Project A: Let QP and SR represent the longest sides of the rectangle, we have;

PQ = SR = 2·x + 1

Given parameters are;

Length of the rectangular plot = 2·x + 1

Width of the rectangular plot = x - 1

Vertices of the rectangular plot are; QPSR

Project A: Let QP and SR represent the longest sides of the rectangle, we have;

PQ = SR = 2·x + 1

Which gives;

SP = QR = x - 1

The length of the fencing, L = SP + PQ + QR = x - 1 + 2·x + 1 + x - 1 = 4·x - 1

The formula for the length of the fencing is, L =<u> 4·x - 1</u>

Project B: The front side is SR

Therefore;

Length of the fancy wall = <u>2·x + 1</u>

Project C:

Area, A = Length × Width

Area of the plot, A = (2·x + 1) × (x - 1) = 2·x² - x - 1

$Area \ of \ the \ paving = \dfrac{1}{3} \times \left(2 \cdot x^2 - x - 1\right) = \underline{ \dfrac{2\cdot x^2 }{3} - \dfrac{x }{3} - \dfrac{1}{3}}$

Learn more about writing equations here:

brainly.com/question/24760633

6 0

2 years ago