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3 years ago

56x+34y=4376 what is x and y

Mathematics

1 answer:

Alja [10]3 years ago

4 0

Answer:

x and y are the input and output of the function. The graph attached is all the values of x and y. For a specific value, more information is needed.

Step-by-step explanation:

56x+34y=4376 is a linear function with two variables. It represents the rule between inputs (x) and outputs (y). To find all its values of x and y, graph the equation.

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Aidan has 20 ft of fence with which to build a rectangular dog run. What is the maximum area he can enclose? Enter your answer i

irina1246 [14]

<h2>Maximum area is 25 m²</h2>

Explanation:

Let L be the length and W be the width.

Aidan has 20 ft of fence with which to build a rectangular dog run.

Fencing = 2L + 2W = 20 ft

L + W = 10

W = 10 - L

We need to find what is the largest area that can be enclosed.

Area = Length x Width

A = LW

A = L x (10-L) = 10 L - L²

For maximum area differential is zero

So we have

dA = 0

10 - 2 L = 0

L = 5 m

W = 10 - 5 = 5 m

Area = 5 x 5 = 25 m²

Maximum area is 25 m²

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2 years ago

How do I prepare for the PSAT? Every time I take a practice test, I get stuck on a problem that is within my topic range, but fo

melamori03 [73]

Answer:

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Step-by-step explanation:

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2 years ago

If S-1=10, what is s?

allsm [11]

Answer:

Step-by-step explanation:

This is because 11-1=10

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2 years ago

Read 2 more answers

Litter such as leaves falls to the forest floor, where the action of insects and bacteria initiates the decay process. Let A be

Travka [436]

Answer:

D = L/k

Step-by-step explanation:

Since A represents the amount of litter present in grams per square meter as a function of time in years, the net rate of litter present is

dA/dt = in flow - out flow

Since litter falls at a constant rate of L grams per square meter per year, in flow = L

Since litter decays at a constant proportional rate of k per year, the total amount of litter decay per square meter per year is A × k = Ak = out flow

So,

dA/dt = in flow - out flow

dA/dt = L - Ak

Separating the variables, we have

dA/(L - Ak) = dt

Integrating, we have

∫-kdA/-k(L - Ak) = ∫dt

1/k∫-kdA/(L - Ak) = ∫dt

1/k㏑(L - Ak) = t + C

㏑(L - Ak) = kt + kC

㏑(L - Ak) = kt + C' (C' = kC)

taking exponents of both sides, we have

$L - Ak = e^{kt + C'} \\L - Ak = e^{kt}e^{C'}\\L - Ak = C"e^{kt} (C" = e^{C'} )\\Ak = L - C"e^{kt}\\A = \frac{L}{k} - \frac{C"}{k} e^{kt}$

When t = 0, A(0) = 0 (since the forest floor is initially clear)

$A = \frac{L}{k} - \frac{C"}{k} e^{kt}\\0 = \frac{L}{k} - \frac{C"}{k} e^{k0}\\0 = \frac{L}{k} - \frac{C"}{k} e^{0}\\\frac{L}{k} = \frac{C"}{k} \\C" = L$

$A = \frac{L}{k} - \frac{L}{k} e^{kt}$

So, D = R - A =

$D = \frac{L}{k} - \frac{L}{k} - \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{kt}$

when t = 0(at initial time), the initial value of D =

$D = \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{k0}\\D = \frac{L}{k} e^{0}\\D = \frac{L}{k}$

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2 years ago

A doctor studies the known cancer patients in a certain town. The probability that a randomly chosen resident has cancer is foun

Alex777 [14]

Let the event 'has cancer' be C, and the event 'works for Ajax' be A.
P(C) = 0.0019. P(A) = 0.28
The events A and C are independent if P(C|A) = P(C).
P(C|A) = 0.0019 = P(C).
Therefore the answer is YES.

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3 years ago