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3 years ago

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The table shows the record low temperatures in Juneau, Alaska for each of the 12 months of the year in degrees F. Record Low Tem

p Month Year (in °F) January -22 1972 February -22 1968 March -15 1972 April 6 1963 May 25 1972 June 31 1971 July 18 1895 August 27 1948 September 23 1972 October 11 1984 November -5 2006 December 1941 -21 According to the data, what is the range of the record low temperatures in Juneau, Alaska?
D. 31°
A. -22°
E. 53°
B. 4.67°
C. 8.5°

1 answer:

nadezda [96]3 years ago

3 0

Answer: -22

Step-by-step explanation: the most repeated number ig

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Help! Prove the equality<br><br>arccos √(2/3) - arccos (1+√6)/(2*√3) = π/6

stealth61 [152]

Answer:

<em>Proof in the explanation</em>

Step-by-step explanation:

<u>Trigonometric Equalities</u>

Those are expressions involving trigonometric functions which must be proven, generally working on only one side of the equality

For this particular equality, we'll use the following equation

$\displaystyle cos(x-y)=cos\ x\ cos\ y+sin\ x\ sin\ y$

The equality we want to prove is

$\displaystyle arccos\ \sqrt{\frac{2}{3}}-arccos\left(\frac{1+\sqrt{6}}{2\sqrt{3}}\right)=\frac{\pi}{6}$

Let's set the following variables:

$\displaystyle x=arccos\ \sqrt{\frac{2}{3}},\ y=arccos(\frac{1+\sqrt{6}}{2\sqrt{3}})$

And modify the first variable:

$\displaystyle x=arccos\ \frac{\sqrt{6}}{3}}=>\ cos\ x= \frac{\sqrt{6}}{3}}$

Now with the second variable

$\displaystyle y=arccos\ \frac{1+\sqrt{6}}{2\sqrt{3}}=>cos\ y=\frac{1+\sqrt{6}}{2\sqrt{3}}=\frac{\sqrt{3}+3\sqrt{2}}{6}$

Knowing that

$sin^2x+cos^2x=1$

We compute the other two trigonometric functions of X and Y

$\displaystyle sin \ x=\sqrt{1-cos^2\ x}=\sqrt{1-(\frac{\sqrt{6}}{3})^2}=\sqrt{1-\frac{6}{9}}=\frac{\sqrt{3}}{3}$

$\displaystyle sin\ y=\sqrt{1-cos^2y}=\sqrt{1-\frac{(\sqrt{3}+3\sqrt{2})^2}{36}}}$

$\displaystyle sin\ y=\sqrt{\frac{36-(3+6\sqrt{6}+18)}{36}}=\sqrt{\frac{15-6\sqrt{6}}{36}}$

Computing

$15-6\sqrt{6}=(3-\sqrt{6})^2$

Then

$\displaystyle sin\ y=\frac{3-\sqrt{6}}{6}$

Now we replace all in the first equality:

$\displaystyle cos(x-y)=\frac{\sqrt{6}}{3}.\frac{\sqrt{3}+3\sqrt{2}}{6}+\frac{\sqrt{3}}{3}.\frac{3-\sqrt{6}}{6}$

$\displaystyle cos(x-y)=\frac{3\sqrt{2}+6\sqrt{3}}{18}+\frac{3\sqrt{3}-3\sqrt{2}}{18}$

$\displaystyle cos(x-y)=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}=cos\ \pi/6$

Thus, proven

5 0

3 years ago

PLEASE HELP ASAP!!! I NEED CORRECT ANSWERS ONLY PLEASE!!!

Keith_Richards [23]

It should be 7.615773106
But rounded should be 7.6

5 0

3 years ago

At the start of an experiment, there are 2,500 bacteria in a colony. The colony grows at a rate of 11% every hour. What will be

nata0808 [166]

Answer:

just multiply the number by 1.11 each time

Step-by-step explanation:

1hour (2500x1.11=2775)

2hour (2775x1.11=3080.25)

3 hour (3080.25x1.11=)

and on and on

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3 years ago

Please show step by step of working out the value of r for which is A minimum and calculate the minimum surface area of the cont

Genrish500 [490]

Answer:

A = 59.63cm^2

Step-by-step explanation:

You have the following function for the surface area of the container:

$A=\pi r^2+\frac{100}{r}$ (1)

where r is the radius of the cross sectional area of the container.

In order to find the minimum surface are you first calculate the derivative of A respect to r, to find the value of r that makes the surface area a minimum.

$\frac{dA}{dr}=\frac{d}{dr}[\pi r^2+\frac{100}{r}]\\\\\frac{dA}{dr}=2\pi r-\frac{100}{r^2}$ (2)

Next, you equal the expression (2) to zero and solve for r:

$2\pi r-\frac{100}{r^2}=0\\\\2\pi r=\frac{100}{r^2}\\\\r^3=\frac{50}{\pi}\\\\r=(\frac{50}{\pi})^{1/3}$

Finally, you replace the previous result in the equation (1):

$A=\pi (\frac{50}{\pi})^{2/3}+\frac{100}{(\frac{50}{\pi})^{1/3}}}$

$A=59.63$

The minimum total surface area is 59.63cm^2

3 0

3 years ago

Find the approximate length of the hypotenuse of a right triangle with leg lengths 8.4cm and 7.6 cm.

GaryK [48]

Answer:

11.3

Step-by-step explanation:

A Right Angled Triangle obeys the Pythagorean Theorem. According to the theorem:

The square of Hypotenuse is equal to sum of squares of both the legs. In equation form we can write this as:

$\textrm{Hypotenuse}^{2}=\textrm{(Leg 1)}^{2}+\textrm{(Leg 2)}^{2}$

We are given the length of the legs to be 8.4 cm and 7.6 cm.

Using these values, we get:

$\textrm{(Hypotenuse)}^{2}=(8.4)^{2}+ (7.6)^{2} \\\\\textrm{(Hypotenuse)}^{2}=128.32$

By taking square root of both sides we can find the length of Hypotenuse.

$\textrm{Hypotenuse}=\sqrt{128.32} \\\\ \textrm{Hypotenuse}=11.3$

Therefore, the approximate length of the hypotenuse will be 11.3 cm, rounded of to one decimal place.

3 0

2 years ago