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LenaWriter [7]
3 years ago
8

Please someone help me with this question

Mathematics
1 answer:
SpyIntel [72]3 years ago
4 0
<h2 /><h2>d  = 7</h2>

<em>-</em><em> </em><em>BRAINLIEST</em><em> answerer</em>

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8(5-32) equal to and explain <br> please help
Gre4nikov [31]
You would use order of operations: PEMDAS
P(parenthesis) E(exponents) MD(multiplication/division) AS(addition/subtraction)
with MD and AS order doesnt matter.
8(5-32) you would start with inside the Parenthesis for "P" so (5-32)=(-27)
next you would go to the E but because you dont have an exponent you go to the next step with is the "MD" you have multiplication so next would be 8(-27) and 8 multiplied by -27 is: 8(-27)= -216
ANSWER: -216
4 0
4 years ago
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4/5 divided by 1/10?
Evgesh-ka [11]

Answer:

the answer to the question is 8

6 0
3 years ago
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Find the distance traveled by a particle with position (x, y) as t varies in the given time interval. x = 5 sin2 t, y = 5 cos2 t
rodikova [14]
\begin{cases}x(t)=5\sin2t\\y(t)=5\cos2t\end{cases}\implies\begin{cases}\frac{\mathrm dx}{\mathrm dt}=10\cos2t\\\frac{\mathrm dy}{\mathrm dt}=-10\sin2t\end{cases}

The distance traveled by the particle is given by the definite integral

\displaystyle\int_C\mathrm dS=\int_0^{3\pi}\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt

where C is the path of the particle. The distance is then

\displaystyle\int_0^{3\pi}\sqrt{100\cos^22t+100\sin^22t}\,\mathrm dt=10\int_0^{3\pi}\mathrm dt=30\pi
6 0
3 years ago
HELP ON THIS PROBLEM!!<br><br> Solve the following 0deg&lt; x&lt; 360deg: cos x= cos 2x
blsea [12.9K]

Answer:

x=0^\circ,120^\circ,240^\circ,360^\circ

Step-by-step explanation:

<u>Use identities to set the equation up as a quadratic</u>

\cos x=\cos 2x;\:0^\circ \leq x \leq 360^\circ\\\\\cos x=2\cos^2 x-1\\\\0=2\cos^2 x-\cos x-1

<u>Make the substitution u=cos(x) and solve the quadratic</u>

<u />0=2u^2-u-1\\\\0=(2u+1)(u-1)

\displaystyle 0=2u+1\\\\0=2\cos x+1\\\\-1=2\cos x\\\\-\frac{1}{2}=\cos x\\\\120^\circ,240^\circ=x

0=u-1\\\\0=\cos x-1\\\\1=\cos x\\\\0^\circ,360^\circ=x

Hence, x=0^\circ,120^\circ,240^\circ,360^\circ

7 0
2 years ago
Need help with math, Lol thank you! :)
attashe74 [19]
That would be A,   22/12 and 27/12
7 0
4 years ago
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