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2 years ago

14

HELP PLEASE! Identify the distance between the points (2,4,12) and (7,4,0), and identify the midpoint of the segment for which t

hese are the endpoints. Round to the nearest tenth, if necessary.

1 answer:

hjlf2 years ago

5 0

Do u have a paper to show the model Of your question..

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In the figure, triangle CAE is an enlargement of triangle CBD with scale factor of 4/3. The area of the smaller triangle is 9cm2

balandron [24]

Answer:

16 cm^2

Step-by-step explanation:

Given

$\triangle CAE$ -- Bigger Triangle

$\triangle CBD$ -- Smaller Triangle

$k = \frac{4}{3}$ --- Scale factor

Area of CBD = 9

Required

Determine the area of CAE

The area of triangle CBD is:

$A_1 = \frac{1}{2}bh$

$\frac{1}{2}bh = 9$

The area of CAE is:

$A_2 = \frac{1}{2}BH$

Where:

$B = \frac{4}{3}b$ and

$H = \frac{4}{3}h$

The above values is the dimension of the larger triangle (after dilation).

So, we have:

$A_2 = \frac{1}{2}*\frac{4}{3}b * \frac{4}{3} * h$

$A_2 = \frac{1}{2}*\frac{4}{3} * \frac{4}{3} *b* h$

$A_2 = \frac{1}{2}*\frac{16}{9} *b* h$

Re-order

$A_2 = \frac{16}{9}*\frac{1}{2}* b* h$

$A_2 = \frac{16}{9}*\frac{1}{2}bh$

Recall that:

$\frac{1}{2}bh = 9$

$A_2 = \frac{16}{9}*9$

$A_2 = 16$

Hence, the area is 16 cm^2

3 0

3 years ago

The distance from the ground of a person riding on a Ferris wheel can be modeled by the equation d equals 20 times the sine of t

kodGreya [7K]

We have the function d, representing the distance from the ground of a person riding on a Ferris wheel:

$d(t)=20\sin (\frac{\pi}{30}t)+10$

If we consider the position of the person at t = 0, which is:

$d(0)=20\sin (\frac{\pi}{30}\cdot0)+10=20\cdot0+10=10$

This position, for t = 0, will be the same position as when the argument of the sine function is equal to 2π, which is equivalent to one cycle of the wheel. Then, we can find the value of t:

$\begin{gathered} \sin (\frac{\pi}{30}t)=\sin (2\pi) \\ \frac{\pi}{30}\cdot t=2\pi \\ t=2\pi\cdot\frac{30}{\pi} \\ t=60 \end{gathered}$

Then, the wheel will repeat its position after t = 60 seconds.

Answer: 60 seconds.

7 0

1 year ago

A box contains 20 light box of which five or defective it for lightbulbs or pick from the box randomly what's the probability th

Snowcat [4.5K]

Answer:

1

Step-by-step explanation:

Given:-

- The box has n = 20 light-bulbs

- The number of defective bulbs, d = 5

Find:-

what's the probability that at most two of them are defective

Solution:-

- We will pick 2 bulbs randomly from the box. We need to find the probability that at-most 2 bulbs are defective.

- We will define random variable X : The number of defective bulbs picked.

Such that, P ( X ≤ 2 ) is required!

- We are to make a choice " selection " of no defective light bulb is picked from the 2 bulbs pulled out of the box.

- The number of ways we choose 2 bulbs such that none of them is defective, out of 20 available choose the one that are not defective i.e n = 20 - 5 = 15 and from these pick r = 2:

X = 0 , Number of choices = 15 C r = 15C2 = 105 ways

- The probability of selecting 2 non-defective bulbs:

P ( X = 0 ) = number of choices with no defective / Total choices

= 105 / 20C2 = 105 / 190

= 0.5526

- The number of ways we choose 2 bulbs such that one of them is defective, out of 20 available choose the one that are not defective i.e n = 20 - 5 = 15 and from these pick r = 1 and out of defective n = 5 choose r = 1 defective bulb:

X = 1 , Number of choices = 15 C 1 * 5 C 1 = 15*5 = 75 ways

- The probability of selecting 1 defective bulbs:

P ( X = 1 ) = number of choices with 1 defective / Total choices

= 75 / 20C2 = 75 / 190

= 0.3947

- The number of ways we choose 2 bulbs such that both of them are defective, out of 5 available defective bulbs choose r = 2 defective.

X = 2 , Number of choices = 5 C 2 = 10 ways

- The probability of selecting 2 defective bulbs:

P ( X = 2 ) = number of choices with 2 defective / Total choices

= 10 / 20C2 = 10 / 190

= 0.05263

- Hence,

P ( X ≤ 2 ) = P ( X =0 ) + P ( X = 1 ) + P (X =2)

= 0.5526 + 0.3947 + 0.05263

= 1

7 0

3 years ago

The volume of this cylinder is 1751 cubic units. What is the volume of a cone that has the same base area and the sam height

siniylev [52]

Answer:

$V_2 = 583.67$

Step-by-step explanation:

Given

Represent the volume of the cylinder with V1 and the cone with V2.

So:

$V_1 = 1751$

Required

Find V2

The volume of a cylinder is:

$V_1 = \pi r^2h$

Because the cone and the cylinder have the same base area and height, the volume of the cone is:

$V_2 = \frac{1}{3}\pi r^2h$

Substitute $V_1$ for $\pi r^2h$ .

$V_2 = \frac{1}{3}\pi r^2h$

$V_2 = \frac{1}{3}V_1$

Substitute 1751 for V1

$V_2 = \frac{1}{3} * 1751$

$V_2 = \frac{1* 1751}{3}$

$V_2 = \frac{1751}{3}$

$V_2 = 583.67$

<em>The volume of the cone is 583.67 cubic units</em>

6 0

3 years ago

.....?????...........

Liula [17]

You need 3a^3 to equal a^3, so you would need at subtract 2a^3.

Since the problem is addition, add - 2a^3:

3a^3 + -2a^3 = 3a^3 - 2a^3 = a^3

Then you have -5b^3 and you need it to equal b^^3, so you would need to add 6b^3:

-5b^3 + 6b^3 = b^3

First box would be -2

Second box would be 6

3 0

3 years ago