Ask question

Ask question

All categories

tekilochka [14]

3 years ago

14

A sector of a circle of radius is 15Cm, and angle 216 degrees is bent to form a cone. What is base radius of the cone and the ve

rtical height of the cone?

1 answer:

-Dominant- [34]3 years ago

3 0

Answer:

Step-by-step explanation:

Radius r = 15

Circumference = 2πr = 30π cm

Circumference of cone = arc length of sector

= 30π × 216°/360°

= 18π cm

Radius of base of cone = 18π/(2π) = 9 cm

Slant height of cone = radius of circle = 15 cm

Height of cone = √(15² - 9²) = 12 cm

You might be interested in

How do I create a radical equation with an extraneous solution quickly? How do I go the same for a radical equation with a non-e

Step2247 [10]

Answer:

why r u wafflin

Step-by-step explanation:

5 0

2 years ago

Brittany buys 2.55 pounds of turkey for $5.96 per pound and 3.7 pounds of

Galina-37 [17]

Answer:

Brittany needs another $3.7405.

Step-by-step explanation:

Per pound Cost of turkey = $5.96 per pound

The amount Brittany buys the turkey = 2.55 pounds

Brittany's cost for turkey = 2.55 × $5.96 = $15.198

Per pound cost for cheese = $3.35 per pound

The amount Brittany buys the cheese = 3.7 pounds

Brittany's cost for cheese = 2.55 × $3.35 = $8.5425

So,

Brittany's total cost = Turkey cost + Cheese cost

= $15.198 + $8.5425

= $23.7405

As brittany gave the clerk 20 dollars.

So, the amount she further needs will be:

$23.7405 - $20 = $3.7405

Therefore, Brittany needs another $3.7405.

8 0

3 years ago

I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur

Butoxors [25]

$\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx$

Setting $x=5\sin\theta$ , you have $\mathrm dx=5\cos\theta\,\mathrm d\theta$ . Then the integral becomes

$\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta$
$\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

Now, $\sqrt{x^2}=|x|$ in general. But since we want our substitution $x=5\sin\theta$ to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means $\theta=\sin^{-1}\dfrac x5$ , which implies that $\left|\dfrac x5\right|\le1$ , or equivalently that $|\theta|\le\dfrac\pi2$ . Over this domain, $\cos\theta\ge0$ , so $\sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta$ .

Long story short, this allows us to go from

$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

to

$\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta$

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

$\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta$

Then integrate term-by-term to get

$\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)$
$=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C$

Now undo the substitution to get the antiderivative back in terms of $x$ .

$=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C$

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

$=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C$

4 0

2 years ago

Read 2 more answers

Can I get another answer please ?

gavmur [86]

G(x) = x^2 + 2x + 6
g(x) = x^2 + 2x + 1 + 5
g(x) = (x+1)^2 + 5

so
f(x) = x^2 to g(x) = (x+1)^2 + 5
shifted to the left 1 unit and up 5 units

answer is A. left 1 unit and up 5 units

4 0

3 years ago

From 1965 to 1996 Canadian Gary Duschi made the world's longest gum wrapper chain. It was 18,721 feet long. About how many miles

jok3333 [9.3K]

It would be about 3.5 miles long so you chose 3 miles or 4 miles i would go with (4 miles)

6 0

3 years ago