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2 years ago

A grasshopper has a length of (5x - 2) inches. A spider has a length of (2x - 1) inches. How much longer is the grasshopper?

Mathematics

1 answer:

KonstantinChe [14]2 years ago

6 0

Answer:

The grasshopper is 3x-1 longer.

$(5x-2)-(2x-1)=5x-2-2x+1=3x-1$

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(5x^2+8x+18)+(11x^2-17x-7)

GuDViN [60]

Answer: 16x^2 - 9x + 11

Step-by-step explanation:

(5x^2+8x+18)+(11x^2-17x-7)

5x^2+8x+18+11x^2-17x-7

16x^2 - 9x + 11

3 0

2 years ago

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A piece of land is shaped like a right triangle. Two people start at the right angle at the same time, and walk at the same spee

sweet [91]

Answer:

1 m/s

Step-by-step explanation:

the area of the triangle = A = ¹/₂ab

where a is the distance covered by the first person and b is the distance covered by the second person. since they walk at the same speed, then a = b, so our formula can be changed to A = ¹/₂x²

now we find the derivative of both sides:

dA/dt = ¹/₂ ⋅ dx²/dt

dA/dt = ¹/₂ ⋅ 2x ⋅ dx/dt

dA/dt = x ⋅ dx/dt = x ⋅ velocity

we are told that x = 5 m, and dA/dt = 5 m²/s

velocity = dA/dt / x

velocity = 5 m²/s / 5 m = 1 m/s

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3 years ago

Which of the following is not a function<br>B is not right

FinnZ [79.3K]

Answer:

The first one

Step-by-step explanation:

They have 2 outputs for the same input

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2 years ago

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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)

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