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____ [38]
3 years ago
13

Convert 51 in base 10 into base 5

Mathematics
1 answer:
weeeeeb [17]3 years ago
6 0

i cant put a link but i looked up "51 in base 5"

and got 201

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What numbers do we need to round to the nearest 100?
bixtya [17]
All the way up till 150
4 0
3 years ago
Please help me with this
juin [17]

Answer:

10π (31.415926)

√101(10.0498756)

9.92749...

0.99853...

Step-by-step explanation:

For 10π, we can simplify π into 3.14 for an approximate amount. 3.14*10=31.4

For the square root of ten, we can estimate it's between 10 and 11 since 10 squared is 100 and 11 squared is 121.

6 0
2 years ago
Read 2 more answers
Melanie invested $9,800 in an account paying an interest rate of 4 1 2 4 2 1 ​ % compounded quarterly. Amelia invested $9,800 in
Gnom [1K]

Step-by-step explanation:

yifhkdtjayixyrsyisjdyjdrtdgjz

5 0
3 years ago
1. Make sure your answers and work is SHOWN please and steps are in order.
JulijaS [17]

Been a while since I've done synthetic division.

1a. Let's assume that's supposed to be an equals sign

p(x) = 2x⁴ - 3x³ - 6x² + 5x + 6

possible rational roots have factors of 6 in the numerator and of 2 in the denominator.  We'll only worry about negative numerators.

Factors of six: 1,2,3,6, and we don't forget -1,-2,-3,-6

Factors of 2: 1,2

Possible rational roots:

(dividing by 1:) 1,-1,2,-2,3,-3,6,-6

(dividing by 2:) 1/2, -1/2  (2/2=1 is a duplicate, don't have to repeat it), 3/2, -3/2

Possible rational roots: 1,-1,2,-2,3,-3,6,-6, 1/2, -1/2, 3/2, -3/2

Synthetic division, trying x=1,

   1 | 2  -3  -6  5  6

             2  -1  -7  -2

       2    -1  -7  -2  4

Got a remainder of 4, so 1 isn't a root;

Trying x=-1  

-1 | 2  -3  -6  5  6

         -2   5  1  -6

     2  -5  -1  6  0

Zero remainder, found a root, x=-1.  This division says

(2x⁴ - 3x³ - 6x² + 5x + 6) / (x + 1) = 2x³ - 5x² - x + 6

Same set of rational roots on the cubic, we continue with x=2

2 | 2 -5 -1 6

         4 -2 -6

    2  -1 -3 0

Another zero remainder, x=2 is a root.  We're left with 2x² - x - 3 = 0, which factors as

(2x - 3)(x + 1) = 0

That's a second factor of x+1.  Our final factorization is

2x⁴ - 3x³ - 6x² + 5x + 6 = (x + 1)²(x-2)(2x - 3)

Fourth degree with a positive leading coefficient so goes to +infinity at both ends.  Double zero at x=-1, so it's tangent there, just touching the x axis, then down through x=3/2 and up through x=2.  

I'll leave the actually sketching and the other two polynomials to you -- that took some time.

4 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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