I got a bunch of questions under my account if someone just wants to do some work

1/5+1/7+3/21 the answer is

inessss [21]

Answer:

Step-by-step explanation:

We need to get all the fractions with a common denominator before we can solve the problem.

The only fractions that have a common multiple are 1/7 and 3/21, so we'll add these first and then add 1/5.

1/7 = 3/21

Add

$\frac{3}{21} + \frac{3}{21} =\frac{6}{21}$

6/21 = 30/105

1/5 = 21/105

$\frac{30}{105} + \frac{21}{105} = \frac{51}{105}$

Simplify

$\frac{51}{105} / \frac{3}{3} =\frac{17}{35}$

<h3>ANSWER: 17/35</h3>

I'm always happy to help :)

4 0

2 years ago

In an indirect proof, you prove an "if-then" statement is true by assuming the statement is false (stating the inverse or conver

ArbitrLikvidat [17]

If a figure has exactly three sides, then it is not a triangle.

5 0

2 years ago

Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species

kotegsom [21]

Answer:

$P(t) = \frac{160000e^{1.36t}}{2000 + 80(e^{1.36t} - 1)}$

Step-by-step explanation:

The logistic equation is the following one:

$P(t) = \frac{KP(0)e^{rt}}{K + P(0)(e^{rt} - 1)}$

In which P(t) is the size of the population after t years, K is the carrying capacity of the population, r is the decimal growth rate of the population and P(0) is the initial population of the lake.

In this problem, we have that:

Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 2,000. This means that $P(0) = 80, K = 2000$ .

The number of fish tripled in the first year. This means that $P(1) = 3P(0) = 3(80) = 240$ .

Using the equation for P(1), that is, P(t) when $t = 1$ , we find the value of r.

$P(t) = \frac{KP(0)e^{rt}}{K + P(0)(e^{rt} - 1)}$

$240 = \frac{2000*80e^{r}}{2000 + 80(e^{r} - 1)}$

$280*(2000 + 80(e^{r} - 1)) = 160000e^{r}$

$280*(2000 + 80e^{r} - 80) = 160000e^{r}$

$280*(1920 + 80e^{r}) = 160000e^{r}$

$537600 + 22400e^{r} = 160000e^{r}$

$137600e^{r} = 537600$

$e^{r} = \frac{537600}{137600}$

$e^{r} = 3.91$

Applying ln to both sides.

$\ln{e^{r}} = \ln{3.91}$

$r = 1.36$

This means that the expression for the size of the population after t years is:

$P(t) = \frac{160000e^{1.36t}}{2000 + 80(e^{1.36t} - 1)}$

4 0

3 years ago

Please help on this one?

tia_tia [17]

On the right side we have st which is the same as s*t or s times t

We want to undo the multiplication that is happening to s, so we have to divide both sides by t

d = s*t

d/t = s*t/t ..... divide both sides by t

d/t = s

s = d/t

Answer: Choice A)

3 0

3 years ago

Read 2 more answers

The quotient of (x⁴+ 5x³ 3x - 15) and (x³ - 3) is a polynomial. What is the quotient?

lesya [120]

Answer:

x+5 (under assumption you meant to do -3x

Step-by-step explanation:

you can use long division.

Take the leading coefficient x^4 and divide it by x^3. This results in x which is going to be the first part of you quotient. Now take that x and multiply it by the divisor (x^3 - 3). This gives you x(x^3 - 3) = x^4 - 3x. Now subtract that x^4 - 3x from the original polynomial and repeat this until you can't divide anymore

$x^4+5x^3-3x-15 - (x^4-3x) = 5x^3-15\\x^3/5x^3=5\\5(x^3-3) = 5x^3-15\\5x^3-15-(5x^3-15) = 0\\\\x+5$

6 0

2 years ago