All categories

3 years ago

What is the volume of the cone?

Mathematics

2 answers:

KATRIN_1 [288]3 years ago

7 0

Answer:

Step-by-step explanation:

Volume of a cone = 1/3 πr^2h

r = diameter/2

V = 1/3 × 22/7 × 22/2 × 22/2 × 22

V = 1/3 × 22/7 × 11 × 11 × 22

V = 58564/21

V = 2788.76 yd

maks197457 [2]3 years ago

4 0

Answer:

Step-by-step explanation:

FORMULA IS .

$\frac{1}{3} \times \pi {r}^{2}h$

r = 22/2=11

h=22

You might be interested in

Last free brainliest of the day <br> answer this question <br> 12x4(6+5)x2

almond37 [142]

The answer is 1056 lmk if you need the work

8 0

2 years ago

Read 2 more answers

I have hours looking for a tutor, help me please

elena-s [515]

This statement is A. True

3 0

1 year ago

Read 2 more answers

I need this answered ASAP !!!! Right answers only

otez555 [7]

This is not a function because they use the same y axis.

8 0

2 years ago

the 2008 year book at middleton middle school had 236 candid pictures of students. the 2007 yearbook had 214 candid pictures of

Nostrana [21]

Use proportions.
214/236=x/100(100 because it's a percentage.
cross multiply:
23600=214x
solve for x:
x= 11.3(rounded)
11.3% increase

4 0

3 years ago

Find the solutions to x⁴-5x²-36=0 and the x-intercepts of the graph of y=x⁴-5x²-36.

disa [49]

Answer:

The solutions $x^4-5x^2-36=0$ are $x=3,\:x=-3,\:x=2i,\:x=-2i$ and the x-intercepts of $y=x^4-5x^2-36$ are $x=3,\:x=-3$

Step-by-step explanation:

Finding the solutions to $x^4-5x^2-36$ means finding the roots, a root is where the function is equal to zero.

The x-intercept is the point at which the graph crosses the x-axis. At this point, the y-coordinate is zero.

To find the roots you need to:

Rewrite the equation with $u=x^2$ and $u^2=x^4$

$u^2-5u-36=0$

Solve by factoring

$\mathrm{Break\:the\:expression\:into\:groups}\\u^2-5u-36=\left(u^2+4u\right)+\left(-9u-36\right)$

$\mathrm{Factor\:out\:}u\mathrm{\:from\:}u^2+4u=u\left(u+4\right)$

$\mathrm{Factor\:out\:}-9\mathrm{\:from\:}-9u-36=-9\left(u+4\right)$

$u^2-5u-36=u\left(u+4\right)-9\left(u+4\right)$

$\mathrm{Factor\:out\:common\:term\:}u+4\\\left(u+4\right)\left(u-9\right)$

$u^2-5u-36=\left(u+4\right)\left(u-9\right)=0$

Using the Zero factor Theorem: if ab = 0 then a = 0 or b = 0 (or both a = 0 and b = 0)

The solutions to the quadratic equation are:

$\:u=-4,\:u=9$

Substitute back $u=x^2$ , solve for x

$x^4-5x^2-36=(u-9)(u+4)=(x^2-9)(x^2+4)$

Apply the difference of squares formula

$x^4-5x^2-36=(x^2-9)(x^2+4)=(x-3)(x+3)(x^2+4)$

$(x-3)(x+3)(x^2+4)=0$

Using the Zero factor Theorem: if ab = 0 then a = 0 or b = 0 (or both a = 0 and b = 0)

The solutions are:

$\:x=3,\:x=-3,\:x=2i,\:x=-2i$

Because two of the solutions are complex roots the only x-intercepts are $x=3,\:x=-3$

3 0

3 years ago