All categories

3 years ago

1)Find the angle of elevation of the sun from the ground when a tree that is

Mathematics

1 answer:

KIM [24]3 years ago

7 0

Answer:

$\theta=41.18^{\circ}$

Step-by-step explanation:

Given that,

The height of the tree, h = 14 ft

The height of the shadow, b = 16 ft

We need to find the angle of elevation of the sun from the ground. Let the angle be θ. We can use trigonometry to find it. So,

$\tan\theta=\dfrac{P}{B}\\\\\tan\theta=\dfrac{14}{16}\\\\\theta=41.18^{\circ}$

So, the required angle of elevation of the sun is equal to $41.18^{\circ}$ .

You might be interested in

Seven over nine<br> to the power of<br> two

anyanavicka [17]

Answer:0.08

Step-by-step explanation:

3 0

2 years ago

Need help due in 10 mins

Marina CMI [18]

Answer:

A) The amount of money deposited monthly

Step-by-step explanation:

y = 35x (slope) (amount deposited each month)+ 250 (y-intercept) (starting amount)

4 0

2 years ago

The altitude of a triangle is increasing at a rate of 1 cm/min while the area of the triangle is increasing at a rate of 2 cm2/m

Mnenie [13.5K]

Answer:

The base is reducing at a rate of 3.4cm per min.

Step-by-step explanation:

Let The Height of the Triangle, h

Area of the triangle = A

Area of a Triangle, $A= \frac{1}{2}bh$

When $A=190 cm^2, h=10cm$

$190= \frac{1}{2}*b*10\\b=38cm$

$\frac{dA}{dt}=\frac{1}{2}h\frac{db}{dt}+ \frac{1}{2}b\frac{dh}{dt}\\\frac{dh}{dt}=1 cm/min, \frac{dA}{dt}=2cm^2/min,$

$2=\frac{1}{2}*10*\frac{db}{dt}+ \frac{1}{2}*38*1\\2=5\frac{db}{dt}+19\\2-19=5\frac{db}{dt}\\-17=5\frac{db}{dt}\\\frac{db}{dt}=-\frac{17}{5} =-3.4 cm/min$

The base is reducing at a rate of 3.4cm per min.

8 0

2 years ago

What is the coefficient of x2y3 in the expansion of (2x + y)5?

Zigmanuir [339]

Option C:

The coefficient of $x^{2} y^{3}$ is 40.

Solution:

Given expression:

$(2 x+y)^{5}$

Using binomial theorem:

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here $a=2 x, b=y$

Substitute in the binomial formula, we get

$(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}$

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}$