All categories

2 years ago

Why do clouds tend to form around 3:00 pm and 6:00 am

1 answer:

4 0

During daytime, the sun heats the land and warms the air that rises. That lowers the pressure, causing a gradient force to bring air from the colder sea. The wind is then onshore. Small cumulus clouds form over the land, often following the coastline very closely

You might be interested in

Solve AABC. Round your answers to the nearest hundredth, if necessary

Aloiza [94]

Answer:

$C=25^{\circ},\\a\approx 10.72,\\b\approx 11.83$

Step-by-step explanation:

The sum of the interior angles of a triangle is 180 degrees. Thus, angle C must be $180-90-65=25^{\circ}$ .

In any triangle, the Law of Sines is given by $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$ .

Therefore, we have:

$\frac{\sin 90^{\circ}}{b}=\frac{\sin 25^{\circ}}{5},\\\\b=\frac{5\sin90^{\circ}}{\sin 25^{\circ}}=11.8310079158\approx \boxed{11.83}$

$\frac{a}{\sin 65^{\circ}}=\frac{5}{\sin25^{\circ}},\\a=\frac{5\sin 65^{\circ}}{\sin25^{\circ}}=10.7225346025\approx \boxed{10.72}$

8 0

2 years ago

Lim<br> x->infinity (1+1/n)

FrozenT [24]

Answer:

$^{ \lim}_{n \to \infty} (1+\frac{1}{n})=1$

Step-by-step explanation:

We want to evaluate the following limit.

$^{ \lim}_{n \to \infty} (1+\frac{1}{n})$

We need to recall that, limit of a sum is the sum of the limit.

So we need to find each individual limit and add them up.

$^{ \lim}_{n \to \infty} (1+\frac{1}{n})=^{ \lim}_{n \to \infty} (1) +^{ \lim}_{n \to \infty} \frac{1}{n}$

Recall that, as $n\rightarrow \infty,\frac{1}{n} \rightarrow 0$ and the limit of a constant, gives the same constant value.

This implies that,

$^{ \lim}_{n \to \infty} (1+\frac{1}{n})= 1 +0$

This gives us,

$^{ \lim}_{n \to \infty} (1+\frac{1}{n})= 1$

The correct answer is D

5 0

3 years ago

Please help asap

Nostrana [21]

I Think The answer is d I hope it helps My friend Message Me if I’m wrong and I’ll change My answer and fix it for you

6 0

1 year ago

Simplify the expression. Assume that all variables represent nonzero real numbers.StartFraction (4 n Superscript 4 Baseline q Su

Alja [10]

Answer:

$\frac{ - 3}{ 256 {q}^{10} {n}^{8} }$

Step by step explanation:

$\frac{ {(4 {n}^{4} {q}^{5})}^{2} {(8 {n}^{4} q)}^{-2} }{ {(- 3 {nq}^{9})}^{ - 1} {(4 {n}^{3} {q}^{9}) }^{3} }$

first we will change the terms with negative superscrips to the other side of the fraction

$\frac{{(4 {n}^{4} {q}^{5})}^{2}{(- 3 {nq}^{9})}^{ 1}}{{(4 {n}^{3} {q}^{9})}^{3} {(8 {n}^{4} q)}^{2} }$

then we will distribute the superscripts

$\frac{ {4}^{2} {n}^{2 \times 4} {q}^{2 \times 5} (- 3) {nq}^{9}}{ {4 }^{3}{n}^{3 \times 3} {q}^{9 \times 3} {8 }^{2}{n}^{4 \times 2} {q}^{2} }$

$\frac{ {4}^{2} {n}^{8} {q}^{10} (- 3) {nq}^{9}}{ {4 }^{3}{n}^{9} {q}^{27} {8 }^{2}{n}^{8} {q}^{2} }$

as when multiplying two powers that have the same base, we can add the exponents and, to divide podes with the same base, we can subtract the exponents

${4}^{2 - 3} {q}^{10 + 9 - 2 - 27} {n}^{8 + 1 - 8 - 9} {8}^{ - 2} { (- 3)}^{1}$