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3 years ago

Write tally marks to show the number 8.

Mathematics

2 answers:

sladkih [1.3K]3 years ago

8 0

Look at the attatchment below

Please make sure to check with classmates for answers, to help each other.

ale4655 [162]3 years ago

7 0

1111 slanted line 111 op

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What are double facts what do they look like

motikmotik

Double facts the even and odd numbers for example 2×2=4

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3 years ago

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About 16\20 of a tomato is water. what percent of a tomato is water?

notka56 [123]

The percent of a tomato is 80%

5 0

2 years ago

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1.73639 rounded to the nearest thousanth

Maurinko [17]

^ is in the thousadths place so look for the number before it, which is 3. 3 is less than 5 so 6 stays as is so it becomes 1.736

8 0

3 years ago

Please see attachment

Dafna11 [192]

Answer:

a) The value of absolute minimum value = - 0.3536

b) which is attained at $x = \frac{1}{\sqrt{2} }$

Step-by-step explanation:

Step(i):-

Given function

$f(x) = \frac{-x}{2x^{2} +1}$ ...(i)

Differentiating equation (i) with respective to 'x'

$f^{l} = \frac{2x^{2} +1(-1) - (-x) (4x)}{(2x^{2}+1)^{2} }$ ...(ii)

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} }$

Equating Zero

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

$\frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

$2 x^{2}-1 = 0$

$2 x^{2} = 1$

$x^{2} = \frac{1}{2}$

$x = \frac{-1}{\sqrt{2} } , x = \frac{1}{\sqrt{2} }$

Step(ii):-

Again Differentiating equation (ii) with respective to 'x'

$f^{ll}(x) = \frac{(2x^{2} +1)^{2} (4x) - 2(2x^{2} +1) (4x)(2x^{2}-1) }{(2x^{2}+1)^{4} }$

put

$x = \frac{1}{\sqrt{2} }$

$f^{ll} (x) > 0$

The absolute minimum value at $x = \frac{1}{\sqrt{2} }$

Step(iii):-

The value of absolute minimum value

$f(x) = \frac{-x}{2x^{2} +1}$

$f(\frac{1}{\sqrt{2} } ) = \frac{-\frac{1}{\sqrt{2} } }{2(\frac{1}{\sqrt{2} } )^{2} +1}$

on calculation we get

The value of absolute minimum value = - 0.3536

Final answer:-

a) The value of absolute minimum value = - 0.3536

b) which is attained at $x = \frac{1}{\sqrt{2} }$

3 0

2 years ago

Nine voltage measurements in a circuit have produced the following values: 12.09, 12.05, 11.94, 11.99, 12.12, 12.05, 12.02, 11.9

Debora [2.8K]

Answer:

we can not reject any value

Step-by-step explanation: From data we can test the highest and the lowest value to evaluate if one of these values are out of certain confidence Interval

If we established CI = 95 % then α = 5 % and α/2 = 0,025

From data we find the mean of the values

μ₀ = 12,03 and σ = 0,07

From z table we find z score for 0,025 is z(c) = ± 1,96

So limits of our CI are:

12,03 + 1,96 = 13,99

12,03 - 1,96 = 10,07

And all our values are within ( 10,07 , 13,99)

So we can not reject any value

4 0

3 years ago