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Ronch [10]
3 years ago
13

Which is greater 8.016 or 8.16

Mathematics
2 answers:
shusha [124]3 years ago
7 0
8.16 because .016 is not as close to a whole as .16
Tanya [424]3 years ago
5 0
The answer is 8.16 lol hope it helped in time
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Click on image and please you can you answer . put the fractions in order please thank you. i will give you 5 stars and mark you
Sergeeva-Olga [200]

Answer:

Mean : 19.997

Mode : 1

Median : 4

Range : 0 - 9

Hope this helps

4 0
3 years ago
HELP!! I'm not sure what goes in the blank.
natta225 [31]

Answer:

10sin^{2} (β)

Step-by-step explanation:

We can find this two ways, first by seeing in the step after it, cosines are canceled out. Since you already have 10sin^{2} (β) on the next step, you can assume that (since only the cosines changed and the cosine next ot the blank was removed), the value is 10sin^{2} (β).

You can also use double angle formulas from the previous step:

(sin(2β) = 2 sin(β) cos(β))and find that:

5 sin (2β) sin(β) = 5 * (2 sin(β) cos(β)) sin(β)) = (10 sin(β) sin(β)) cos(β) =

10sin^{2} (β) cos(β)

But since cos(β) is already present, we can see that the answer is 10sin^{2} (β)

4 0
2 years ago
What equation in slope intercept form represents the line that passes through the two point? (6.9,5.9), (10.9,-2.1)
den301095 [7]

Answer:

y = 2x + b

I dont know what b is but the slope is 2 I think

6 0
3 years ago
What is the value of x?<br> 4x − 6 = 18
tensa zangetsu [6.8K]
You need to get rid of -6 by adding 6, you also need to add 6 to 18.

4x = 24

Now you need to get rid of the 4 attached to the variable, you do this by dividing 4 since it is multiplying, also dividing 4 and 24

x = 6

So the value of x is 6

Hope this helps! :)
8 0
3 years ago
Read 2 more answers
Lim x--&gt; 0 (e^x(sinx)(tax))/x^2
Tom [10]

Make use of the known limit,

\displaystyle\lim_{x\to0}\frac{\sin x}x=1

We have

\displaystyle\lim_{x\to0}\frac{e^x\sin x\tan x}{x^2}=\left(\lim_{x\to0}\frac{e^x}{\cos x}\right)\left(\lim_{x\to0}\frac{\sin^2x}{x^2}\right)

since \tan x=\dfrac{\sin x}{\cos x}, and the limit of a product is the same as the product of limits.

\dfrac{e^x}{\cos x} is continuous at x=0, and \dfrac{e^0}{\cos 0}=1. The remaining limit is also 1, since

\displaystyle\lim_{x\to0}\frac{\sin^2x}{x^2}=\left(\lim_{x\to0}\frac{\sin x}x\right)^2=1^2=1

so the overall limit is 1.

4 0
3 years ago
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