The lateral areas of two similar cones are 175π and 252π respectively. What is the ratio of their radii?

Please help hurry! ty

MAXImum [283]

Answer:

....

Step-by-step explanation:

3 0

2 years ago

Simplify start-fraction 2 over start root 5 end root end-fraction

Lerok [7]

Answer:

Hope this helps

Step-by-step explanation:

Please give brainliest

7 0

2 years ago

PLZ HELP ME QUICKLY!!!!!!!!

topjm [15]

Answer:

D. 8 units

Step-by-step explanation:

I chose that answer because -4 is 4 units away from 0 then you need to get to 4 so you add 4 more units and 4 + 4 is 8 so that means it's 8 units away

4 0

3 years ago

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B) there is a line with positive slope that is tangent to both circles. find the equation of this line, and (c) determine the po

IgorC [24]

Given that there is a line with positive slope that is tangent to both circles $x^2 + y^2 = 1$ and $(x-3)^2 + y^2 = 4$ .

Let the point of tangency in the first circle be (a, b) and the point of tangency in the second circle be (c, d), then the slope of the tangent to the first circle is given by

$2x+\left.2y \frac{dy}{dx}\right|_{(a, b)} =0 \\ \\ \Rightarrow \left.\frac{dy}{dx}\right|_{(a, b)} =- \frac{x}{y} =- \frac{a}{b}$

and the slope of the second circle is given by

$2(x-3)+\left.2y \frac{dy}{dx}\right|_{(c, d)} =0 \\ \\ \Rightarrow \left.\frac{dy}{dx}\right|_{(a, b)} =- \frac{x-3}{y} =- \frac{c-3}{d}$

From the first circle equation, we have

$x^2 = 1 - y^2 \\ \\ \Rightarrow a^2 = 1 - b^2$

and from the second equation, we have

 $(x-3)^2 = 4 - y^2 \\ \\ (c-3)^2 = 4 - d^2$ 

Since the two slopes above are for the same line, then

 $\frac{a}{b}= \frac{c-3}{d} \\ \\ \Rightarrow ad=b(c-3) \\ \\ \Rightarrow a^2d^2=b^2(c-3)^2 \\ \\ \Rightarrow(1-b^2)d^2=b^2(4-d^2) \\ \\ \Rightarrow d^2-b^2d^2=4b^2-b^2d^2 \\ \\ \Rightarrow d^2=4b^2 \\ \\ \Rightarrow d=\pm2b$

Thus, we have

$\frac{a}{b}= \frac{c-3}{2b} \\ \\ \Rightarrow a= \frac{c-3}{2}$

7 0

2 years ago

Henry is standing on a bridge over a creek. If he could throw a stone straight up into the air with a velocity of 50 feet per se

Naddika [18.5K]

Henry is 20 feet above the water. You want to know whether stone can reach
60 feet above the water. So what you're really asking is: Calculate whether
the stone can reach 40 feet above Henry, and we can forget about the creek ?

Call Henry's elevation zero, and the height of the stone at any time after
the toss 'H'.

Way back among the pages in your Physics book that are clean and shiny
because they have never yet been exposed to air or sunlight, you will find
the formula for the height of an object in free-fall:

Height = H₀ + V₀t + 1/2 A t²

H₀ = the object's height when it was released
V₀ = the object's speed when it was released, negtive if downward
A = the object's acceleration, negative if downward
t = time since the object was released

In the case that involves Henry on the bridge . . .

H₀ = 0
V₀ = +50 ft/sec
A = -32 ft/sec² (acceleration due to gravity)

We want to know if the height of the object can ever be +40 feet.
We can plug all the numbers into the equation, and solve it. Since the equation
is written in terms of ' t ', any solution we get will be a 'time'. That's not what
we're looking for, but if there's any real solution, then we'll know that it's possible.

40 = 50t + 16t²

Subtract 40 from each side:

16t² + 50t - 40 = 0

Just to make the numbers more manageable, divide each side by 2 :

8t² + 25t - 20 = 0

Plug this into the quadratic formula:

t = (1/16) x (-25 plus or minus the square root of [625 - 640] )

Do you see that 'square root of -15 in there ?

The ' -15 ' is called the 'discriminant' of our quadratic equation, and
since it's negative, our equation has no real solutions ... there's no
such thing as the real square root of a negative number.

So the answer to the question is: No. The stone never reaches a height
of 40 feet above Henry, or 60 feet above the creek.

Whew!
===============================================

A slightly easier way to do it:

Henry throws the stone upward at 50 ft/sec.
The acceleration of gravity is 32 ft/sec² downward.

The stone keeps rising for (50/32) = 1.5625 second, until its upward speed
has shrunk to zero, and then it starts falling.

How high is it when it stops rising ?

Its upward speed was 50 when Henry tossed it, and zero when it stopped rising.
Its average speed on the way up was (1/2)(50 + 0) = 25 ft/sec upward.

It has that average speed for 1.5625 seconds.
How far does it climb in that time ?

H = (25 ft/sec) x (1.5625 sec) = 39.0625 feet.

That's pretty close, but not quite 40 feet above Henry.
So the answer to the question is: No.

7 0

3 years ago

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