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3 years ago

What figure will result when the triangle is rotated around RE?

Mathematics

1 answer:

Oduvanchick [21]3 years ago

3 0

Answer:

The figure will result is cone ⇒ B

Step-by-step explanation:

When a right triangle is rotated about its vertical leg 360°, then it formed a cone its radius is the horizontal leg of the triangle and its height is the vertical lege of the triangle

In the given figure

∵ Triangle TRE is a right triangle

∴ The horizontal leg is RE

∴ The vertical leg is TR

∵ The triangle is rotated around the horizontal leg RE 360°

∴ It formed a cone its radius = TR and its height = RE

∴ The figure will result is the cone

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3 years ago

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I need help finding functions

velikii [3]

Answer:

y= 7
y= -3
y= -4
y= -8

Step-by-step explanation:

y= -5x+1

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3 years ago

show that the sum of the x- and y- intercepts of any tangent line to the curve sqrt(x) sqrt(y) = sqrt(c) is equal to c.

expeople1 [14]

The answer is √x + √y = √c 
=> 1/(2√x) + 1/(2√y) dy/dx = 0 
=> dy/dx = - √y/√x 

Let (x', y') be any point on the curve 
=> equation of the tangent at that point is 
y - y' = - (√y'/√x') (x - x') 

x-intercept of this tangent is obtained by plugging y = 0 
=> 0 - y' = - (√y'/√x') (x - x') 
=> x = √(x'y') + x' 

y-intercept of the tangent is obtained by plugging x = 0 
=> y - y' = - (√y'/√x') (0 - x') 
=> y = y' + √(x'y') 

Sum of the x and y intercepts 
= √(x'y') + x' + y' + √(x'y') 
= (√x' + √y')^2 
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3 years ago

Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 3.0 km/h due east. Runner B is initia

ArbitrLikvidat [17]

Answer:

The distance of the two runners from the flagpole is 0.0882 km

Step-by-step explanation:

We are going to use the minus sign when the runner is on the west and the plus sign when the runner is on the east.

Then, Taking into account the runner A is 6 km west and is running with a constant velocity of 3 Km/h east, the distance of the runner A from the flagpole is given by the following equation:

Xa = -6 Km + (3 Km/h)* t

Where Xa is the position of the runner A from the flagpole and t is the time in hours.

At the same way the distance of the runner B, Xb, from the flagpole is given by the following equation:

Xb = 7.4 Km - (3.8 Km/h)*t

Then, the two runners cross their path when Xa is equal to Xb, so if we solve this equation for t, we get:

Xa = Xb

-6 + (3*t) = 7.4 - (3.8*t)

(3.8*t) + (3*t) = 7.4 + 6

(6.8*t) = 13.4

t = 13.4/6.8

t = 1.9706

Therefore, at time t equal to 1.9706 hours, both runners cross their path. The distance of the two runners from the flagpole can be calculated replacing the value of t in equation for Xa or in equation for Xb as:

Xa = -6 Km + (3 Km/h)* t

Xa = -6 Km + (3 Km/h)*(1.9706 h)

Xa = -0.0882 Km

That means that both runners are 0.0882 Km west of a flagpole.

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Ganezh [65]

Answer:

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