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lara [203]
2 years ago
7

Solve the systems equations for x and y and check: y=2x+10 x=3y

Mathematics
1 answer:
andriy [413]2 years ago
6 0
For Y=2x+10 X=y/2 -5. Second x/3
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Five times the sum of a number and 6 is 48
MissTica
5(n+6)=48

5n+30=48

5n=18

n=18/5
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3 years ago
Which is the equation of a parabola with vertex (0, 0), that opens to the left and
andrezito [222]

Answer:

Step-by-step explanation:

I'm going to guess that the square  exponents are on the top y's   only?   then answer b) looks good

8 0
3 years ago
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Please answer this quick -8х (-4 )
Dafna1 [17]

Answer:

32x

Step-by-step explanation:

Distribute the -8x to the (-4)

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3 years ago
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The perimeter of a semicircle is 5.14 feet. What is the semicircle’s area?
hammer [34]
\bf \textit{circumference of a semi-circle}\\\\
C=\pi r\qquad 
\begin{cases}
r=radius\\
-----\\
C=5.14
\end{cases}\implies 5.14=\pi r\implies \boxed{\cfrac{5.14}{\pi }=r}\\\\
-------------------------------\\\\
\textit{area of a semi-circle}\\\\
A=\cfrac{\pi r^2}{2} \implies A=\cfrac{\pi }{2}r^2\implies A=\cfrac{\pi }{2}\left( \boxed{\cfrac{5.14}{\pi }} \right)^2
\\\\\\
A=\cfrac{\pi }{2}\implies \cfrac{5.14^2}{\pi ^2}\implies A=\cfrac{26.4196}{2\pi }\implies A=\cfrac{13.2098}{\pi }
5 0
3 years ago
Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"
vampirchik [111]
\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h

Employ a standard trick used in proving the chain rule:

\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h

The limit of a product is the product of limits, i.e. we can write

\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)

The rightmost limit is an exercise in differentiating \sqrt x using the definition, which you probably already know is \dfrac1{2\sqrt x}.

For the leftmost limit, we make a substitution y=\sqrt x. Now, if we make a slight change to x by adding a small number h, this propagates a similar small change in y that we'll call h', so that we can set y+h'=\sqrt{x+h}. Then as h\to0, we see that it's also the case that h'\to0 (since we fix y=\sqrt x). So we can write the remaining limit as

\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}

which in turn is the derivative of \tan y, another limit you probably already know how to compute. We'd end up with \sec^2y, or \sec^2\sqrt x.

So we find that

\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}
7 0
3 years ago
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