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3 years ago

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Simplify. (5+squartrt3)2 25 + sqrtrt10 10 + 28sqrtrt3 28 + 10sqrtrt3

1 answer:

SSSSS [86.1K]3 years ago

5 0

Answer:

Step-by-step explanation:

If the 2 is an exponent, it needs to be written as ^2.

(5+√3)² = 5² + 10√3 + 3 = 28 + 10√3

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How do I do this activity, help is appreciated

Kitty [74]

This is fairly easy, all you need to do is write the ordered pairs for each graph, for these there are 5 ordered pairs for both.

In order to create the ordered pair all you need is this 'formula' of (x, y)

In the first graph 'input' would be the x value and output is the y value. So all you do is plug them into the formula.

(-10,-20)
(-5,-10)
(0,0)
(5,10)
(10,20)

Those are all the ordered pairs for graph A. Now just do the same thing for graph B.

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4 years ago

Consider the transformation T: x = \frac{56}{65}u - \frac{33}{65}v, \ \ y = \frac{33}{65}u + \frac{56}{65}v

irina1246 [14]

Answer:

Step-by-step explanation:

$T: x = \frac{56}{65}u - \frac{33}{65}v, \ \ y = \frac{33}{65}u + \frac{56}{65}v$

A)

$\frac{d(x,y)}{d(u,v)} =\left|\begin{array}{ccc}x_u&x_v\\y_u&y_v\end{array}\right|$

$=(\frac{56}{65} )^2+(\frac{33}{65} )^2\\\\=\frac{(56)^2+(33)^2}{(65)^2} \\\\=\frac{4225}{4225} \\\\=1$

B )

$S:-65 \leq u \leq 65, -65 \leq v \leq 65$

$T(65,65)=(x=\frac{56}{65} (65)-\frac{33}{65} (65),\ \ y =\frac{33}{65} (65)+\frac{56}{65} (65)\\\\=(23,89)$

$T(-65,65)=(-56-33,\ \ -33+56)\\\\=(-89,23)$

$T(-65,-65) = (-56+33,-33-56)\\\\=(-23,-89)$

$T(65,-65)=(56+33, 33-56)\\\\=(89,-23)$

C)

$\int \!\! \int_{T(S)} \ x^2 + y^2 \ {dA}$

$=\int\limits^{65}_{v=-65} \int\limits^{65}_{u=-65}(x^2+y^2)(\frac{d(x,y)}{d(u,v)} du\ \ dv$

Now

$x^2+y^2=(\frac{56}{65} u-\frac{33}{65} v)^2+(\frac{33}{65} u+\frac{56}{65} v)^2$

$[(\frac{56}{65} )^2+(\frac{33}{65}) ^2]u^2+[(\frac{33}{65} )^2+(\frac{56}{65}) ^2]v^2$

$=\frac{(65)^2}{(65)^2} u^2+\frac{(65)^2}{(65)^2} v^2=u^2+v^2$

$\int \!\! \int_{T(S)} \ x^2 + y^2 \ {dA}$

$=\int\limits^{65}_{v=-65} \int\limits^{65}_{u=-65}(u^2+v^2) du\ \ dv$

$=\int\limits^{65}_{-65}\int\limits^{65}_{-65}u^2du \ \ dv+\int\limits^{65}_{-65}\int\limits^{65}_{-65}v^2du \ \ dv$

By symmetry of the region

$=4\int\limits^{65}_0 \int\limits^{65}_0u^2 du \ \ dv + u\int\limits^{65}_0 \int\limits^{65}_0v^2 du \ \ dv$

$= 4(\frac{u^3}{3} )^{65}_{0}(v)_0^{65}+(\frac{v^3}{3} )^{65}_{0}(u)_0^{65}\\\\=4[\frac{(65)^4}{3} +\frac{(65)^4}{3} ]$

$=\frac{8}{3} (65)^4$

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3 years ago

Please help me due today easy question

Kitty [74]

Answer:

$24.00 dollars. The problem is confusing, so sorry if this is wrong.

Step-by-step explanation:

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3 years ago

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What are the x-values of the quadratic equation (x-2)^2-9=0?

skelet666 [1.2K]

Answer: x=−1 or x=5

Step-by-step explanation:

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Amaria and her Mom want to use fewer boxes and put different of brownies into the same box. How many whole boxes do they fill? W

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We need to find the numebr of brownies.

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