All categories

3 years ago

What’s the surface area of a cylinder with a height of 6yd and width of 4yd

1 answer:

5 0

Answer: 100.53096 (round if you need to)

Steps:
Surface area of cylinder formula = 2πrh + 2 πr^2

2πrh + 2 πr^2
2(π)(2)(6) + 2(π)(2^2)
2(π)(12) + 2(π)(4)
= 100.53096

You might be interested in

Which number from the set {0,2,3,4} makes this inequality true?

sdas [7]

Answer:

Step-by-step explanation:

6x - 10 > 2x + 2

4x - 10 > 2

4x > 12

x > 3

7 0

3 years ago

(assume no variable equals 0) PLEASE HELP

saw5 [17]

$(3n^2 + 1) + (8n^2 - 8)\\= 3n^2 + 1 + 8n^2 - 8\\= 11n^2 - 7$

5 0

3 years ago

This question, please answer it

kobusy [5.1K]

I don’t really see a question-

5 0

3 years ago

Read 2 more answers

Y= - x+2 Y=3x-4<br> Estimate the solution to the system of equations

IrinaVladis [17]

Answer:

$x=\frac{3}{2}\\ y=\frac{1}{2}$

Step-by-step explanation:

$y=-x+2\\y=3x-4$

Let's solve one of them for x.

$y=3x-4$

Add 4

$y+4=3x$

Divide by 3.

$x=\frac{y+4}{3}$

Now, plug the value of y in the formula, or you can plug the value of x in the other equation. I'll take this one.

$x=\frac{y+4}{3}$

$x=\frac{(-x+2)+4}{3}$

$x=\frac{-x+2+4}{3}$

$x=\frac{-x+6}{3}$

Multiply by 3 to get rid of the denominator.

$3x=-x+6$

add x

$3x+x=6$

Combine like terms;

$4x=6$

Divide by 4.

$x=\frac{6}{4}$

Simplify.

$x=\frac{3}{2}$

Now that you found the value of x, replace it in any of the equations to find y.

$y=-x+2\\y=-(\frac{3}{2})+2\\ y=-\frac{3}{2}+2$

$y=\frac{-3+2*2}{2} \\y=\frac{-3+4}{2} \\y=\frac{1}{2}$

Proof:

$y=3x-4\\\frac{1}{2}=3(\frac{3}{2})-4\\\frac{1}{2}=(\frac{9}{2})-4$

$\frac{1}{2}=\frac{9-4*2}{2}$

$\frac{1}{2}=\frac{9-8}{2}$

$\frac{1}{2}=\frac{1}{2}$

8 0

3 years ago

I need help for this, thanks

gulaghasi [49]

(a) Yes all six trig functions exist for this point in quadrant III. The only time you'll run into problems is when either x = 0 or y = 0, due to division by zero errors. For instance, if x = 0, then tan(t) = sin(t)/cos(t) will have cos(t) = 0, as x = cos(t). you cannot have zero in the denominator. Since neither coordinate is zero, we don't have such problems.

---------------------------------------------------------------------------------------

(b) The following functions are positive in quadrant III:

tangent, cotangent

The following functions are negative in quadrant III

cosine, sine, secant, cosecant

A short explanation is that x = cos(t) and y = sin(t). The x and y coordinates are negative in quadrant III, so both sine and cosine are negative. Their reciprocal functions secant and cosecant are negative here as well. Combining sine and cosine to get tan = sin/cos, we see that the negatives cancel which is why tangent is positive here. Cotangent is also positive for similar reasons.

5 0

3 years ago