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VMariaS [17]
3 years ago
12

The diagram shows the universal set U = {parallelograms}. Set A represents parallelograms with four congruent sides, while Set B

represents parallelograms with four congruent angles. How many of the parallelograms fall into the category A B?

Mathematics
2 answers:
poizon [28]3 years ago
6 0

Answer:

Step-by-step explanation:

The Answer is 26 on e2020

alexdok [17]3 years ago
5 0

Answer:

U  ={ Parallelograms}

A= { Parallelogram with four congruent sides}={ Rhombus,Square}

B ={ Parallelograms with four congruent angles} ={ Rectangle, Square}

So, AB= { Square}

So among all the parallelograms "Square" is the only parallelogram which has all congruent sides as well as all congruent angles.



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What is the domain of the function graphed below?
Vlad1618 [11]

Answer:

The domain is the set of x values for which the function resides in.

In this graph, it is unclear as to if the function goes on to infinity, but if it does, the answer is quite easily:

(-2, 4] and [7, ∞)

since the function starts on the left at -2 then continues to the right, with a pause, then indefinitely after.

3 0
2 years ago
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Reuben is making a replica of the US flag for his school project.
Leno4ka [110]
P1=2.5+2.5+4.75+4.75=14.5 cm

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2.5 cm - 5 in
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8 0
3 years ago
Mile is obsessed with her new book she started. She read 185 pages in 2 days. If she keeps up this pace about how many pages wil
inna [77]

Answer:

462.5 pages

Step-by-step explanation:

You find out how many books she reads in 1 day then you can multiply it by 5 to get 5 days.

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8 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

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dA/(L - Ak) = dt

Integrating, we have

∫-kdA/-k(L - Ak) = ∫dt

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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)

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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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lina2011 [118]

Answer:

5

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6 0
3 years ago
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