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4 years ago

What is a sphere made up of?

1 answer:

soldier1979 [14.2K]4 years ago

7 0

In my opinion, the correct answer among the choices listed above is option B. A sphere made up of many circular cross sections. When you cut through a plane at any angle or side, you will see that the plane made will always be a circle.

You might be interested in

What’s the slope of (-7,2) and (1,4)

mamaluj [8]

Answer:

1/4

Step-by-step explanation:

Slope = (change in the y-values)/(change in the x-values)

Slope = (4 - 2)/(1 - (-7) )

Slope = (4 - 2)/(1 + 7)

Slope = 2/8

Slope = 1/4

8 0

3 years ago

Find the surface area of the composite figure. SA = [ ? ] cm^2

slava [35]

Answer:

644

Step-by-step explanation:

SA=[2×(5×20 + 5×6 + 20×6)] +

[2×(4×12 + 4×6 + 12×6)] -

[ 2× 12×6]

= [2×(250)]+[2×(144)] - [144]

= 500+288-144

= 644 cm²

6 0

3 years ago

As a bowl of soup cools, the temperature of the soup is given by the twice-differentiable function H for 0<img src="https://tex.

Aleks04 [339]

The rate of change of temperature with time at a point in time is given by

the derivative of the function for the temperature of the soup.

The correct responses are;

a) H'(5) is approximately -2.6 degrees Celsius per minute.

b) Yes

c) The equation for the line tangent is y = -3.6·t + 90.8
The approximate value of C(5) is 72.8 °C

d) The rate of change of the temperature of the soup at t = 3 minutes is -3.6 degrees Celsius per minute.

Reasons:

a) From the data in the table, we have;

The approximate value of H'(5) is given by the average value of the rate of

change of the temperature with time between points, t = 3, and t = 8

Therefore;

$\displaystyle H'(5) = \mathbf{\frac{H(8) - H(3)}{8 - 3}}$

Which gives;

$\displaystyle H'(5) = \frac{80 - 67}{8 - 3} = \mathbf{2.6}$

Therefore, H'(5) = -2.6°C per minute

b) Given that the function is twice differentiable over the interval, 0 ≤ t ≤ 12, the function for the change in temperature is continuous in the interval 0 ≤ t ≤ 12

At t = 0, H(0) = 90 °C

At t = 12, H(12) = 58 °C

58 °C < 60 < 90 °C

Therefore, there exist a temperature, of 60 °C between 90° C and 58 °C

c) The given derivative of C is, $C'(t) = \mathbf{-3.6 \cdot e^{-0.05 \cdot t}}$

At t = 3, we have;

$The \ slope \ at \ t = 3 \ is \ C'(3) = -3.6 \cdot e^{-0.05 \times 3} \approx -3.1$

Therefore, we have;

y - 80 ≈ -3.1 × (x - 3)

The equation for the tangent is; y = -3.6 × (x - 3) + 80

y = -3.6·x + 10.8 + 80 = -3.6·x + 90.8

The equation for the tangent is; y = -3.6·x + 90.8

The value of C(5) is approximately, C(5) ≈ -3.6 × 5 + 90.8 = 72.8

C(5) ≈ 72.8°

d) Based on the the model above, the rate at which the temperature of the

soup is changing at t = 3 minutes is -3.6 degrees per minute.

Learn more about calculus and concepts here:

brainly.com/question/20336420

8 0

3 years ago

X = 10 x = 2 x = two divided by five. x = 11

KengaRu [80]

Answer:

x = 10

Step-by-step explanation:

You can try the answers to see which works. (The first one does.)

Or, you can solve for the variable:

Divide by 75

... (1/5)^(x/5) = 3/75 = 1/25

Recognize that 25 = 5^2, so ...

... (1/5)^(x/5) = (1/5)^2

Equating exponents, you have

... x/5 = 2

... x = 10 . . . . . multiply by 5

_____

You can also start by taking logarithms:

... log(75) +(x/5)log(1/5) = log(3)

... (x/5)log(1/5) = log(3) -log(75) = log(3/75) = log(1/25) . . . . simplify the log

... x/5 = log(1/25)/log(1/5) = 2 . . . . . simplify (or evaluate) the log expression

... x = 10 . . . . . multiply by 5

_____

"Equating exponents" is essentially the same as taking logarithms.

5 0

3 years ago

Determine the general solution of: sin 3x - sin x = cos2x

SpyIntel [72]

Sin³ x-sin x=cos ² x

we know that:
sin²x + cos²x=1 ⇒cos²x=1-sin²x
Therefore:

sin³x-sin x=1-sin²x
sin³x+sin²x-sin x-1=0

sin³x=z

z³+z²-z-1=0

we divide by Ruffini method:
1   1 -1 -1
1 1 2 1 z=1
-------------------------------------
1   2 1 0
-1   -1 -1 z=-1
--------------------------------------
1 1 0 z=-1

Therefore; the solutions are z=-1 and z=1

The solutions are:
if z=-1, then
sin x=-1   ⇒x= arcsin -1=π+2kπ (180º+360ºK) K∈Z

if z=1, then
sin x=1   ⇒ x=arcsin 1=π/2 + 2kπ (90º+360ºK) k∈Z

π/2 + 2kπ U   π+2Kπ=π/2+kπ k∈Z ≈(90º+180ºK)

Answer: π/2 + Kπ or 90º+180ºK K∈Z
Z=...-3,-2,-1,0,1,2,3,4....

6 0

3 years ago