For what value of x is the rational expression below undefined<br> 2x+4 / x-7

Chris shoots an arrow up into the air. The height (in feet) of the arrow is given by the function h(t) = - 16t2 + 64t + 112 wher

iris [78.8K]

Answer:

After 2 seconds

Step-by-step explanation:

First step is find an equation of velocity:

h(t)=-16t2+64t+112. The velocity function is derivated from position function, so we should do that.

v(t)= h´(t)= -16*2*t+64 ⇒ v(t)= -32t+64.

Now we know the arrow reach the maximun height when this velocity is equal to zero when t>0.

0=-32t+64 ⇒ 32t=64, split each member for 32 ⇒ $\frac{32t}{32}$ = $\frac{64}{32}$

t=2. That means after 2 sec. the arrow reach the maximum height

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4 years ago

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20 points and brainliest <br> I’m in quiz in need it asap <br> Number 4

iren [92.7K]

Answer and step-by-step explanation:

The polar form of a complex number $a+ib$ is the number $re^{i\theta}$ where $r = \sqrt{a^2+b^2}$ is called the modulus and $\theta = tan^-^1 (\frac ba)$ is called the argument. You can switch back and forth between the two forms by either remembering the definitions or by graphing the number on Gauss plane. The advantage of using polar form is that when you multiply, divide or raise complex numbers in polar form you just multiply modules and add arguments.

(a) let's first calculate moduli and arguments

$r_1 = \sqrt{(-2\sqrt3)^2+2^2}=\sqrt{12+4} = 4\\ \theta_1 = tan^-^1(\frac{2}{-2\sqrt3}) =-\pi/6\\r_2=\sqrt{1^2+1^2}=\sqrt2\\ \theta_2 = tan^-^1(\frac 11)= \pi/4$

now we can write the two numbers as

$z_1=4e^{-i\frac \pi6}; z_2=e^{i\frac\pi4}$

(b) As noted above, the argument of the product is the sum of the arguments of the two numbers:

$Arg(z_1\cdot z_2) = Arg(z_1)+Arg(z_2) = -\frac \pi6 + \frac \pi4 = \frac\pi{12}$

(c) Similarly, when raising a complex number to any power, you raise the modulus to that power, and then multiply the argument for that value.

$(z_1)^1^2=[4e^{-i\frac \pi6}]^1^2=4^1^2\cdot (e^{-i\frac \pi6})^1^2=2^2^4\cdot e^{-i(12)\frac\pi6}\\=2^2^4 e^{-i\cdot2\pi}=2^2^4$

Now, in the last step I've used the fact that $e^{i(2k\pi+x)} = e^i^x ; k\in \mathbb Z$ , or in other words, the complex exponential is periodic with $2\pi$ as a period, same as sine and cosine. You can further compute that power of two with the help of a calculator, it is around 16 million, or leave it as is.