Multiply the following binomials (X+5) (X-5)

Evaluate the line integral for x^2yds where c is the top hal fo the circle x62 _y^2 = 9

natulia [17]

Parameterize $C$ by

$\mathbf r(t)=\langle x(t),y(t)\rangle=\langle3\cos t,3\sin t\rangle$

where $0\le t\le\pi$ . Then the line integral is

$\displaystyle\int_Cx^2y\,\mathrm dS=\int_{t=0}^{t=\pi}x(t)^2y(t)\left\|\frac{\mathrm d\mathbf r(t)}{\mathrm dt}\right\|\,\mathrm dt$
$=\displaystyle\int_{t=0}^{t=\pi}(3\cos t)^2(3\sin t)\sqrt{(-3\sin t)^2+(3\cos t)^2}\,\mathrm dt$
$=\displaystyle3^4\int_{t=0}^{t=\pi}\cos^2t\sin t\,\mathrm dt$

Take $u=\cos t$ , then

$=\displaystyle-3^4\int_{u=1}^{u=-1}u^2\,\mathrm du$
$=\displaystyle3^4\int_{u=-1}^{u=1}u^2\,\mathrm du$
$=\displaystyle2\times3^4\int_{u=0}^{u=1}u^2\,\mathrm du$
$=54$

6 0

3 years ago

if alpha and beta are the zeros of the quadratic polynomial FX = x square - 5 x + 4 find the value of one by alpha + 1 by beta-2

Kryger [21]

Answer:

suggestions use math scanner or mathaway or algebra calculator they will provide the steps and answer

5 0

3 years ago

If a cat with short brown hair (llBb) is crossed with a cat with long black hair (Llbb), what is the likelihood that there will

Advocard [28]

Bb = brown hair phenotype

bb = black hair phenotype

we see that B = brown hair is the dominant gene while b = black hair is the recessive gene

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ll = short hair, this is a lowercase L

Ll = long hair

In this case, l = short hair is the recessive gene while L = long hair is the dominant gene.

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Check out the diagram below to see the dihybrid crosses for each of the hair color and hair length genotypes. From there, I show all of the genotypes possible for any offspring. Those genotypes are

LlBb = long brown hair (comes up 4 times)
Llbb = long black hair (comes up 4 times)
llBb = short brown hair (comes up 4 times)
llbb = short black hair (comes up 4 times)

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If we are talking about 1 offspring, then it is not possible for him/her to have both "long black hair AND short brown hair" as the instructions state. So let's consider two offspring instead.

The probability of having long black hair is 4/16 = 1/4 as there are 4 cases of Llbb out of 16 cases total. Similarly, the probability of getting llBb is 4/16 = 1/4 as well, for pretty much the same reasons.

The probability of having both events occur is (1/4)*(1/4) = 1/16

The multiplication of probabilities is valid because the two events are independent.

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<h3>Final Answer: A) 1/16</h3>