The Pythagorean Theorem and Irrational numbers

(1 point) A fish tank initially contains 15 liters of pure water. Brine of constant, but unknown, concentration of salt is flowi

Drupady [299]

Answer:

a. $\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c$

b. $x(t) = 6\frac{2}{3} \cdot c \cdot t$

c. $c = \dfrac{3}{8} \ g/L$

Step-by-step explanation:

a. The volume of water initially in the fish tank = 15 liters

The volume of brine added per minute = 5 liters per minute

The rate at which the mixture is drained = 5 liters per minute

The amount of salt in the fish tank after t minutes = x

Where the volume of water with x grams of salt = 15 liters

dx = (5·c - 5·c/3)×dt = 20/3·c = $6\frac{2}{3} \cdot c \cdot dt$

$\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c$

b. The amount of salt, x after t minutes is given by the relation

$\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c$

$dx = 6\frac{2}{3} \cdot c \cdot dt$

$x(t) = \int\limits \, dx = \int\limits \left ( 6\frac{2}{3} \cdot c \right) \cdot dt$

$x(t) = 6\frac{2}{3} \cdot c \cdot t$

c. Given that in 10 minutes, the amount of salt in the tank = 25 grams, and the volume is 15 liters, we have;

$x(10) = 25 \ grams(15 \ in \ liters) = 6\frac{2}{3} \times c \times 10$

$6\frac{2}{3} \times c =\dfrac{25 \ grams }{10}$

$c =\dfrac{25 \ g/L }{10 \times 6\frac{2}{3} } = \dfrac{25 \ g/L}{10 \times \dfrac{20}{3} } =\dfrac{3}{200} \times 25 \ g/L= \dfrac{75}{200} \ g/L = \dfrac{3}{8} \ g/L$

$c = \dfrac{3}{8} \ g/L$

4 0

2 years ago

The normal distribution is:

Alex_Xolod [135]

Answer:

The normal distribution is a probability function that describes how the values of a variable are distributed. It is a symmetric distribution where most of the observations cluster around the central peak and the probabilities for values further away from the mean taper off equally in both directions.

Step-by-step explanation:

8 0

2 years ago

One-third of the students at the local community college live in the county. If there are 2037 students enrolled, how many do no

densk [106]

Answer:

ow many students are enrolled in community colleges? ... How does the population of community college students break down by race/ethnicity and income? ... What percentage of community college students obtain a college credential? ... How many community college students complete their remedial requirements?

Step-by-step explanation:

Provisional National Center for Education Statistics (NCES) data on unduplicated year-round enrollment (as opposed to fall enrollment only) indicate that 8.2 million undergraduates were enrolled in public two-year colleges in 2018–19.

In fall 2018, 5.6 million students were enrolled in public, two-year colleges. About 2 million were full-time students, and 3.6 million were part-time.

The National Student Clearinghouse Research Center estimates that 5.4 million students were enrolled in public two-year colleges in fall 2019, down 1.4% from fall 2018.

However, these studies underestimate the number of community college students, as about 100 community colleges offer a small number of bachelor’s degree programs and thus are listed in NCES as four-year institutions. According to a CCRC analysis correcting for this misclassification, 6.7 million students were enrolled at community colleges in fall 2017, and nearly 10 million students enrolled at a community college at some point during the 2017-18 academic year.

7 0

2 years ago

A person standing close to the edge on top of a 24-foot building throws a ball vertically upward. The quadratic function h(t)=−1

Nutka1998 [239]

Answer:

a) maximum height is 156.25 ft

b) 6 seconds

Step-by-step explanation:

a) The maximum height can be found by finding the y-coordinate of the vertex. The vertex is the highest point in a parabola opened downward.

I start with by finding the t-coordinate of the vertex.

The t-coordinate of the vertex is $\frac{-b}{2a}$ where the expression $-16t^2+92t+24$ will need to be compared to $at^2+bt+c$ .

We see that $a=-16,b=92,c=24$ .

So the t-coordinate of the vertex is $\frac{-b}{2a}=\frac{-92}{2(-16)}=\frac{-92}{-32}=\frac{23}{8}$ .

We can find the h, the height, that corresponds to this t by using the equation $h=-16t^2+92t+24$ where $t=\frac{23}{8}$ .

So inserting 23/8 for $t$ .

$-16(23/8)^2+92(23/8)+24$

Plugging into calculator gives you 625/4.

So the maximum height is 625/4 or 156.25.

b) Let's find how long it takes the ball to hit the ground. If the ball is on the ground, then the distance between the ball and the ground is 0. So we are looking to solve h(t)=0 for t.

So this is equation we are solving for t:

$-16t^2+92t+24=0$

These numbers are big but since all the terms have a common factor we can make them slightly smaller.

That is, we are going to divide both sides by -4 and see

$4t^2-23t-6=0$

It looks like it could be possible to factor.

a=4

b=-23

c=-6

We need to find two numbers that multiply to be ac and add up to be b.

a*c=4(-6)=-24

b=-23

So -24=-24*1 and -23=-24+1.

We are going to replace -23t with -24t+1t giving up something that should be factorable by grouping.

$4t^2-23t-6=0$

$4t^2-24t+1t-6=0$

Group the first 2 together and group the last two together:

$(4t^2-24t)+(1t-6)=0$

Factor what you can from each pair:

$4t(t-6)+1(t-6)=0$

Now each term has a common factor of (t-6) so factor that out giving you:

$(t-6)(4t+1)=0$

Set both factors equal to 0 giving you

t-6=0 or 4t+1=0

t=6 or 4t =-1

t=6 or t=-1/4

So it hit the ground in 6 seconds.

8 0

2 years ago

The length of a shadow from the vertical pole (see picture to the right), which is 7m high, is 4m. What is the degree measure of

Liono4ka [1.6K]

Answer:

The answer to your question is: 60.3°

Step-by-step explanation:

Data

height = 7 m

shadow = 4 m

angle = ?

Process

tangent Ф = opposite side / adjacent side

tangent Ф = 7 / 4

tan Ф = 1.75

Ф = 60.3°

6 0

2 years ago