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SIZIF [17.4K]
3 years ago
9

Find the first five terms of the sequence described.

Mathematics
1 answer:
kipiarov [429]3 years ago
5 0

Answer:

I dont think i dont know this but its 10?

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In what quadrant is the terminal side<br> of -323°?
dexar [7]

Answer:

Quadrant 4 i believe plz tell me if im wrong

Step-by-step explanatio

6 0
3 years ago
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Find the area of this circle. Use 3 for a.A = 7r2=11 in[?] in?
liberstina [14]

SOLUTION

We want to find the area of the circle in the picture given.

We have been given the formula as

\begin{gathered} A=\pi r^2 \\ We\text{ are told to take }\pi\text{ as 3} \\  \end{gathered}

From the circle, the radius r = 11 in.

The area of the circle becomes

\begin{gathered} A=\pi r^2 \\ A=\pi\times r^2 \\ A=3\times11^2 \\ A=3\times11\times11 \\ A=363in^2 \end{gathered}

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7 0
1 year ago
What’s the area of the trapezoid?
myrzilka [38]

Answer:

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Step-by-step explanation:

2x2=4

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7 0
3 years ago
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You have 42,784 grams of a radioactive kind of curium. If its half-life is 18 years, how much will be left after 72 years?
s344n2d4d5 [400]

Answer:

2,674.14 g

Step-by-step explanation:

Recall that the formula for radioactive decay is

N = N₀ e^(-λt)

where,

N is the amount left at time t

N₀ is the initial amount when t=0, (given as 42,784 g)

λ = coefficient of radioactive decay

  = 0.693 ÷ Half Life

  = 0.693 ÷ 18

  = 0.0385

t = time elapsed (given as 72 years)

e = exponential constant ( approx 2.7183)

If we substitute these into our equation:

N = N₀ e^(-λt)

= (42,787) (2.7183)^[(-0.0385)(72)]

= (42,787) (2.7183)^(-2.7726)

=  (42,787) (0.0625)

= 2,674.14 g

6 0
3 years ago
Really need help!!! quadratic equations UGH
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4 0
3 years ago
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