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3 years ago

Name the three-dimensional figure that is formed by folding the net.

Mathematics

2 answers:

wariber [46]3 years ago

8 0

Answer:

A triangular prism

Step-by-step explanation:

The two bases, which are the two parallel faces of the prism, are triangles. This means the prism is a triangular prism.

Ivanshal [37]3 years ago

3 0

A triangular Prism I would say but it does have rectangles

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Alex was biking along a 48 mile trail. First half of the time he was biking twice as fast as the second half of the time. If he

nydimaria [60]

Answer:

The speed in the first part is 16 mph.

Step-by-step explanation:

The total distance is 48 miles. The total time is 4 hours.

In the second part of the trip, he was traveling at speed s for distance d for 2 hours.

In the first part of the trip he was traveling at twice the speed, or 2s, for distance 48 - d, for 2 hours.

speed = distance/time

distance = speed * time

First part of trip:

48 - d = 2s * 2

d + 4s = 48 Equation 1

Second part of the trip:

d = s * 2

d = 2s Equation 2

Equations 1 and 2 form a system of equations in two unknowns, d and s.

d + 4s = 48

d = 2s

Substitute 2s for d in equation 1.

2s + 4s = 48

6s = 48

s = 8

The speed in the second part is 8 mph.

The speed in the first part is 2s = 2(8) = 16.

The speed in the first part is 16 mph.

Check:

d = 2s = 2(8) = 16

48 - d = 48 - 16 = 32

The second part is 16 miles. The first part is 32 miles.

16 miles at 8 mph takes 2 hours.

32 miles at 16 mph takes 2 hours.

2 hours + 2 hours = 4 hours.

Our answer is correct.

Answer: The speed in the first part is 16 mph.

7 0

4 years ago

Circle 1 is centered at (−4, 5) and has a radius of 2 centimeters. Circle 2 is centered at (2, 1) and has a radius of 6 centimet

IgorC [24]

Basically, you have two circles. You are asked to take circle 1 and "move it" so that it is on top of circle 2. This process of moving is called a translation and can be thought of as sliding. You do this by ensuring that the two have the same center. So, starting at (-4,5) how do you have to move to end up at (2,1)?

To do this we need to move right 6 as the x-coordinate goes from -4 to 2. We also need to move down 4 as the y-coordinate goes from 5 to 1. So we add 6 to the x-coordinate and subtract 4 from the y-coordinate. The transformation rule is (x+6, y-4).

Once you do this the circles have the same center. Next you wish to dilate circle 1 so it ends up being the same size at circle 2. That means you stretch it out in such a way that it keeps its shape. Circle 1 has a radius of 2 centimeters and circle 2 has a radius of 6 centimeters. That is 3x bigger. So we dilate by a factor of 3.

Translations and dilations (along with reflections and rotations) belong to a group known as transformations.

3 0

3 years ago

Do all triangles have a hypotenuse, or only right triangles have a hypotenuse?

Nadusha1986 [10]

Only Right triangles

8 0

3 years ago

Read 2 more answers

What are the possible outcomes?

vesna_86 [32]

Answer:

A= X,B

B= L,G

C= M,W

The total number of combinations is 12

6 0

3 years ago

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponenti

melomori [17]

Answer:

a) 0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

b) Capacity of 252.6 cubic feet per second

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second).

This means that $m = 100, \mu = \frac{1}{100} = 0.01$

(a) Find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.)

We have that:

$P(X > x) = e^{-\mu x}$

This is P(X > 190). So

$P(X > 190) = e^{-0.01*190} = 0.1496$

0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

(b) What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.08?

This is x for which:

$P(X > x) = 0.08$

$e^{-0.01x} = 0.08$

$\ln{e^{-0.01x}} = \ln{0.08}$

$-0.01x = \ln{0.08}$

$x = -\frac{\ln{0.08}}{0.01}$

$x = 252.6$

Capacity of 252.6 cubic feet per second

5 0

3 years ago