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Margaret [11]
3 years ago
12

Any help will be really appreciated

Mathematics
2 answers:
RSB [31]3 years ago
5 0

Answer: B

Step-by-step explanation:

matrenka [14]3 years ago
3 0
The answer is B

5x2=10. 10+1=11
7x2=14. 14+1=15
9x2=18. 18+1=19
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In your own words define “reciprocal” and describe how to find the reciprocal of a fraction. Use 3/6 as an example in your expla
Fittoniya [83]

Answer:

In mathematics, a multiplicative inverse or reciprocal for a number x, denoted by 1/x or x⁻¹, is a number which when multiplied by x yields the multiplicative identity, 1. The multiplicative inverse of a fraction a/b is b/a. For the multiplicative inverse of a real number, divide 1 by the number.

Step-by-step explanation:

To find the reciprocal, divide

1

by the number given.

1

3

6

Simplify.

Tap for fewer steps...

Cancel the common factor of

3

and

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.

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Factor

3

out of

3

.

1

3

(

1

)

6

Cancel the common factors.

Tap for fewer steps...

Factor

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out of

6

.

1

3

⋅

1

3

⋅

2

Cancel the common factor.

1

3

⋅

1

3

⋅

2

Rewrite the expression.

1

1

2

Multiply the numerator by the reciprocal of the denominator.

1

⋅

2

Multiply

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by

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2

7 0
3 years ago
Read 2 more answers
Which could be the missing data item for the given set of data if the median of the complete data set is 14?
Goshia [24]
10,10,11,11,12,12,14,15,16,18,18,19

10,10,10,11,11,12,12,14,15,16,18,18,19
10,10,11,11,11,12,12,14,15,16,18,18,19
10,10,11,11,12,12,12,14,15,16,18,18,19
10,10,11,11,12,12,14,15,16,18,18,18,19

I THINK it's D
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guapka [62]

Answer:

The difference is that the rate is just one thing and the the set of data values are more than one value.They are in a set.

Sorry there is no explanation to this this is not literally math.

4 0
3 years ago
What is 360 divided into 11 steps by steps
Zolol [24]
Answer:

Hope I helped you! :)

4 0
3 years ago
\lim _{x\to 0}\left(\frac{2x\ln \left(1+3x\right)+\sin \left(x\right)\tan \left(3x\right)-2x^3}{1-\cos \left(3x\right)}\right)
Vinvika [58]

\displaystyle \lim_{x\to 0}\left(\frac{2x\ln \left(1+3x\right)+\sin \left(x\right)\tan \left(3x\right)-2x^3}{1-\cos \left(3x\right)}\right)

Both the numerator and denominator approach 0, so this is a candidate for applying L'Hopital's rule. Doing so gives

\displaystyle \lim_{x\to 0}\left(2\ln(1+3x)+\dfrac{6x}{1+3x}+\cos(x)\tan(3x)+3\sin(x)\sec^2(x)-6x^2}{3\sin(3x)}\right)

This again gives an indeterminate form 0/0, but no need to use L'Hopital's rule again just yet. Split up the limit as

\displaystyle \lim_{x\to0}\frac{2\ln(1+3x)}{3\sin(3x)} + \lim_{x\to0}\frac{6x}{3(1+3x)\sin(3x)} \\\\ + \lim_{x\to0}\frac{\cos(x)\tan(3x)}{3\sin(3x)} + \lim_{x\to0}\frac{3\sin(x)\sec^2(x)}{3\sin(3x)} \\\\ - \lim_{x\to0}\frac{6x^2}{3\sin(3x)}

Now recall two well-known limits:

\displaystyle \lim_{x\to0}\frac{\sin(ax)}{ax}=1\text{ if }a\neq0 \\\\ \lim_{x\to0}\frac{\ln(1+ax)}{ax}=1\text{ if }a\neq0

Compute each remaining limit:

\displaystyle \lim_{x\to0}\frac{2\ln(1+3x)}{3\sin(3x)} = \frac23 \times \lim_{x\to0}\frac{\ln(1+3x)}{3x} \times \lim_{x\to0}\frac{3x}{\sin(3x)} = \frac23

\displaystyle \lim_{x\to0}\frac{6x}{3(1+3x)\sin(3x)} = \frac23 \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}\frac{1}{1+3x} = \frac23

\displaystyle \lim_{x\to0}\frac{\cos(x)\tan(3x)}{3\sin(3x)} = \frac13 \times \lim_{x\to0}\frac{\cos(x)}{\cos(3x)} = \frac13

\displaystyle \lim_{x\to0}\frac{3\sin(x)\sec^2(x)}{3\sin(3x)} = \frac13 \times \lim_{x\to0}\frac{\sin(x)}x \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}\sec^2(x) = \frac13

\displaystyle \lim_{x\to0}\frac{6x^2}{3\sin(3x)} = \frac23 \times \lim_{x\to0}x \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}x = 0

So, the original limit has a value of

2/3 + 2/3 + 1/3 + 1/3 - 0 = 2

6 0
3 years ago
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