All categories

3 years ago

Add brackets to make this statement correct.. 12 - 2 x 3 + 1 = 4

Mathematics

2 answers:

Talja [164]3 years ago

3 0

Answer:

$12 - (2 \times 3) + 1 = 4$

LekaFEV [45]3 years ago

3 0

Answer:

12 - (2x3) +1=4

Step-by-step explanation:

Sana makatulong

You might be interested in

Last question on the quiz

Sveta_85 [38]

Answer: 152 degrees.

Step-by-step explanation: Since angle 1 and angle 5 are alternate exterior angles, they are congruent. Thus, angle 5 is also equal to 152 degrees.

6 0

3 years ago

Population Growth A lake is stocked with 500 fish, and their population increases according to the logistic curve where t is mea

Alexus [3.1K]

Answer:

a) Figure attached

b) For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

c) $p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}$

And if we find the derivate when t=1 we got this:

$p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114$

And if we replace t=10 we got:

$p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404$

d) $0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And then:

$0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)$

$0 =19e^{-\frac{t}{5}} -1$

$ln(\frac{1}{19}) = -\frac{t}{5}$

$t = -5 ln (\frac{1}{19}) =14.722$

Step-by-step explanation:

Assuming this complete problem: "A lake is stocked with 500 fish, and the population increases according to the logistic curve p(t) = 10000 / 1 + 19e^-t/5 where t is measured in months. (a) Use a graphing utility to graph the function. (b) What is the limiting size of the fish population? (c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months? (d) After how many months is the population increasing most rapidly?"

Solution to the problem

We have the following function

$P(t)=\frac{10000}{1 +19e^{-\frac{t}{5}}}$

(a) Use a graphing utility to graph the function.

If we use desmos we got the figure attached.

(b) What is the limiting size of the fish population?

For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

(c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months?

For this case we need to calculate the derivate of the function. And we need to use the derivate of a quotient and we got this:

$p'(t) = \frac{0 - 10000 *(-\frac{19}{5}) e^{-\frac{t}{5}}}{(1+e^{-\frac{t}{5}})^2}$

And if we simplify we got this:

$p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}$

And if we simplify we got:

$p'(t) =\frac{38000 e^{-\frac{t}{5}}}{(1+19e^{-\frac{t}{5}})^2}$

And if we find the derivate when t=1 we got this:

$p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114$

And if we replace t=10 we got:

$p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404$

(d) After how many months is the population increasing most rapidly?

For this case we need to find the second derivate, set equal to 0 and then solve for t. The second derivate is given by:

$p''(t) = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And if we set equal to 0 we got:

$0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And then:

$0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)$

$0 =19e^{-\frac{t}{5}} -1$

$ln(\frac{1}{19}) = -\frac{t}{5}$

$t = -5 ln (\frac{1}{19}) =14.722$

7 0

3 years ago

Hazel has band practice every other day and volunteers every seven days. if she had a band practice and volunteered on oct 23, w

LuckyWell [14K]

Cjdfb hjkbftbbui yuftg

3 0

3 years ago

17. Evaluate each pair of expressions. a. (-3)-8 and -3-8 b. (-3)-9 and -3-9

Semmy [17]

A = -11
B= -12

This is because of how calculations work in negatives. Because they are both negative they work similarly to addition

5 0

3 years ago

Find the sum of the first 12 terms of the sequence. Show all work for full credit. 1, -4, -9, -14, . . . (2 points)

My name is Ann [436]

Answer:

$S_n=$ -348

Step-by-step explanation:

We are given the following arithmetic sequence and we are to find the sum of its first 12 terms:

1, -4, -9, -14, . . .

For that, we will use the formula for the sum of the arithmetic mean:

$S_n=\frac{n}{2} (a_1+a_n)$

We know the value of the first term ( $a_n$ ) but we need to find the value of $a_{12}$ . So we will use the following formula:

$a_{12}=a_1+(n-1)d$

$a_{12}=(-4)+(12-1)(5)$

$a_{12}=-59$

Substituting these values in the sum formula to get:

$S_n=\frac{12}{2} (1+(-59))$

$S_n=$ -348

4 0

3 years ago

Read 2 more answers