In right triangle ABC, angle A and angle B are complementary angles. The value of cos A is 9/41. What is the value of sin B?

Which of the following functions gives the radius, r(v), of a conical artifact that is 20 inches tall as a function of its volum

irinina [24]

<h3>Radius as function of volume is:</h3>

$r(v) = \sqrt{\frac{3v}{20 \pi }}\ inches$

Solution:

The volume of cone is given as:

$v = \frac{1}{3}\pi r^{2} h$

Where,

r is the radius

h is the height

From given,

height = 20 inches

From formula,

$v = \frac{1}{3}\pi r^{2} h$

Rearrange , so that r is alone in left side of equation

$3v = \pi r^2 h\\\\\pi r^2 h = 3v\\\\r^2 = \frac{3v}{\pi h}\\\\Take\ square\ root\ on\ both\ sides\\\\r = \sqrt{\frac{3v}{\pi h}}$

Substitute h = 20

$r = \sqrt{\frac{3v}{20 \pi }}$

Thus, radius as function of volume is:

$r(v) = \sqrt{\frac{3v}{20 \pi }}\ inches$

4 0

2 years ago

If 3.0 grams of strontium-90 in a rock sample remained in 1989, approximately how many grams of strontium-90 were present in the

kirill [66]

Answer:

11.29 grams of strontium-90 were present in the original rock sample in 1933.

Step-by-step explanation:

Strontium has a half-life of 28.8 years. Therefore,

1989 - 1933 = 56 years.

56 / 28.8 = 1.94 half-lives

Thus, the quantity of the radioisotope remaining will double for each half-life elapsed. moving backwards in time.

Therefore, moving backwards in time by 1.94 half-lives, the quantity remaining will double by 1.94 times.

Thus, the amount remaining in 1933 is

3.0 × (1.94)² = 11.29 grams

5 0

2 years ago

A sample of tritium-3 decayed to 94.5% of its original amount after a year.

Zolol [24]

(a) If $y(t)$ is the mass after $t$ days and $y(0) = A$ then $y(t) = Ae^{kt}.$

$y(1) = Ae^k = 0.945A \implies e^k = 0.945 \implies k = \ln 0.945$

$\text{Then } Ae^{(\ln 0.945)t} = \frac{1}{2}A\ \Leftrightarrow\ \ln e^{(\ln 0.945)t} = \ln \frac{1}{2}\ \Leftrightarrow\ ( \ln 0.945) t = \ln \frac{1}{2} \Leftrightarrow \\ \\
t = - \frac{\ln 5}{\ln 0.945} \approx 12.25 \text{ years}$

(b)
$Ae^{(\ln 0.945) t} = 0.20 A \ \Leftrightarrow\ (\ln 0.945) t \ln \frac{1}{5}\ \implies \\ \\
t = - \frac{\ln 5}{\ln 0.945} \approx 28.45\text{ years}$

3 0

2 years ago

Which statements about the hyperbola are true? Check all that apply.

mariarad [96]

Answer:

True options: 1, 2 and 5

Step-by-step explanation:

From the given diagram, you can see that the center of the hyperbola is placed at the origin, so first option is true (see attached diagram for definition of center, vertices, foci, i.e.)

There are two vertices of the hyperbola, they are placed at (-6,0) and (6,0), so second option is true.

The transverse axis is the segment connecting vertices, this segment is horizontal, so option 3 is false.

The foci are not placed within the rectangular reference box, so this option is false.

The directrices are vertical lines with equations $x=\pm \dfrac{a}{e}$ , so this option is true.

8 0

2 years ago

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What is the y-intercept of the function,represented by the table of values below?

Hitman42 [59]

Answer:

8

Step-by-step explanation:

So the y-intercept is not given by your table because there is no x that is listed as 0.

But don't fret; we can still find it.

Let's see if the function is linear by seeing if we have the same slope per two points in the table.

For the first pair ( the points (-2,16) and (1,4) ), x increased by 3 and the y decreased by 12 so the slope there is -12/3=-4.

Now looking at the next pair ( the points (1,4) and (2,0) ), x increased by 1 while y decreased by 4 so the slope is -4/1=-4.

So the function appears to be linear.

So the slope-intercept form of a line is y=mx+b where m is slope and b is y-intercept.

We already found the slope from earlier which is m=-4.

So the equation so far is y=-4x+b.

Now to find b, the y-intercept, we need to use a point (x,y) on the line along with y=-4x+b.

Let's see my favorite on the list of points is (2,0).

y=-4x+b with (x,y)=(2,0)

0=-4(2)+b

0=-8+b

8=b

So the y-intercept is 8.

8 0

2 years ago

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