hectictar

5% of the original population = 7

0.05p = 7

p = 7 / 0.05 = 140

The original population = 140 people. 7 people died. So the current population = 140 - 7 = 133

The question is: what is 150% of 4 ?

150% of 4 = 150% * 4 = 1.5 * 4 = 6   (miles)

To solve for t...

400,000 = 2^t/5

Take the log of both sides.

log(400,000) = log(2^t/5)

Rewrite as

log(400,000) = (t/5) * log(2)

Divide both sides by log(2).

log(400,000) / log(2) = t/5

Multiply both sides by 5.

5log(400,000) / log(2) = t

93.048 ≈ t

mass of first box + mass of second box = total mass of the two boxes

9 kilograms + 7 kilograms = 16 kilograms

16 kilograms = 16 × 1000 grams = 16 000 grams

x² - 48x - 2 = (x + a)(x +b)

a + b = -48 → b = -48 - a

ab = -2

a(-48 - a) = -2

-48a - a² = -2

a² + 48a = 2

a² + 48a + 576 = 2 + 576

(a + 24)² = 578

a = sqrt(578) - 24 = 17sqrt(2) - 24

b = -17sqrt(2) - 24

(x + 17sqrt(2) - 24)(x - 17sqrt(2) - 24)

Here's a third explanation lol 

How many pairs of apples are there in a batch of 10 apples? 10 ÷ 2 = 5 pairs of apples

How many \(\frac12\times\frac12=\frac14\) square feet tiles are there in \( 11\frac12\times9\frac12 = \frac{437}{4} \) square feet?

\(\frac{437}{4} \div \frac14 =\frac{437}{4}\times\frac41=437\) one-fourth square feet, or tiles

Triangle ABC is a 45º 45º 90º triangle where the side across from the 90º angle is \(6\sqrt2\)

So the sides across from the 45º angles are \( \frac{6\sqrt2}{\sqrt2}=6\)

BC = 6

Triangle BCD is a 30º 60º 90º triangle where the side across from the 90º angle is 6

So the side across from the 30º angle is 6/2 = 3

Type it in like this: " log(x,b) ",

where b is the base and x is the thing you want the log_b of.

So, If you wanted to do

log₅ 2   type: " log(2,5) "

volume of cone = (1/3) * (π * radius²) * (height)

radius = 8/2 = 4  cm

height = 7  cm

Plug in:

volume of cone = (1/3) * (π * 4²) * 7 = (1/3)(16π)(7) = 112π / 3 ≈ 117.286 cm³

I have never used it before either! I just googled a way to do it and that was one way I found :)

I also have tried to figure out a way to make arcs but I couldn't.

\(\stackrel{\cap}{AB}\qquad \leftarrow\) That's about as close as I can get...Lol !