All categories

2 years ago

Please help me find what “a” is!!!

Mathematics

1 answer:

Len [333]2 years ago

3 0

Answer:

Step-by-step explanation:

Use the Pythagorean Theorem which states

a² + b² = c²

then plug in your given terms

a² + 5.6² = 10.6²

find the squares

a² + 31.36 = 112.36

subtract 31.36 from 112.36 to find your a²

a² = 81

then find the square root of 81, which is 9 becuase 9 x 9 is 81

You might be interested in

All books in a store are being discounted by 40%. let x represent the regular price of any book in the store which expression de

antoniya [11.8K]

To find the sale price we can set up a proportion. 25/100 = x/48. We keep the 100 and 48 both on the bottom of the fraction since they represent the "whole". x represents the discount off the full price of the dress. To solve this proportion, we cross multiply, yielding 48 * 25 = 100x. Alternatively, if you notice that 25/100 ...

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

A family has two cars. the first car has a fuel efficiency of 30 miles per gallon of gas and the second has a fuel efficiency of

jeka57 [31]

X : the gas used by first car in 1 particular week
y : the gas used by second car in 1 particular week
$x + y = 60 \\ x = 60 - y$
$30x + 35y = 1975 \\ 30 \times (60 - y) + 35y = 1975 \\ ... \\ y = 35 \\ x = 60 - y = 60 - 35 = 25$

5 0

3 years ago

A rectangle is 6m wide and 11m long. How long is the diagonal of the rectangle?

Svet_ta [14]

Answer:

≈12.53

Step-by-step explanation:

7 0

3 years ago

Check image please help

Stells [14]

Answer:

Step-by-step explanation:

"The total number of dollars in fives and tens" tells you that these denominations are being added together. Because a five dollar bill is worth $5 and a ten dollar bill is worth $10, then the number of $5 bills is represented by 5f and a the number of $10 bills is represented by 10t. Therefore,

T = 5f + 10t

3 0

2 years ago