Anna wants to make 4 sets of curtains. Each set requires 5 1/8 yards of fabric. How much fabric does she need?

Mathematics

2 answers:

hammer [34]3 years ago

7 0

Answer: $20\dfrac{1}{2}\text{ Yards}$

Step-by-step explanation:

Given : Anna wants to make 4 sets of curtains

The requirement of fabric for each set = $5\dfrac{1}{8}\text{ yards}$

Here, $5\dfrac{1}{8}\text{ yards}=\dfrac{5\times8+1}{8}\text{ yards}=\dfrac{41}{8}\text{ yards}$

For 4 sets , the total requirement of fabric will be :_

$4\times\dfrac{41}{8}\text{ Yards}\\\\=\dfrac{41}{2}\text{ Yards}\\\\=20\dfrac{1}{2}\text{ Yards}$

Hence, she needs $20\dfrac{1}{2}\text{ Yards}$ fabric.

Nesterboy [21]3 years ago

4 0

20 yards
take 5 1/8 and make that into 40/8 and then you will divide 40 into 8 and it = 5 so then you do 4x5=20

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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

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

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