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2 years ago

Find the missing length indicate

Mathematics

1 answer:

mote1985 [20]2 years ago

4 0

Answer:

2.99973 but i would put 3 just incase

Step-by-step explanation:

just search on the internet "right triangle calculator" and click one of the links

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Simplify. 6/2 + 4/2

antiseptic1488 [7]

The answer is 5. You get 5 by added the 6 and 4 together to get 10 and the 2 is lowest you can go so you keep that the same.

8 0

3 years ago

Read 2 more answers

Me puedes ayudar es para hoy porfavor

Inga [223]

Answer:

mira a las base de las figuras y busca cual base matchs con la figuras de la figura prinsipal

Step-by-step explanation:

como el cono va con la del toda la izquierda de la de la mitad

5 0

3 years ago

An automobile tire has a diameter of about 1.7. What is the circumference

lilavasa [31]

Answer:

r = 0.85 ft
C = 5.3407075111026 ft
A = 2.2698006922186 ft²

Agenda:

r = radius
C = circumference
A = area
π = pi = 3.1415926535898
<span>√ = square root
</span>
Formula:
Circumference of a circle:

C = 2πr = πd

3 0

3 years ago

A rocket is launched from the top of a 99-foot cliff with an initial velocity of 122 ft/s.

Harman [31]

Remember that c is the initial height. Since we the rocket is in a 99-foot cliff, c=99. Also, we know that the velocity of the rocket is 122 ft/s; therefore v=122
Lets replace the values into the the vertical motion formula to get:
$0=-16 t^{2} +122t+99$
Notice that the rocket hits the ground at the bottom of the cliff, which means that the final height is 99-foot bellow its original position; therefore, our final height will be h=-99
Lets replace this into our equation to get:
$-99=-16 t^{2} +122t+99$
$-16 t^{2} +122+198=0$

Now we can apply the quadratic formula $t= \frac{-b+or- \sqrt{ b^{2} -4ac} }{2a}$ where a=-16, b=122, and c=198
$t= \frac{-122+or- \sqrt{ 122^{2}-(4)(-16)(198) } }{(2)(-16)}$
$t= \frac{-122+ \sqrt{27556} }{-32}$ or $t= \frac{-122- \sqrt{27556} }{-32}$
$t= \frac{-122+166}{-32}$ or $t= \frac{-122-166}{-32}$
$t= \frac{-11}{8}$ or $t=9$

Since the time can't be negative, we can conclude that the rocket hits the ground after 9 seconds.

5 0

3 years ago

Read 2 more answers

Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

-----------------------

Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

6 0

3 years ago