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2 years ago

8

The double cone is intersected by a vertical plane passing through the point where the tips of the cones meet. What is the shape

of the cross section formed? HELP PLEASE ITS FOR PLATO

2 answers:

SOVA2 [1]2 years ago

8 0

Answer:

B.

Step-by-step explanation:

answer: B. one triangle tip on top and invert above it with the top touching

Leni [432]2 years ago

6 0

Answer:

B.

Step-by-step explanation:

The double cone is a cone on top of another cone. The bottom cone has the circular base on the bottom and the tip on top. The upper cone is upside down, and the two tips touch. Since the vertical plane goes through the tips of both cones, the cross section must have a shape that gets to a point at the middle of the height.

Answer: B. One triangle with the tip on top and an inverted triangle above it with the tips touching.

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The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillat

Greeley [361]

Answer:

1) $L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

2) $T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

3) $T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

Step-by-step explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

$L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

Part 2

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

Part 3

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

Replacing we got:

$T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

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2 years ago

A museum has a 1/1000 scale model of the Grand Canyon, complete with a miniature river in the bottom. Explain how you could use

defon

Provide the approximated measurement to associate the volume of water.

3 0

2 years ago

Darcie wants to crochet a minimum of 3 blankets to donate to a homeless shelter. Darcie crochets at a rate of 1/15 of a blanke

Gnesinka [82]

Answer:

Step-by-step explanation:

Given that Darcie wants to donate minimum 3 blankets to donate to a homeless shelter. No of days left =60

No of days to complete one blanket = 1/(1/15) = 15 days

Hence in 60 days she has to complete ≥3 blankets

No of days she has to utilize for this will be ≥3(15)=45

No of days she can skip crocheting ≤15

Let x be the no of days crocheted and y no of days skipped

Then x+y≤60

See the graph attached as x axis for days crocheted and y for days skipped

No of blankets can be either 3 or 4.

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2 years ago

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A tile mosaic in the shape of a right triangle is used as the corner for a rectangular pathway. The hypotenuse of the mosaic is

Anit [1.1K]

Answer:

The lengths of the sides are 2.2 ft and 4.5 ft

Step-by-step explanation:

Let

a -----> the length of one side of the mosaic

b ----> the length of the other side of the mosaic

c ----> the hypotenuse of the mosaic

we know that

Applying the Pythagoras Theorem

$c^{2}=a^{2}+b^{2}$ -----> equation A

we have

$a=2b$ -----> equation B

$c=5\ ft$ -----> equation C

substitute equation B and equation C in equation A and solve for b

$5^{2}=(2b)^{2}+b^{2}$

$25=5b^{2}$

$b=\sqrt{5}=2.2\ ft$

Find the value of a

$a=2b$ ------> $a=2\sqrt{5}=4.5\ ft$

therefore

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2 years ago

Write a real world problem that can be represented by the equation 3/4c=21

Black_prince [1.1K]

Answer: Real world problem is "A student have c toffee he distribute $\frac{3}{4}$ th part of those toffees to his friends. He gave total 21 toffees to his friend".

Explanation:

Let a student have c number of toffees in his bag.

It is given that he distribute $\frac{3}{4}$ th part of those toffees to his friends.

The $\frac{3}{4}$ th part of c toffees is,

$\frac{3c}{4}$

The total number of distributed toffees is 21.

$\frac{3c}{4}=21$

It is the same as given equation.

If we change the equation in words it means the $\frac{3}{4}$ th part of a number c is 21.

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2 years ago

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