The diagram shows squares 1, 2, and 3 constructed on the
sides of a right triangle.
3
2
Which statement about the squares must be true?
A. (Perimeter of 1) + (Perimeter of 2) = (Area of 3)
B. (Area of 1) + (Area of 2) = (Area of 3)
C. (Perimeter of 1) + (Perimeter of 2) = (Perimeter of 3)
D. (Area of 1) + (Area of 2) = (Perimeter of 3)
SUBMIT

Answers

Answer 1

The statement that must be true is:  (Area of 1) + (Area of 2) = (Area of 3). This statement accurately reflects the relationship between the areas of the squares in the given diagram.

The correct answer is option B.

To determine which statement about the squares must be true, let's analyze the given diagram and the properties of squares.

In the diagram, square 1 is constructed on one side of the right triangle, square 2 is constructed on another side of the right triangle, and square 3 is constructed on the hypotenuse of the right triangle.

Statement A: (Perimeter of 1) + (Perimeter of 2) = (Area of 3)

This statement relates to the perimeters of squares 1 and 2 being equal to the area of square 3. However, this is not necessarily true. The perimeters of squares 1 and 2 are related to the lengths of their sides, while the area of square 3 is related to the length of its side. The perimeters and areas are different measures, so this statement is not necessarily true.

Statement B: (Area of 1) + (Area of 2) = (Area of 3)

This statement compares the areas of squares 1 and 2 to the area of square 3. Since squares have all sides equal in length, the areas of squares 1 and 2 are equal to the area of square 3. Therefore, this statement must be true

Statement C: (Perimeter of 1) + (Perimeter of 2) = (Perimeter of 3)

This statement relates to the perimeters of squares 1 and 2 being equal to the perimeter of square 3. However, since the squares have different side lengths, their perimeters are different. Therefore, this statement is not necessarily true.

Statement D: (Area of 1) + (Area of 2) = (Perimeter of 3)

This statement compares the areas of squares 1 and 2 to the perimeter of square 3. As mentioned earlier, the areas of squares 1 and 2 are equal to the area of square 3, which means they are not necessarily equal to the perimeter of square 3. Hence, this statement is not necessarily true.

Therefore, the correct statement is option B.

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Related Questions

assignment questions

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Answer:

which on a is the question

use equivalent ratios in higher terms to eliminate decimals from the following ratio 3.04 to 6 and ratio is ?​

Answers

The equivalent ratio in higher terms to eliminate decimals from the ratio 3.04 to 6 is 76 to 150.

To eliminate decimals from the ratio 3.04 to 6, we can multiply both terms of the ratio by a common factor that will result in whole numbers.

First, let's convert 3.04 to a fraction:

3.04 = 3 + 0.04 = 3 + 4/100 = 3 + 1/25 = 75/25 + 4/100 = 76/25

Now, the ratio 3.04 to 6 can be written as:

3.04/6 = 76/25 / 6

To eliminate the decimal, we can multiply both the numerator and denominator by 25:

(76/25) * 25 / (6 * 25) = 76 / 150

Therefore, the equivalent ratio in higher terms to eliminate decimals from the ratio 3.04 to 6 is 76 to 150.

By multiplying both terms by 25, we effectively scale up the ratio to eliminate the decimal and create whole numbers. This allows us to express the ratio in higher terms without decimals. The final ratio 76 to 150 represents the same relationship as the original ratio 3.04 to 6, but in whole number form.

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A block lies on a horizontal frictionless surfaces.A horizontal force of 100N is applied to the block giving rise to an acceleration of 3.0 .(a) determine the mass of the block . (b) calculate the distance of the block will travel in the force applied for 10s .(c) calculate the speed of the block after the force has been applied for 10s​

Answers

Answer:

Explanation:

A) The mass of the block is 33.3 Kg.

We know that F=ma

m=F/a

Where,

F⇒Force

m⇒mass

a⇒acceleration

Given,

F=100N

a=3m/s^2

m=100/3

m= 33.3Kg.

Therefore the mass of the block is 33.3 kg.

B) Distance traveled in 10s is 150m.

According to Newton's laws of motion,

s=ut+1/2at^2

s⇒displacement

u⇒initial velocity

t⇒time

Given,

u=0

t=10s

a=3m/s^2

s=0+1/2*3*100

s=150m.

Therefore distance traveled by the block after 10s is 150m.

C)The speed of the block after 10s will be 30m/s.

We know that

v^2=u^2+2as

v⇒Final velocity

u⇒initial velocity

v^2=0+2*3*150

v^2=900

v=30m/s.

Therefore the velocity of the block will be 30m/s after 10s.

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A rope pulls a Tesla out of mud. The guy pulls a force F⊥ of 300N, and theta = 4.2°. The tension force T is ___ Newton.

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A rope pulls a Tesla out of mud. The guy pulls a force F⊥ of 300N, and theta = 4.2°. The tension force T is 298.44__ Newton.

The problem describes a Tesla that is stuck in the mud and needs to be pulled out using a rope. the guy pulls a force F⊥ of 300N and that the angle between the rope and the horizontal plane is θ = 4.2°. The goal is to find the tension force T exerted by the rope.To solve for T, we'll need to use trigonometry. We can break the force vector into its horizontal and vertical components as follows:

Fx = F⊥ cosθ and Fy = F⊥ sinθ.

Since the rope is pulling the Tesla horizontally, the horizontal component of the force will be the tension force T. So we have:

T = Fx = F⊥ cosθ = (300 N) cos(4.2°) ≈ 298.44 N

Taking the cosine of the angle is necessary since it's the adjacent side that we're interested in, which is the horizontal component of the force. Therefore, the tension force exerted by the rope is approximately 298.44 N.

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A cup of coffee cools from 120-115°f in 0.5 minutes in a room of 65°f. how long will it take to cool from 105°f _90°f in the same room use newton's law of Cooling​

Answers

1.5 minutes to get to 90 degrees f from 105 degrees f (it is just basic adding

In Figure below, m₁=2.00kg and m₂-4.00kg. Consider the pulley to be frictionless. (a) If m₂ is released, what will its acceleration be? (b) What is the tension in the string? m₂ 55 m₂​

Answers

(a) When m₂ is released, its acceleration will be approximately -3.27 m/s².

(b) The tension in the string is approximately -13.08 N.

To determine the acceleration of m₂ when it is released and the tension in the string, we need to consider the forces acting on the system.

(a) Acceleration of m₂:

Since the pulley is assumed to be frictionless, the tension in the string is the same on both sides of the pulley. We can consider the system consisting of m₁ and m₂ as one body. The net force acting on this system is the difference between the weight of m₁ and the weight of m₂:

Net force = m₁g - m₂g

Applying Newton's second law, F = ma, where F is the net force and a is the acceleration, we have:

m₁g - m₂g = (m₁ + m₂)a

Rearranging the equation to solve for the acceleration, we get:

a = (m₁g - m₂g) / (m₁ + m₂)

Substituting the given values, m₁ = 2.00 kg and m₂ = 4.00 kg, and the acceleration due to gravity, g = 9.8 m/s², we can calculate the acceleration:

a = ((2.00 kg)(9.8 m/s²) - (4.00 kg)(9.8 m/s²)) / (2.00 kg + 4.00 kg)

a = (19.6 N - 39.2 N) / 6.00 kg

a = -19.6 N / 6.00 kg

a = -3.27 m/s²

Therefore, when m₂ is released, its acceleration will be approximately -3.27 m/s². The negative sign indicates that the acceleration is in the opposite direction of the gravitational force.

(b) Tension in the string:

The tension in the string can be determined by considering the forces acting on m₂. The net force on m₂ is equal to its mass multiplied by its acceleration:

Net force = m₂a

Substituting the given values, m₂ = 4.00 kg and a = -3.27 m/s², we can calculate the tension:

Tension = (4.00 kg)(-3.27 m/s²)

Tension = -13.08 N

Therefore, the tension in the string is approximately -13.08 N. The negative sign indicates that the tension acts in the opposite direction of the weight of m₂.

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Cobalt-60 (Co) is often used as a radiation source in medicine. It has a half-life of 5.25 years. 4.1. Explain what is meant by the underlined sections in the statement above. [5] Using her knowledge and understanding of nuclear physics, a student was asked to answer the following problem about cobalt-60: How long after a new sample is delivered will its activity have decreased (a) to about one-eighth its original value? (b) to about one-third its original value? Give your answers to two significant figures. The student was also provided with the following information: The activity is proportional to the number of undecayed atoms (AN/At = AN) 4.2. Explain what is meant by the information above provided to the student. [5]​

Answers

From the question;

1) It takes 15.75 years to decrease to 1/8

2) It takes 8.36 years to decrease to 1/3

What is half life?

Half-life is the length of time it takes for a chemical to degrade or go through a particular process. It frequently refers to the length of time it takes for half of a radioactive substance to decay into a stable form in the context of radioactive decay.

We know that;

[tex]N/No = (1/2)^t/t1/2[/tex]

No = initial amount

N = amount at time t

t = Time taken

t1/2 = half life

[tex]1/8 = (1/2)^t/5.252^-3 = 2^-t/5.25[/tex]

t = 15.75 years

Again;

[tex]1/3 = (1/2)^t/5.25[/tex]

ln0.33 = t/5.25ln0.5

t = ln0.33/ln0.5 * 5.25

t = 8.36 years

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Select the correct answer.
Object A attracts object B with a gravitational force of 5 newtons from a given distance. If the distance between the two objects is reduced in half, what will
be the changed force of attraction between them?
O A.
B.
C.
D.
O E.
2.5 newtons
10 newtons
15 newtons
20 newtons
25 newtons
© 2023 Edmentum. All rights reserved.
Reset
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The distance between the two objects is reduced in half, that will be the changed force of attraction between them is option (E) 2.5 newtons.

According to Newton's law of universal gravitation, the force of attraction between two objects is inversely proportional to the square of the distance between them.

If the distance between Object A and Object B is reduced to half, the force of attraction will increase by a factor of 2² = 4. This means that the new force of attraction will be one-fourth of the original force.

Given that the original force of attraction is 5 newtons, the changed force of attraction between Object A and Object B, when the distance is reduced in half, will be 5 newtons / 4 = 1.25 newtons, which can be rounded to 2.5 newtons.The correct answer is option e.

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A sample of vegetable oil with density 913 kg/m3 is found to have a mass of 0.0365 kg. Find the volume of this sample

Answers

The volume of the vegetable oil is  0.00003998 m³.

The density of vegetable oil,

ρ = 913 kg/m³

The mass of vegetable oil,

m = 0.0365 kg

To find: The volume of the vegetable oil, V Solution: The density of any substance is defined as the mass of the substance per unit volume.

The formula for density is:

ρ = m/V

where, ρ is the density of the substancem is the mass of the substance V is the volume of the substance We can rearrange the above formula to find the volume of the substance:

V = m/ρSubstituting the given values of mass and density in the above formula,

We get:

V = 0.0365 kg / 913 kg/m³ = 0.00003998 m³ (approx)

Therefore, the volume of the vegetable oil is approximately 0.00003998 m³.

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The rainbow of visible colors in the electromagnetic spectrum varies continuously from the longest wavelengths (the reddest colors) to the shortest wavelengths (the deepest violet colors) our eyes can detect. Wavelengths near 655 nm are perceived as red. Those near 515 nm are green and those near 475 nm are blue. Calculate the frequency of light (in Hz) with a wavelength of 655 nm, 515 nm, and 475 nm.
HINT
(a)
655 nm
Hz
(b)
515 nm
Hz
(c)
475 nm
Hz

Answers

Answer:

The frequency of light can be calculated using the formula:

`c = λv`

Where `c` is the speed of light in a vacuum, `λ` is the wavelength of light, and `v` is the frequency of light.

The speed of light in a vacuum is `3.00 × 10^8 m/s`.

To convert the wavelength from nanometers to meters, we need to divide by `1 × 10^9`.

Thus, the frequency of light with a wavelength of 655 nm is:

`v = c/λ`

`v = (3.00 × 10^8 m/s)/(655 × 10^-9 m)`

`v = 4.58 × 10^14 Hz`

Therefore, the frequency of light with a wavelength of 655 nm is `4.58 × 10^14 Hz`.

Similarly, the frequency of light with a wavelength of 515 nm is:

`v = c/λ`

`v = (3.00 × 10^8 m/s)/(515 × 10^-9 m)`

`v = 5.83 × 10^14 Hz`

Therefore, the frequency of light with a wavelength of 515 nm is `5.83 × 10^14 Hz`.

Finally, the frequency of light with a wavelength of 475 nm is:

`v = c/λ`

`v = (3.00 × 10^8 m/s)/(475 × 10^-9 m)`

`v = 6.32 × 10^14 Hz`

Therefore, the frequency of light with a wavelength of 475 nm is `6.32 × 10^14 Hz`.

So, the frequency of light with a wavelength of 655 nm is `4.58 × 10^14 Hz`, the frequency of light with a wavelength of 515 nm is `5.83 × 10^14 Hz` and the frequency of light with a wavelength of 475 nm is `6.32 × 10^14 Hz`.

The boiling point of nitrogen is -196°C. What is that on the absolute temperature scale?

Answers

Answer:

The boiling point of nitrogen on the absolute temperature scale is 77.15 K

Explanation:

Temperature in Kelvin(Absolute temperature) = Temperature in Celcius + 273.15.

The rainbow of visible colors in the electromagnetic spectrum varies continuously from the longest wavelengths (the reddest colors) to the shortest wavelengths (the deepest violet colors) our eyes can detect. Wavelengths near 655 nm are perceived as red. Those near 515 nm are green and those near 475 nm are blue. Calculate the frequency of light (in Hz) with a wavelength of 655 nm, 515 nm, and 475 nm.

Answers

The frequency of light with a wavelength of 655 nm is[tex]4.57 x 10^14 Hz[/tex] and 515 nm is [tex]5.82 x 10^14[/tex] Hz and  475 nm is[tex]6.31 x 10^14 Hz[/tex]

The equation that links the speed of light to wavelength and frequency is

c = λν

Where, c = speed of lightλ = wavelengthν = frequency c is a constant of 2.998 x 10^8 m/s.

Calculating the frequency of light with a wavelength of

655 nm:λ = 655 nm = [tex]6.55 x 10^-7m[/tex]

Using the above equation, we get

c = λνν = c/λ = [tex](2.998 x 10^8 m/s)/(6.55 x 10^-7m)ν = 4.57 x 10^14 Hz[/tex]

Therefore, the frequency of light with a wavelength of 655 nm is 4.57 x [tex]10^14 Hz.[/tex]

Calculating the frequency of light with a wavelength of 515 nm:λ = 515 nm = [tex]5.15 x 10^-7m[/tex]

Using the above equation, we get

c = λνν = c/λ =[tex](2.998 x 10^8 m/s)/(5.15 x 10^-7m)ν = 5.82 x 10^14 Hz[/tex]

Therefore, the frequency of light with a wavelength of 515 nm is 5.82 x [tex]10^14 Hz[/tex].

Calculating the frequency of light with a wavelength of 475 nm:λ = 475 nm = [tex]4.75 x 10^-7[/tex]m Using the above equation, we get

c = λνν = c/λ = [tex](2.998 x 10^8 m/s)/(4.75 x 10^-7m)ν = 6.31 x 10^14 Hz[/tex]

Therefore, the frequency of light with a wavelength of 475 nm is 6.31 x [tex]10^14 Hz.[/tex]

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Part 3 Waves on a string-with a loose end The reflected
wave interferes with the original wave and creates standing wave composed of
nodes and antinodes if the frequency is just right: Instead of a node an antinode
will always exist at the loose end: (This happens because the phase of the wave
is not inverted upon reflection from loose end and therefore always constructively
interfere at that position:) Draw and measure the frequency of the 1st harmonic
(node near driver end followed by an antinode on loose end) Settings: amplitude:
0.05 cm tension: high damping: none turn on: Loose End What fraction of a
wavelength is this? Hz Click Restart' to observe the standing wave. 2. Predict the
frequencies of several higher harmonics: Use the wave simulator to test each of
your calculated harmonics Draw and label the standing waves for each of the
harmonics you discovered: Divide each higher harmonic by the first harmonic:
Are the higher harmonics even-number or odd-number multiples of the first
harmonic?

Answers

The first harmonic of the standing wave on a string with a loose end represents half a wavelength.

The fraction of a wavelength represented by the first harmonic is 1/2.

The higher harmonics of a standing wave on a string with a loose end are odd-number multiples of the first harmonic.

1. The first harmonic of a standing wave on a string with a loose end occurs when there is a node near the driver end and an antinode at the loose end. To measure the frequency of the first harmonic, we need to determine the fraction of a wavelength represented by this standing wave.

The first harmonic of the standing wave on a string with a loose end represents half a wavelength.

The first harmonic of a standing wave on a string with a loose end consists of a node near the driver end and an antinode at the loose end. This configuration creates the simplest standing wave pattern.

In a standing wave, a node is a point where the amplitude of the wave is always zero, representing a point of minimum displacement. An antinode, on the other hand, is a point of maximum displacement, where the amplitude is at its highest.

Since the loose end does not invert the phase of the wave upon reflection, the reflected wave and the original wave constructively interfere at the loose end, resulting in an antinode.

In the first harmonic, there is exactly half a wavelength between the node near the driver end and the antinode at the loose end.

Therefore, the fraction of a wavelength represented by the first harmonic is 1/2.

2. To predict the frequencies of higher harmonics, we can use the relationship that the frequency of each harmonic is a multiple of the frequency of the first harmonic. The higher harmonics can be calculated as follows:

Second Harmonic: The second harmonic consists of two nodes and one additional antinode compared to the first harmonic. The fraction of a wavelength for the second harmonic is 1/2 * 2 = 1. Thus, the second harmonic has a frequency that is twice that of the first harmonic.

Third Harmonic: The third harmonic consists of three nodes and two additional antinodes compared to the first harmonic. The fraction of a wavelength for the third harmonic is 1/2 * 3 = 1.5. Thus, the third harmonic has a frequency that is three times that of the first harmonic.

Fourth Harmonic: The fourth harmonic consists of four nodes and three additional antinodes compared to the first harmonic. The fraction of a wavelength for the fourth harmonic is 1/2 * 4 = 2. Thus, the fourth harmonic has a frequency that is four times that of the first harmonic.

In general, the higher harmonics of a standing wave on a string with a loose end are odd-number multiples of the first harmonic.

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Select the correct answer.
A car traveling south is 200 kilometers from its starting point after 2 hours. What is the average velocity of the car?
O A.
B.
100 kilometers/hour south
200 kilometers/hour
200 kilometers/hour north
O C.
O D. 100 kilometers/hour

Answers

The average velocity of the car is 100 kilometers/hour south. This means that, on average, the car is traveling 100 kilometers per hour in the south direction relative to its starting point.

To determine the average velocity of the car, we need to calculate the displacement and divide it by the time taken. Velocity is defined as the rate of change of displacement with respect to time.

In this case, the car is traveling south, and its displacement is 200 kilometers from its starting point after 2 hours.

The average velocity is given by the formula:

Average velocity = Displacement / Time

The displacement is 200 kilometers south, and the time is 2 hours. Therefore, we have:

Average velocity = 200 kilometers south / 2 hours

Simplifying the calculation:

Average velocity = 100 kilometers/hour south

Hence, the correct answer is B. 100 kilometers/hour south. This indicates that the car's average velocity is 100 kilometers per hour towards the south direction.

It's important to note that velocity is a vector quantity and includes both magnitude (speed) and direction. In this case, the direction is specified as south, which indicates that the car is moving towards the south relative to its starting point.

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