Ask question

Ask question

All categories

3 years ago

7

1. Building on the Essential Question What would have a greater effect on the volume of a cone: doubling its radius or doubling

its height? Explain .
u

1 answer:

seraphim [82]3 years ago

7 0

Answer:

The answer is in this link !

https://williamreid.weebly.com/uploads/5/8/1/4/58144319/hrtjndvsef.pdf

not a tricky link like the others lol

Step-by-step explanation:

You might be interested in

Options: for A-G

Sindrei [870]

Answer:

a. Definition of segment Bisector

b. Line segment KM

c. Given

d. Definition of segment bisector

e. Reflexive property of congruence

f. SAS

g. CPCTC (Corresponding parts of congruent triangles are congruent)

3 0

3 years ago

Aaron estimated that 148% of 333 is 495

maksim [4K]

Answer:

492.84

Step-by-step explanation:

333 x 148÷100

= 492.84

7 0

3 years ago

Read 2 more answers

A cube has a volume of 64 cubic meters what is the length of each edge

Murljashka [212]

Answer:

The length of each side is 4 m

Step-by-step explanation:

The volume of a cube is given by

V = s^3 where s is the side length

64 = s^3

Take the cube root of each side

64 ^ (1/3) = s^3 ^ (1/3)

4 = s

The length of each side is 4 m

7 0

3 years ago

Read 2 more answers

In the triangle pictured, let A, B, C be the angles at the three vertices, and let a,b,c be the sides opposite those angles. Acc

Troyanec [42]

Answer:

Step-by-step explanation:

(a)

Consider the following:

$A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°$

Use sine rule,

$\frac{b}{a}=\frac{\sinB}{\sin A}
\\\\=\frac{\sin{\frac{\pi}{3}}
}{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}$

Again consider,

$\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
\\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Thus, the angle B is function of A is, $B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Now find $\frac{dB}{dA}$

Differentiate implicitly the function $\sin{B}=\sqrt{\frac{3}{2}}\sin{A}$ with respect to A to get,

$\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}$

b)

When $A=\frac{\pi}{4},B=\frac{\pi}{3}$ , the value of $\frac{dB}{dA}$ is,

$\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}$

c)

In general, the linear approximation at x= a is,

$f(x)=f'(x).(x-a)+f(a)$

Here the function $f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

At $A=\frac{\pi}{4}$

$f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}$

And,

$f'(A)=\frac{dB}{dA}=\sqrt{3}$ from part b

Therefore, the linear approximation at $A=\frac{\pi}{4}$ is,

$f(x)=f'(A).(x-A)+f(A)\\\\=f'(\frac{\pi}{4}).(x-\frac{\pi}{4})+f(\frac{\pi}{4})\\\\=\sqrt{3}.[x-\frac{\pi}{4}]+\frac{\pi}{3}$

d)

Use part (c), when $A=46°$ , B is approximately,

$B=f(46°)=\sqrt{3}[46°-\frac{\pi}{4}]+\frac{\pi}{3}\\\\=\sqrt{3}(1°)+\frac{\pi}{3}\\\\=61.732°$

8 0

3 years ago

What is the perimeter of the figure to the nearest tenth of a millimeter

torisob [31]

The answer would be 18mm because 6+6+3+3 considering that there are 4 sides.

6 0

3 years ago

Read 2 more answers