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2 years ago

I got g=(3d-2)^2-5 Simplified : g=9d-1 Did I get it correct?

Mathematics

2 answers:

SVEN [57.7K]2 years ago

4 0

Answer:

Step-by-step explanation:

$g=(3d-2)^2-5$

$g=(3d-2)(3d-2)-5$

<em>remember to use distributive property when squaring a value with 2 or more terms.</em>

g = $9d^2-12d+4-5$

g = $9d^2-12d-1$

Alex787 [66]2 years ago

4 0

Answer:

Close but no, g=3d^2-1

Step-by-step explanation:

Since d cannot be determined, we cannot square just the three, we have to square x too so 3d^2

I hope this helps u pls give a brainliest and a thx

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2 years ago

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, <em>x</em> + <em>y</em> + 2<em>z</em> = 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, 4<em>x</em> + 4<em>y</em> + <em>z</em> = 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 <em>x</em>- and <em>y</em>-intercepts, but the second plane's <em>z</em>-intercept is higher, so Plane 2 acts as the roof of the bounded region.

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

So the volume of the region is

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

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