Which one of this is the best deal? EXPLAIN. No links or you will get reported and you will get your acc deleted

ig hanganaahajssislaalalla

Answer:

a=76 b=83

Step-by-step explanation:

for a. we divide 6688(area) by 88 to find the width bc 88*w=6688

6688/88 is 76 so a=76

Since 312 is perimeter we have to subtract 73 twice and then divide that by 2 bc 73+73+L+L=312

(312-73-73)/2=83 so b=83

Answer:

Here, I graphed this especially for you.

Check attachment.

Answer:

r=10

Step-by-step explanation:

I used a cone calculator.

Step-by-step explanation:

for 1

x = 11 using tan

y = 22 using sin

for no 2

x= 16 using cos

y= 16root 3 using sin

I believe with this u can do the others

have fun....

Answer:

r and s is (2,6) s and t is (-4,-6) r and t are (4,2)

Step-by-step explanation:

igu

Answer:

a.

Step-by-step explanation:

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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:

yeah same for me but with ice cream

Answer:

1. 23/4  ,  2. 5.555...   , 3. 5 1/2   ,    4. √28

you want to make them all into decimals so you can compare then easier

23/4 would be 5.75

√28 (just put it in a calculator)  would be 5.29

5 1/2  would be 5.5  (make  it into a improper fraction 11/2 then do 11/2 and you get 5.5)

and you have 5.555.

so...

5.75

5.29

5.5

5.55555

first is 5.75 it's bigger because of the .(75)

then 5.555 because it's .(55)

then 5.5 because it's .(50)

and the last one 5.(29)

Answer:

27 - 6 × 4 is not 84 ( Filler because I need it to be 20 characters long)

Answer:

96

Step-by-step explanation:

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

Answer:

"The interviewer for the study did not record the responses of the subjects accurately." I believe this is most pertinent. B, or the boss of the subjects, could have been an answer, but if all members associated in the study exhibit the similar level of response bias based upon superior employers, then the study remains somewhat accurate, but C, or inaccurate record of responses, would remain biased entirely.

Step-by-step explanation:

Answer:

It's C on edge.

Step-by-step explanation:

Did assignment.

Answer:

$14.00

Step-by-step explanation:

Here's why. So, if 5 pounds of bananas is $3.50 per pound. All you need to do is do 3.50 * 4 and you'll get your answer, which is $14.00 .

It's pretty easy. If you have a problem like this just multiple the price with the number that they provide you with. :D  

Answer:

4ab

Step-by-step explanation:

4(-3b)(+a)+8ab÷2a×4a

4(-3ab)+8ab÷2a×4a

-12ab+8ab÷2a×4a

-12ab+4b×4a

-12ab+16ab

16ab-12ab = 4ab

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.

Learn more about distance;

https://brainly.com/question/15172156

Answer:

the answer of the question is 25

Answer:

8a:15b:6c

a:b:c = 8:15:6

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

Answer:

$57.40

Step-by-step explanation:

Hope this helps!!

Answer:

you have the table of the 12in the photo

Step-by-step explanation:

Answer:

12×1=12

12×2=24

12×3=36

12×4=48

12×5=60

12×6=72

12×7=84

12×8=96

12×9=108

12×10=120

In your question

the table of 12 is correct

hope it is helpful to you ☺️☺️☺️☺️☺️

Answer:

n ≥ 13

≥

13

Step-by-step explanation:

Answer: the answer is B

Step-by-step explanation:

Given statement solution is :- The set of ordered pairs (-1, -3) and (5, 5) determine a line with a slope of 4/3, according to the given statement's answer.

To determine the slope of a line given two ordered pairs (x1, y1) and (x2, y2), we use the formula:

slope is equal to y2 - y1 / x2 - x1.

Let's calculate the slopes for each set of ordered pairs:

A. (5, -3) (2, 3)

slope = (3 - (-3)) / (2 - 5)

= 6 / (-3)

= -2

B. (3, 10) (-3, -2)

slope = (-2 - 10) / (-3 - 3)

= -12 / (-6)

= 2

C. (3, -5) (-3, -7)

slope = (-7 - (-5)) / (-3 - 3)

= (-2) / (-6)

= 1/3

D. (-1, -3) (5, 5)

slope = (5 - (-3)) / (5 - (-1))

= 8 / 6

= 4/3

Option D's set of ordered pairs, (-1, -3) and (5, 5), results in a line with a slope of 4/3.

For such more questions on Line Slopes: Identify Correctly

https://brainly.com/question/3493733

#SPJ8

Answer:

the answer is in the photo above

Answer:

Input:

(6 + a)×5 = 5 (6 + 8)

Result:

5 (a + 6) = 70

Answer:

15x

Step-by-step explanation:

perimeter is the sum of all sides

3x + 5x + 7x = 15x

Answer: 38/7

Step-by-step explanation:

5 3/7

5×7+3

35+3 = 38

38/7

Answer:

Step-by-step explanation:

1) multiply the three dimension

V = 7 x 2 x 4 = 56 cm^3

2)  find the base area

base area = (9 x 3)/2 = 13,5 km^2

multiply the base area for the height

V = 13,5 x 5 = 67,5 km^3

3) find the trapezoid area

trapezoid area = [(base 1 + base 2) x h]/2 = [(4+6) x 1.7]/2 = 8.5 cm^2

mutiply the trapezoid area for the height that the priam would assume if the base will touch the ground

V = 8.5 x 4 = 34 cm^3

4) find the base area

base area = radius^2 x pi = 2^2 x pi = 12,57 yd^2

multiply the base area for the height

V= 12.57 x 6 = 75.40 yd^3

5) find the base area

base area = 9^2 x pi = 81 x pi = 254.47 ft^2

multiply the base area for the height

V = 254.47 x 5 = 1272.35 ft^3

6)  multiply the three dimensions

V = 12 x 11 x 9 = 1188 yd^3

7) find the trapezoid area

trapezoid area = [(12+6) x 6.3]/2 = [20 x 6,3]/2 = 63 in^2

mutiply the trapezoid area for the height that the priam would assume if the base will touch the ground

V = 63 x 16 = 1008 in^3

8) find the base area

base area = (leg1 x leg 2)/2 = (4 x 3)/2 = 6 m^2

multiply the base area for the height

V = 6 x 9 = 54 m^3

Answer:

hope this helps you out good luck with your math