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3 years ago

Four big water bottles can hold 8 gallons of water. How much water can ten big water bottles hold?

Mathematics

2 answers:

denis-greek [22]3 years ago

6 0

Ten big water bottles can hold 18 gallons of water

professor190 [17]3 years ago

5 0

Ten big water bottles can hold 20 gallons of water.

When writing unit rates, y goes over x. However, in the problem above, x represents the water bottles and y represents the gallons of water.

4/8 (x, y) -> 8/4 (y, x)

8/4 \ 4/4 = 2/1

Now, you know the constant rate of change. There are 2 gallons of water for every 1 big water bottle. Simply multiply 2 by 10.

Since you are looking for how many gallons of water in ten water bottles, multiply 2 gallons of water (I assume you now know that there are 2 gallons of water in every 1 big bottle) by the number of water bottles. In this case, 10.

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i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.

Natali [406]

Answer:

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = -5\sqrt 2$

Step-by-step explanation:

Given

$N = (-7,-1)$ --- terminal side of $\varnothing$

Required

Determine the values of trigonometric functions of $\varnothing$ .

For $\varnothing$ , the trigonometry ratios are:

$sin\ \varnothing = \frac{y}{r}$ $cos\ \varnothing = \frac{x}{r}$ $tan\ \varnothing = \frac{y}{x}$

$cot\ \varnothing = \frac{x}{y}$ $sec\ \varnothing = \frac{r}{x}$ $csc\ \varnothing = \frac{r}{y}$

Where:

$r^2 = x^2 + y^2$

$r = \sqrt{x^2 + y^2$

In $N = (-7,-1)$

$x = -7$ and $y = -1$

So:

$r = \sqrt{(-7)^2 + (-1)^2$

$r = \sqrt{50$

$r = \sqrt{25 * 2$

$r = \sqrt{25} * \sqrt 2$

$r = 5 * \sqrt 2$

$r = 5 \sqrt 2$

<u>Solving the trigonometry functions</u>

$sin\ \varnothing = \frac{y}{r}$

$sin\ \varnothing = \frac{-1}{5\sqrt 2}$

Rationalize:

$sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$sin\ \varnothing = \frac{-\sqrt 2}{5*2}$

$sin\ \varnothing = \frac{-\sqrt 2}{10}$

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{x}{r}$

$cos\ \varnothing = \frac{-7}{5\sqrt 2}$

Rationalize

$cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}$

$cos\ \varnothing = \frac{-7\sqrt 2}{10}$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{y}{x}$

$tan\ \varnothing = \frac{-1}{-7}$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = \frac{x}{y}$

$cot\ \varnothing = \frac{-7}{-1}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{r}{x}$

$sec\ \varnothing = \frac{5\sqrt 2}{-7}$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = \frac{r}{y}$

$csc\ \varnothing = \frac{5\sqrt 2}{-1}$

$csc\ \varnothing = -5\sqrt 2$

3 0

3 years ago

A cube with each face a different colour will fit exactly into a box in which it is packaged. In how many ways can the cube be p

dmitriy555 [2]

Answer:

362880 times

I hope this helps

4 0

2 years ago

At noon, ship A is 180 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 25 km/h. How fast is

weeeeeb [17]

Answer:

Step-by-step explanation:

Gotta love these motion problems!!! This one is especially tricky! At noon, ship A is 180 km west of ship B. If ship A is traveling east, it is closing its distance to ship B (if ship B didn't move). But ship B is moving north at the same time. We need to find the distance each can travel in 4 hours using the d = rt formula. For ship A. We know it's moving at 35 km/h, so in 4 hours it can move d = 35(4) which is 140 km. Remember that is closing the distance to ship B. So it is 180 - 140 directly south of ship B. This problem will turn out to be a right triangle problem of sorts. The distance of 40 km serves as the base of this right triangle.

With ship B moving north at 25 km/hr, it can travel d = 25(4) which is 100 km. This serves as the height of the triangle. We are asked to find the rate at which the distance is changing 4 hours from their starting points. We know how far each can travel in those 4 hours, so in order to find the rate of change (which is the same thing as the slope!), we take the height and divide it by the base because that is the same thing as taking the rise over the run. 100/40 is 10/4 which is a rate of change of 2.5 km/hr

5 0

3 years ago

Mi primo llego de estados unidos y me regalo 24 dolares si el dolar esta a 14.65,?cuantos equivale aproximadamente en pesos es c

Norma-Jean [14]

Yo hablo un poco español, pero el solucion esta b) 351.6 pesos. 24x14.65 = 351.6

4 0

3 years ago

What is the distance between (-3, -4) and (2,2)

LenKa [72]

Answer:

I did for you but I don't know you will understand my HANDWRITING or not .