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3 years ago

11

Industrial chemicals are stored in 55-gallon drums. The radius of the drum is 12 inches, and the height is 32 inches. Determine

the surface area of the drum.

1 answer:

KIM [24]3 years ago

8 0

Answer:

The surface area is 3317.52 inch^2

Step-by-step explanation:

The drum is in the shape of a cylinder, therefore..

Area of the drum

=2πrh+2πr^2

=2*π*12*32+2·π*12^2

= 3317.52 inch^2

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What is the y value of the vertex 4x^2+8x-8

Oksana_A [137]

Answer:

The y-value of the vertex is $-12$

Step-by-step explanation:

we know that

The equation of a vertical parabola into vertex form is equal to

$f(x)=a(x-h)^{2}+k$

where

(h,k) is the vertex of the parabola

In this problem we have

$f(x)=4x^{2}+8x-8$ -----> this a vertical parabola open upward

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)+8=4x^{2}+8x$

Factor the leading coefficient

$f(x)+8=4(x^{2}+2x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$f(x)+8+4=4(x^{2}+2x+1)$

$f(x)+12=4(x^{2}+2x+1)$

Rewrite as perfect squares

$f(x)+12=4(x+1)^{2}$

$f(x)=4(x+1)^{2}-12$

The vertex is the point $(-1,-12)$

The y-value of the vertex is $-12$

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3 years ago

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What is the length of arc QR? <br> A. 1/3 π<br> B. 10/3 π<br> C. 5/3 π<br> D. 20/3 π

podryga [215]

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Find the length of the curve y = integral from 1 to x of sqrt(t^3-1)

Arlecino [84]

$y=\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt$

By the fundamental theorem of calculus,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm d}{\mathrm dx}\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt=\sqrt{x^3-1}$

Now the arc length over an arbitrary interval $(a,b)$ is

$\displaystyle\int_a^b\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx=\int_a^b\sqrt{1+x^3-1}\,\mathrm dx=\int_a^bx^{3/2}\,\mathrm dx$

But before we compute the integral, first we need to make sure the integrand exists over it. $x^{3/2}$ is undefined if $x$ , so we assume $a\ge0$ and for convenience that $a$ . Then

$\displaystyle\int_a^bx^{3/2}\,\mathrm dx=\frac25x^{5/2}\bigg|_{x=a}^{x=b}=\frac25\left(b^{5/2}-a^{5/2}\right)$

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3 years ago

Complete the input-output table for the function y = 3x.<br><br> Input-Output table

Jobisdone [24]

Answer:

Y: 0, x:0

Y:1, x: 3

Y: 2, x: 6

Y: 3, x:9

Step-by-step explanation:

Plug in the x to get the y

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3 years ago

One endpoint of a line segment has coordinates represented by (x+4,1/2y). The midpoint of the line segment is (3,−2).

lesya692 [45]

Answer:

$(-x+2, -\frac{1}{2}y-4})$

Step-by-step explanation:

We know that one of the endpoints of the line segment is (x+4, 1/2y)

The midpoint of the line segment is (3, -2).

And we want to find the other coordinates in terms of x and y.

To do so, we can use the midpoint formula:

$M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Since we know that the midpoint is (3, -2), let's substitute that for M:

$(3, -2)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Let's solve for each coordinate individually:

X-Coordinate:

We have:

$3=\frac{x_1+x_2}{2}$

We know that one of the endpoints is (x+4, 1/2y). So, let's let (x+4, 1/2y) be our (x₁, y₁). Substitute x+4 for x₁. This yields:

$3=\frac{(x+4)+x_2}{2}$

Solve for our second x-coordinate x₂. Multiply both sides by 2:

$6=x+4+x_2$

Subtract 4 from both sides:

$2=x+x_2$

Subtract x from both sides. Therefore, the x-coordinate of our second point is:

$x_2=-x+2$

Y-Coordinate:

We have:

$-2=\frac{y_1+y_2}{2}$

Substitute 1/2y for y₁. This yields:

$-2=\frac{\frac{1}{2}y+y_2}{2}$

Solve for y₂. Multiply both sides by 2:

$-4=\frac{1}{2}y+y_2$

Subtract 1/2y from both sides. So:

$y_2=-\frac{1}{2}y-4$

Therefore, the other coordinate expressed in terms of x and y is:

$(-x+2, -\frac{1}{2}y-4})$

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2 years ago