Are the triangles similar? If so, what is the scale factor which converts RPQ to VAQ

A) No.
B) Yes; 1/2
C) Yes; 2/3
D) Yes; 2/1

Answer:

54 unit sq

Step-by-step explanation:

Answer:

54 i think

Step-by-step explanation:

Answer:

I think its A

Step-by-step explanation:

because it's a tighter angle than 29,46,and 134 degrees i might not be right but i hope it is :)

Answer:

118 yards

Step-by-step explanation:

Tamia is getting balloons for her father's birthday party

She wants each balloon to be 6 feet long

Tamia wants to get 59 balloons

Therefore the yards of string needed can be calculated as follows

= 59×6

= 354 feet to yards

= 118 yards

Hence the number of yards needed is 118 yards

Kevin knit 9 scarves, and each scarf was 49.2 inches long.

Converting this length to feet, we find that each scarf is 4.1 feet long.

Let's use the variable "L" to represent the length of each scarf in inches. Since Kevin knit 9 scarves, we can write the equation for the total length of all scarves as:

[tex]\(L_{\text{total}} = L \times 9\)[/tex]

We are given that the total length of all scarves is 442.8 inches. So, we can write:

[tex]\(442.8 \text{ inches} = L \times 9\)[/tex]

To find the length of each scarf, we need to isolate "L" on one side of the equation. To do that, we can perform the following steps:

Divide both sides of the equation by 9 to get:

[tex]\(\frac{{442.8 \text{ inches}}}{{9}} = \frac{{L \times 9}}{{9}}\)[/tex]

Simplifying, we get:

[tex]\(49.2 \text{ inches} = L\)[/tex]

Now, we have found the length of each scarf in inches (49.2 inches). To convert this length to feet, we know that 1 foot is equal to 12 inches. So, to find the length in feet, we divide the length in inches by 12:

[tex]\(L_{\text{feet}} = \frac{{L_{\text{inches}}}}{{12}}\)[/tex]

Substituting the value of [tex]\(L_{\text{inches}}\)[/tex] with 49.2 inches:

[tex]\(L_{\text{feet}} = \frac{{49.2 \text{ inches}}}{{12}}\)[/tex]

Now, let's perform the division:

[tex]\(L_{\text{feet}} = 4.1 \text{ feet}\)[/tex]

So, each scarf is 4.1 feet long.

Kevin knit 9 scarves, and each scarf was 49.2 inches long.

Converting this length to feet, we find that each scarf is 4.1 feet long.

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ig hanganaahajssislaalalla

Answer:

(a) K = 2T

K + 6 = 4T

(b) 2T + 6 = 4T

6 = 4T - 2T

6 = 2T

6/2 = 2T/2

T = 3

When T = 3, K = 6

Answer:

Height of building = 86.9 ft

Step-by-step explanation:

I've drawn a triangle image to depict this question and I have attached it.

From the attached image, we can see that ABC is the height of the building.

We can see the distance from your eye to the ground indicated and also the distance from the horizontal distance from the building to you indicated.

We can use pythagoras theorem to find AB.

AB/57 = tan 55°

AB = 57 × 1.4281

AB = 81.4 ft

Now, total height of the building is AB + BC.

BC = 5.5 ft since B is parallel to your eye and height from there to the ground is 5.5 ft.

Thus;. Height of building = 81.4 + 5.5 = 86.9 ft

Answer:

Here, I graphed this especially for you.

Check attachment.

Answer:

The answer is 7

Step-by-step explanation:

Q11-4=7

Hope this helped!!!

Answer:

652

Step-by-step explanation:

Answer: 60

Step-by-step explanation:

6n2

n=5

6(5)2

6×10 = 60

Answer:

b. (6, 4)

Step-by-step explanation:

i hope this helps :)

Answer:

B. (6,4)

Step-by-step explanation:

The point is inside the triangle.

Answer:

x=5/3 or x=-4

Step-by-step explanation:

Answer:

hi

Step-by-step explanation:

3(x+4)=-18

3 × X + 3× 4=-18

3x+12=-18

3x=-18-12

3x= -30

x= -30/3

x= -10

hope it helps

have a nice day

Answer:

Letter A is the answer pls

9514 1404 393

Answer:

  [tex]x=\dfrac{4\sqrt{3}}{3};\ y=\dfrac{8\sqrt{3}}{3}[/tex]

Step-by-step explanation:

The side ratios of a 30°-60°-90° triangle are ...

  1 : √3 : 2

In your triangle, they are ...

  x : 4 : y

In order to get these values, we need to multiply the reduced ratio by 4/√3 = (4√3)/3. Then we have ...

  x : 4 : y = (4√3)/3 : 4 : (8√3)/3

and we can see that ...

Answer:

a. The distance the train travelled in the first hour is approximately 28.3 miles

b. The location of the train at 5:00 p.m. is 53 miles east, and 46 miles west

c. The location of the train at any given time by the function, f(t) = (-8 + 24·t, -10 + 15·t)

d. The train does not collide with the cyclist when the bike goes over the train tracks

Step-by-step explanation:

a. The given information on the train's motion are;

The location south the train is spotted = 10 miles south and 8 miles west

The time the observer spotted the train = 2:00 pm

The location the train is spotted at 3:00 p.m. = 5 miles north and 16 miles east

Therefore, the difference between the two times the train was spotted, t = 3:00 p.m. - 2:00 p.m. = 1 hour

Making use of the coordinate plane for the two locations the train was spotted, we have;

The initial location of the train = (-10, -8)

The final location of the train = (5, 16)

Therefore the distance the train travelled in the first hour is given by the formula for finding the distance, 'd', between two points, (x₁, y₁) and (x₂, y₂) as follows;

[tex]d = \sqrt{\left (x_{2}-x_{1} \right )^{2}+\left (y_{2}-y_{1} \right )^{2}}[/tex]

Therefore;

[tex]d = \sqrt{\left (5-(-10) \right )^{2}+\left (16-(-8) \right )^{2}} = 3 \cdot\sqrt{89}[/tex]

The distance the train travelled in the first hour, d = 3·√89 ≈ 28.3 miles

b. The speed of the train, v = (Distance travelled by the train)/Time

∴ v ≈ 28.3 miles/(1 hour) = 28.3 miles per hour

The speed of the train in the first hour, v ≈ 28.3 mph

The direction of the train, θ, is given by the arctangent of the slope, 'm', of the path of the train;

[tex]\therefore The \ slope \ of \ the \ path \ of \ the \ train, \, m =tan(\theta) = \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]

∴ m = tan(θ) = (5 - (-10))/(16 - (-8)) = 0.625

c. Distance = Velocity × Time

At 5:00 p.m., we have;

The time difference, Δt = 5:00 p.m. - 3:00 p.m. = 2 hours

The distance, d₁ = (28.3 mph × 2 hours = 56.6 miles

Using trigonometry, we have the horizonal distance travelled, 'Δx', in the 2 hours is given as follows;

Δx = d₁ × cos(θ)

∴ Δx = 56.6 × cos(arctan(0.625)) ≈ 48

The increase in the horizontal position of the train, relative to the point (5, 16), Δx ≈ 48 miles

The vertical distance increase in the two hours, Δy is given as follows;

Δy = 56.6 × sin(arctan(0.625)) ≈ 30

The increase in the vertical position of the train, relative to the point (5, 16), Δy ≈ 30 miles miles

Therefore; the location of the train at 5:00 p.m. = ((5 + 48), (16 + 30)) = (53, 46)

The location of the train at 5:00 p.m. = 53 miles east, and 46 miles west

c. The function, 'f', that would give the train's position at time-t is given as follows;

The P = f(28.3·t, θ)

Where;

28.3·t = √(x² + y²)

θ = arctan(y/x)

Parametric equations

y - 5 = 0.625·(x - 16)

∴ y = 0.625·x - 10 + 5

The equation of the train's track is therefore, presented as follows;

y = 0.625·x - 5

d = 28.3·t

The y-component of the velocity, [tex]v_y[/tex] = 3*√89 mph × sin(arctan(0.625)) = 15 mph

Therefore, we have;

y = -10 + 15·t

The x-component of the velocity, vₓ = 3*√89 mph × cos(arctan(0.625)) = 24 mph

Therefore, we have;

x = -8 + 24·t

The location of the train at any given time, 't', f(t) = (-8 + 24·t, -10 + 15·t)

d. The speed of the cyclist next to the observer at 2:00 p.m., v = 10 mph

The distance of the cyclist from the track = The x-intercept = 5/0.625 = 8

The distance of the cyclist from the track = 8 miles

The time it would take the cyclist to react the track, t = 8 miles/10 mph = 0.8 hours

The location of the train in 0.8 hours, is f(0.8) = (-8 + 24×0.8, -10 + 15×0.8)

∴ f(0.8) = (11.2, 2)

At the time the cyclist is at the track along the east-west axis, at the point (8, 0), the train is at the point (11.2, 2) therefore, the train does not collide with the cyclist when the bike goes over the train tracks.

If the cyclist is travelling at 10 mph the train does not collide with the cyclist when the bike goes over the train tracks.

The location south the train is spotted = 10 miles south and 8 miles west

The time the observer spotted the train = 2:00 pm

The location of the train is spotted at 3:00 p.m. = 5 miles north and 16 miles east

Thus, the difference between the two times the train was spotted,

t = 3:00 p.m. - 2:00 p.m. = 1 hour

The distance is the product of velocity and time.

Distance = Velocity × Time

The initial location of the train = (-10, -8)

The final location of the train = (5, 16)

Therefore the distance the train travelled in the first hour

[tex]d = \sqrt{(x_{2} -x_{1} )^{2} +(y_{2} -y_{1} )^{2} } \\d = \sqrt{(5 -(-10) )^{2} +(16-(-8) )^{2} }[/tex]

d = 3√89

The distance the train travelled in the first hour, d = 28.3 miles

b. The speed of the train, v = (Distance travelled by train)/Time

∴ v ≈ 28.3 miles/(1 hour) = 28.3 miles per hour

The speed of the train in the first hour, v ≈ 28.3 mph

The direction of the train, θ, is given by the arc tangent of the slope, 'm', of the path of the train;

∴ m = tan(θ) = (5 - (-10))/(16 - (-8))

m = 0.625

c. Distance = Velocity × Time

At 5:00 p.m., we have;

The time difference, Δt = 5:00 p.m. - 3:00 p.m. = 2 hours

The distance, d₁ = (28.3 mph × 2 hours = 56.6 miles

Using trigonometry, we have the horizontal distance travelled, 'Δx', in the 2 hours is given as follows;

Δx = d₁ × cos(θ)

∴ Δx = 56.6 × cos(arctan(0.625)) ≈ 48

The increase in the horizontal position of the train, relative to the point (5, 16), Δx ≈ 48 miles

The vertical distance increase in the two hours, Δy is given as follows;

Δy = 56.6 × sin(arctan(0.625)) ≈ 30

The increase in the vertical position of the train, relative to the point (5, 16), Δy ≈ 30 miles miles

Therefore; the location of the train at 5:00 p.m. = ((5 + 48), (16 + 30)) = (53, 46)

The location of the train at 5:00 p.m. = 53 miles east, and 46 miles west

c. The function, 'f', that would give the train's position at time-t is given as follows;

The P = f(28.3·t, θ)

28.3·t = √(x² + y²)

θ = arctan(y/x)

Parametric equations

y - 5 = 0.625·(x - 16)

∴ y = 0.625·x - 10 + 5

The equation of the train's track is, therefore, presented as follows;

y = 0.625·x - 5

d = 28.3·t

The y-component of the velocity,

= 3 × √89 mph × sin(arctan(0.625)) = 15 mph

Therefore, we have

y = -10 + 15·t

The x-component of the velocity,

vₓ = 3 × √89 mph × cos(arctan(0.625))

= 24 mph

Therefore, we have;

x = -8 + 24·t

The location of the train at any given time, 't',

f(t) = (-8 + 24·t, -10 + 15·t)

d. The speed of the cyclist next to the observer at 2:00 p.m., v = 10 mph

The distance of the cyclist from the track = The x-intercept

                                                                   = 5/0.625 = 8

The distance of the cyclist from the track = 8 miles

The time it would take the cyclist to reach the track, t = 8 miles/10 mph = 0.8 hours

The location of the train in 0.8 hours, is f(0.8)

= (-8 + 24×0.8, -10 + 15×0.8)

∴ f(0.8) = (11.2, 2)

therefore, If the cyclist is travelling at 10 mph the train does not collide with the cyclist when the bike goes over the train tracks.

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Answer:

A. -25

Step-by-step explanation:

sana nakatulong :-)

Answer:

yeah same for me but with ice cream

Answer:

See Explanation

Step-by-step explanation:

The question is incomplete, as the required clocks are not given.

To answer this question, I will give an illustration of how to calculate the difference between times.

Note that: I assume you can read the time on a clock

Assume that:

[tex]Left\ clock = 2:15pm[/tex]

[tex]Right\ clock = 3:05pm[/tex]

The difference is calculated as:

[tex]Difference = Right\ clock - Left\ clock[/tex]

[tex]Difference = 3:05pm - 2:15pm[/tex]

From 2:15pm to 3:05pm is 50 minutes.

So:

[tex]Difference = 50\ minutes[/tex]

Another assumption:

[tex]Left\ clock = 2:15pm[/tex]

[tex]Right\ clock = 3:32pm[/tex]

The difference is calculated as:

[tex]Difference = Right\ clock - Left\ clock[/tex]

[tex]Difference = 3:32pm - 2:15pm[/tex]

For ease of calculation, we split the time to 2.

(1) From 2:15pm to 3:15pm

(2) From 3:15pm to 3:32pm

(1) From 2:15pm to 3:15pm: This is 1 hour

(2) From 3:15pm to 3:32pm: This is 17 minutes (i.e. 32 - 15)

So, the difference is:

[tex]Difference = 1hr + 17min[/tex]

[tex]Difference = 1\ hour\ 17minutes[/tex]

Answer:

96

Step-by-step explanation:

12.50 + 18.50 + 20 + 20 + 13 + 12 = 96

Answer: B

Step-by-step explanation: Hope this help :D

Answer:

$72.80

Step-by-step explanation:

1.) Find 60% of 45.50

0.60 * 45.50 = 27.3

2.) Add this to the cost

27.30 + 45.50 = 72.80

Answer:

volume = 10,082.5 in³

Step-by-step explanation:

radius = 26/2 = 13

volume = πr²L = (3.14)(13²)(19) = 10,082.54 in³

Answer: 10.2

Step-by-step explanation:

b = 15² - 11²

= 225 - 121

= 104

= √104

= 10.2

Answer:

30 ft

Step-by-step explanation:

[tex]C=\frac{n^2}{5} +175\\\\355=\frac{n^2}{5} +175\\\\180=\frac{n^2}{5}\\\\n^2=900\\\sqrt{n^2} =\sqrt{900} \\n=30[/tex]

Answer:

khing chong

Step-by-step explanation:

Answer: (6,-4) music

Step-by-step explanation:

Answer:

$480

Step-by-step explanation:

40 × 12 = 480