What is the standard form to number 3

HELP ASAP!!!!!! 34 ÷ (14 − 5) × 2

DIA [1.3K]

Answer:

B

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

A student concluded that

quester [9]

Answer:

#1 is C)

#2 is b)

#3 is C)

#4 is B)

#5 is c)

here's the whole quiz brainlist!!!!

Step-by-step explanation:

1. Which statement about x in the equation a - x = 2a is true?

A. x is equal to twice a

B. x must be greater than a

C. x must be equal to the opposite of a

D. x does not have any real number values

2. A student concluded that 8x - 12 = 4( 1/2x - 6 ) has infinity many solutions. Which statement best describes the students conclusion?

A. The conclusion is incorrect because the equation has no solutions

B. The conclusion is incorrect because there is exactly one solution to the equation

C. The conclusion is correct because there are exactly two solutions to the equation

D. The conclusion is correct because when simplified, both sides of the equation are equivalent

3. A student conducted that 0.5( 6x + 4 ) = 3x + 4 has no solution. Which statement best describes the student conclusion?

A. The conclusion is incorrect because there are two solutions to the equation

B. The conclusion is incorrect because there is exactly one solution to the equation

C. The conclusion is correct because the coefficient before the variable is equivalent

D. The conclusion is correct because, when simplified, both sides of the equation are equivalent

4. Three times the difference of a number x and seven is twenty-three minus the sum of three times a number x and two. What is the value of x?

A. 5

B. 7

C. No solution

D. Infinitely many solutions

5. A piece of wood that measures 175 in. long was cut into three pieces. The second piece of wood is 3 in. more than 3 times the length of the first piece of wood. The third piece of wood is 8 in. less than 4 times the length of the first piece of wood. What is the length of the longest piece of wood?

A. 70 in.

B. 77 in.

C. 82 in.

D. 85 in.

7 0

3 years ago

100 POINTS IF YOU ANSWER CORRECTLY Rotate these coordinates 180 around the origin and then translate 6 units to the left.

snow_tiger [21]

Original Coordinates: (-2, -8) (-6, -8) (-11, -1) (-7, -3)

A 180 degree rotation around the origin flips the signs of x and y, so you get:

(2, 8) (6, 8) (11, 1) (7, 3)

Then a translation to the left 6 units just subtracts 6 from all the x values:

(-4, 8) (0, 8) (5, 1) (1, 3)

Those are your points! Hope this helped!

8 0

4 years ago

Read 2 more answers

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

The class collected $1690, if there were only 13 students that collected money,what is the average amount each student collected

Svet_ta [14]

Divide 1690 by 13, and you get 130, so $130.

5 0

4 years ago