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3 years ago

PLEASE HELP ME ITS DUEEE TODAYY!!!!!!!!!

Mathematics

1 answer:

vredina [299]3 years ago

7 0

height =h=16inch

Answer:

volume of cuboid =l×b×h

3840in³=20×12×h

h=3840/240=16inch

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If a brick balances evenly with three quarters of a pound and three quarters of a brick what is the weight of a whole brick?

frez [133]

Answer:

three quarters of a pound and thee quarters of a brick?

Step-by-step explanation:

You didn't state any answer choices

6 0

3 years ago

Please help thank you! The second question has 2 answers!

Akimi4 [234]

1 is C since the points are not equidistant from the edges of the angle

2: is A and E

4 0

4 years ago

If y varies directly with x and y = 15 when x = 3, find the equation that relates x and y

Zielflug [23.3K]

Answer:

y = 5x

Step-by-step explanation:

The equation for direct variation is y = kx, where k is the factor of variation. So we know that y and x must fit into this formula.

They've given us a y and x value to solve for k.

y = kx

15 = 3k

15/3 = k

k = 5

Plug k back into our equation.

y = 5x

6 0

3 years ago

Read 2 more answers

Find the volume of the region between the planes x plus y plus 2 z equals 2 and 4 x plus 4 y plus z equals 8 in the first octant

Alex787 [66]

Find the intercepts for both planes.

Plane 1, x + y + 2z = 2:

$y=z=0\implies x=2\implies (2,0,0)$

$x=z=0\implies y=2\implies(0,2,0)$

$x=y=0\implies 2z=2\implies z=1\implies(0,0,1)$

Plane 2, 4x + 4y + z = 8:

$y=z=0\implies4x=8\implies x=2\implies(2,0,0)$

$x=z=0\implies4y=8\impliesy=2\implies(0,2,0)$

$x=y=0\implies z=8\implies(0,0,8)$

Both planes share the same x- and y-intercepts, but the second plane's z-intercept is higher, so Plane 2 acts as the roof of the bounded region.

Meanwhile, in the (x, y)-plane where z = 0, we see the bounded region projects down to the triangle in the first quadrant with legs x = 0, y = 0, and x + y = 2, or y = 2 - x.

So the volume of the region is

$\displaystyle\int_0^2\int_0^{2-x}\int_{\frac{2-x-y}2}^{8-4x-4y}\mathrm dz\,\mathrm dy\,\mathrm dx=\displaystyle\int_0^2\int_0^{2-x}\left(8-4x-4y-\frac{2-x-y}2\right)\,\mathrm dy\,\mathrm dx$

$=\displaystyle\int_0^2\int_0^{2-x}\left(7-\frac72(x+y)\right)\,\mathrm dy\,\mathrm dx=\int_0^2\left(7(2-x)-\frac72x(2-x)-\frac74(2-x)^2\right)\,\mathrm dx$

$=\displaystyle\int_0^2\left(7-7x+\frac74 x^2\right)\,\mathrm dx=\boxed{\frac{14}3}$

3 0

3 years ago

I’m stuck on this problem help

Anika [276]

An easy way to think of this when you solve this, is to remember that all the angles in a triangle add up to 180 degrees. Therefore, the equation would be
(3x + 10) + (2x + 10) + 45 = 180. By the way, since the angles D and B are across from each other they are congruent. Now, let’s start solving it:
(3x + 10) + (2x + 10) + 45 = 180
(5x + 20) + 45 = 180
5x + 65 = 180
5x = 115
x = 23

8 0

4 years ago