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PilotLPTM [1.2K]
3 years ago
5

Find the value of x in each of the given figures.

Mathematics
1 answer:
Kaylis [27]3 years ago
8 0

Answer:

Step-by-step explanation:

1) Given figure is a parallelogram

Area = 144 cm²

base  * altitude = 144

16 * x = 144

x =144/16

x = 9 cm

2)2) Trapezium

Area = \frac{a+b}{2}*h      ; {a and b are parallel sides of trapezium}

( 37 +27 /2) * x  = 480 cm²

64/2 * h = 480

32 *  h = 480

h = 480/32

h = 15 cm

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Answer: y = 2x^4

The graph represents a parabolic-looking shape so it's either a quadratic or some even degree polynomial. This rules out y = -2x^3 and y = 2x^3 as they are cubics with odd degrees

The degree of a polynomial is the largest exponent. It determines the end behavior. In this case, both ends are rising upward together. Because they are rising up in the positive y direction, this means that we cannot have y = -2x^4 as the answer. For example, if x = 2 then y = -2*x^4 = -2*2^4 = -32, but the point (x,y) = (2,-32) isn't on the blue curve. So we can rule y = -2x^4 out.

The fact that y = 2x^4 has a positive leading coefficient tells us that the endpoints point upward.  

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G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =
Svetach [21]
Presumably you should be doing this using calculus methods, namely computing the surface integral along \mathbf r(u,v).

But since \mathbf r(u,v) describes a sphere, we can simply recall that the surface area of a sphere of radius a is 4\pi a^2.

In calculus terms, we would first find an expression for the surface element, which is given by

\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv

\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k
\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j
\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k
\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u

So the area of the surface is

\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u
=-2\pi a^2(\cos\pi-\cos 0)
=-2\pi a^2(-1-1)
=4\pi a^2

as expected.
6 0
3 years ago
The height of a tree was 12 feet. after a year, the tree's height increased 25%. what is the height of the tree after a year
g100num [7]
15ft because 12x.25=3 so 3ft is 25% of the 12ft, add the 3+12=15
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3 years ago
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