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3 years ago

What is the range of the function in this table

Mathematics

2 answers:

professor190 [17]3 years ago

8 0

Answer:

Choice D is the correct answer.

Step-by-step explanation:

We have given a table of values of x and y.

We have to find the range of the function.

Range is defined as the set of all output values of function.

The values of y are called outputs and the values of x are called input.

Hence, the values of y are 1,4,4 and 2.

Range = {1,2,4} which is the answer.

Nuetrik [128]3 years ago

3 0

Answer:

Option D is correct

(1,2,4)

Step-by-step explanation:

Given is a table as

x y

2 1

3 4

4 4

5 2

------------------------

Here we are asked to find the range

Range is the values of y which have preimages as x

Here we find that the items y=1,4,2 have preimages in x

Hence range = (1,4,4,2)

Write in order and write only once the repeated values to get range as

Range = {1,2,4}

Option D is right

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324 soldiers queued up such that the number of queues is equal to the number of soldiers in each queue. find the number of queue

Yanka [14]

I think so answer will be 164.

8 0

3 years ago

Find the equations of the tangent lines to x2+y2=1 at x=0, x=1 and x=35.

Ivanshal [37]

Assuming the equation is x^2 + y^2 = 1, then that's a circle, with radius 1, centered on the origin [0,0].

So there are two tangents at x = 0. They are y = 1, and y = -1 (horizontal lines).

There is one tangent at x = 1. It is x = 1 (a vertical line).

There is no tangent at x = 35, because the original equation has no solution at x = 35.

4 0

3 years ago

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)

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