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disa [49]
2 years ago
10

Determine the perimeter of a rectangle that is 10 inches long and 9 inches wide. inches

Mathematics
2 answers:
Vikki [24]2 years ago
7 0
Perimeter formula: l + w + l +w. L represents length and w represents width. So the equation is 10 + 9 + 10+9. The answer is 58 inches. Oh
Alex777 [14]2 years ago
4 0

Answer:

38

Step-by-step explanation:

2(l + b)

2(10 + 9)

2(19)

2*19

38

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What is the value of x in the equation 6(х + 1) — 5х = 8 + 2(x — 1)?
Licemer1 [7]

Multiply the first bracket by 6.

Multiply the second bracket by 8

6(x+1)-5x= 8+2(x-1)

6(x)+6(1)-5x= 8+2(x)+2(-1)

6x+6-5x= 8+2x-2

6x-5x+6= 8-2+2x

x+6= 6+2x

Move +2x to the other side. Sign changes from +2x to -2x.

x-2x+6= 6+2x-2x

-x+6= 6

Move +6 to the other side. Sign changes from +6 to -6.

-x+6-6= 6-6

-x=0

Multiply by -1 for -x and 0

-x(-1)= 0(-1)

x= 0

Answer: x= 0

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
2 years ago
Tim has $20 to buy snacks for 12 people in an office. Each person will get one snack. Tim is buying bags of pretzels that cost $
Greeley [361]

Answer:

  8 bags of pretzels

Step-by-step explanation:

Let x represent the number of bags of pretzels Tim buys. We assume he spends exactly $20 on exactly 12 bags of snack food. Then his purchase is ...

  1.50x + 2.00(12-x) = 20.00

  -0.50x +24.00 = 20.00 . . . eliminate parentheses, collect terms

  -0.50x = -4.00 . . . . . . . . . . .subtract 24

  x = 8 . . . . . . . . . . divide by -0.50

Tim will buy 8 bags of pretzels.

4 0
3 years ago
Can someone plzz help me with these 2 questions. thanks!
Deffense [45]

9514 1404 393

Answer:

  4. height: 5√3 cm; area: 25√3 cm²

  5. 30√2 ft ≈ 42.43 ft

Step-by-step explanation:

4. The height of an equilateral triangle is (√3)/2 times the side length, so is ...

  height = (√3)/2 × (10 cm) = 5√3 cm

The area is given by the formula ...

  A = 1/2bh

  A = 1/2(10 cm)(5√3 cm) = 25√3 cm²

__

5. The diagonal of a square is √2 times the side length, so the distance from 3rd to 1st base is ...

  (30 ft)√2 ≈ 42.43 ft

__

The length of the diagonal of a square is something you should be familiar with. If you're not, you can figure the distance using the Pythagorean theorem.

  d = √(30² +30²) = √1800 ≈ 42.43 . . . feet

6 0
2 years ago
A + x + p + 8 = r<br> solve for x<br><br> all help is appreciated!!
Norma-Jean [14]

a+x+p+8 =r

(to solve isolate x)

x=r-(a+p+8)

x= r-a-p-8

8 0
3 years ago
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