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3 years ago

ANYONE??

Mathematics

1 answer:

RideAnS [48]3 years ago

5 0

Volume of a quadrangular pyramid=(1/3)bh
b=base
h=height

b=area of the base=area of a square=8.4 ft * 8.4 ft=70.56 ft²

Pythagoras theorem:
hypotenuse²=leg₁² + leg₂²

data:
hypotenuse=9.6 ft
leg₁=height=h
leg₂=8.4 ft /2=4.2 ft

(9.6 ft)²= h² + (4.2 ft)²
92.16 ft²=h²+17.64 ft²
h²=92.16 ft²-17.64 ft²
h²=74.52 ft²
h=√(74.52 ft²)=8.63 ft.

Volume of this quadrangular pyramid=(1/3)(70.56 ft²)(8.63 ft)=202.9 ft³≈202.3 ft³

Answer: ≈202.3 ft³

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Step-by-step explanation:

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For continuous compounding the number of compounding periods, $n$ , becomes infinitely large.

Therefore, the formula as we discussed above would become:

$P\left(t\right)=P_0\times e^{\left(i\:\times \:t\right)}$

Part c) How much money will Natalie have after 2 years?

Using the formula

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dmitriy555 [2]

Given:

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$\dfrac{7}{9},\dfrac{-7}{3},7,-21,63,...$

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The 9th term of the given geometric sequence.

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We have,

$\dfrac{7}{9},\dfrac{-7}{3},7,-21,63,...$

Here, the first term is:

$a=\dfrac{7}{9}$

The common ratio is:

$r=\dfrac{a_2}{a_1}$

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The nth term of a geometric sequence is:

$a_n=ar^{n-1}$

Where, a is the first term and r is the common ratio.

Substitute $a=\dfrac{7}{9},r=-3,n=9$ to find the 9th term.

$a_9=\dfrac{7}{9}(-3)^{9-1}$

$a_9=\dfrac{7}{9}(-3)^{8}$

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