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2 years ago

13

Katie is watching a helicopter flying at an altitude of 5 miles. If the angle of elevation is pi/3 , the distance between Katie

and the helicopter is approximately miles.

1 answer:

Art [367]2 years ago

3 0

From SOH CAH TOA, you know that
.. Sin = Opposite/Hypotenuse
.. sin(π/3) = (5 miles)/Hypotenuse
.. Hypotenuse = (5 miles)/sin(π/3) = 5/(√3/2) miles ≈ 5.774 miles

The straight-line distance from Katie to the helicopter is about 5.77 miles.

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The star aldebaran, visible from earth, lies about 65 light years away from us. What is the minimum amount of time that would be

makkiz [27]

A beam of light (travelling at the speed of light) would take 65 years to go and 65 years to come back from aldebaran, i.e. 130 years. You can send and receive a radio message (travelling at the speed of light, being electromagnetic radiation) taking so 130 years.

8 0

2 years ago

The square of a number decreased by 3 times the number 28 find all possible values for the number

stealth61 [152]

Question:

The square of a number decreased by 3 times the number is 28 find all possible values for the number

Answer:

The possible values of number are 7 and -4

Solution:

Given that the square of a number decreased by 3 times the number is 28

To find: all possible values of number

Let "a" be the unknown number

From given information,

square of a number decreased by 3 times the number = 28

$a^2 - 3a = 28$

$a^2 - 3a - 28 = 0$

Let us solve the above quadratic equation

$\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0$

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Using the above formula,

$\text { For } a^{2}-3 a-28=0 \text { we have } a=1, b=-3, c=-28$

$\begin{aligned}&a=\frac{-(-3) \pm \sqrt{(-3)^{2}-4(1)(-28)}}{2 \times 1}\\\\&a=\frac{3 \pm \sqrt{9+112}}{2}\\\\&a=\frac{3 \pm \sqrt{121}}{2}=\frac{3 \pm 11}{2}\\\\&a=\frac{3+11}{2} \text { or } a=\frac{3-11}{2}\\\\&a=7 \text { or } a=-4\end{aligned}$

Thus the possible values of number are 7 and -4

5 0

2 years ago

Model Exponential Growth Question :A sample of bacteria is growing at a continuously compounding rate. The sample triples in 10

Iteru [2.4K]

Answer:

The number of bacteria $B$ after $d$ days is given by

$B = B_0 (3)^{\frac{1}{10} d}$

where $B_0$ is the initial number of bacteria.

Step-by-step explanation:

The number of bacteria $B$ in the sample triples every 10 days, this means after the first 10th day, the number of bacteria is

$B = B_0 *3,$

where $B_0$ is the initial number of bacteria in the sample.

After the 2nd 10th days, the number of bacteria is

$B = (B_0 *3)*3$

after the 3rd day,

$B =( B_0 *3*3)*3$

and so on.

Thus, the formula we get for the number of bacteria after the <em>n</em>th 10-days is

$B = B_0 (3)^n$

where $n$ is is the <em>n</em>th 10-days.

Since, $n$ is 10 days, we have

$d =10n$

or

$n =\dfrac{1}{10}$

Substituting that into $B = B_0 (3)^n$ , we get:

$\boxed{ B = B_0 (3)^{\frac{1}{10} d}}$

8 0

2 years ago

The velocity v of blood that flows in a blood vessel with radius r and length ℓ at a distance r from the central axis is v(r) =

pogonyaev

$v(r)=4P\eta\ell(R^2-r^2)$

(assuming $R$ is the radius of the vessel, and that $0\le r\le R$ )

The average velocity of the blood is given by

$\displaystyle\frac1{R-0}\int_{r=0}^{r=R}v(r)\,\mathrm dr$
$=\displaystyle\frac{4P\eta\ell}R\int_{r=0}^{r=R}(R^2-r^2)\,\mathrm dr$
$=\dfrac{4P\eta\ell}R\left(\dfrac23R^3\right)$
$=\dfrac{8P\eta\ell}3R^2$

7 0

2 years ago

A worker is paid rs. 2130 for 6 days .if his total wage during a month is rs. 9230 find the number of days he worked in the mont

Katyanochek1 [597]

Answer:

26 days

Step-by-step explanation:

If 6 days = rs. 2130

then 1 day = (2130 × 1) ÷ 6

Therefore 1 day = rs 355

For rs. 9230, the number of days he worked:

= 9230 ÷ 355

= 26 days

4 0

2 years ago