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faltersainse [42]
3 years ago
15

Can you please help me

Mathematics
2 answers:
nika2105 [10]3 years ago
4 0
I believe it’s 0.5 :)
Ksju [112]3 years ago
3 0

Answer:

0.5

Step-by-step explanation:

The slope of the line is 1/2, so the answer is 0.5

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Plssss helpppp ASAPP<br> will mark BRAINLIEST!!!
Snezhnost [94]

Answer:

I would say the last one

Step-by-step explanation:

The third works the same way but 46/17 is hard to calculate without a calculator

5 0
3 years ago
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PLEASE HELP IT’S DUE TMR!!
zepelin [54]

Answer:

(a) 1.0792 x 10³

(b) 3.769 x 10³

(c) 9.985 x 10³

This should be the answer to the problems, the rounding part got me confused for a second.

Significant digits are numbers after the decimal that aren't 0. So the 1.0792, the 0 isn't a significant digit.

6 0
3 years ago
If an item that costs $45 sold for a 15% discount, how much do you spend on it? Commission is a % of sales.
Stells [14]

Answer:

You pay $38.25 after the discount is applied

Step-by-step explanation:

First take 15% of 45 by multiplying 45 by 0.15 (which is 15% as a decimal)

0.15 * 45 = 6.75

Subtract 6.75 from $45 and then you get $38.25!

Hope this helps! Good luck on your courses!

8 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
27. Is the product of 5 and 2 the same as the sum of 5<br>and 2? Explain.​
Fantom [35]

Answer:

no

Step-by-step explanation:

The answer is o because product means multiplication and sum means addition, 5*2=10 v.s. 5+2=7

3 0
3 years ago
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