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2 years ago

(x² + 6x − 16)(x − 2) = ax³ + bx² + cx + d

Mathematics

1 answer:

stellarik [79]2 years ago

3 0

B is the right answer to this question

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Please help me I really need help

ElenaW [278]

The answer is B. 83,425

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2 years ago

If A(x,1),B(-3,7), C(-5,9), and D(5,4) find the value of x so that AB||CD

Mashutka [201]

I’m not sure sorryyyy hope someone helps

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2 years ago

The total number of fungal spores can be found using an infinite geometric series where a1 = 8 and the common ratio is 4. Find t

GalinKa [24]

Answer:

$S_{n} = \frac{2(4^{n}-1) }3} if r>1$

Step-by-step explanation:

Explanation:-

Geometric sequence:-

The geometric sequence is of the form

$a,ar,ar^{2},ar^{3}$ ……… be an infinite sequence

here first term is 'a' and ratio is 'r '

In this geometric sequence 'n' t h term is $t_{n} = a r^{n-1}$ ....(1)

Given data a1 term is '8 ' and ratio is '4'

substitute n=1 in equation(1)

$t_{1} = a r^{1-1}$ = 8

$ar=8$ ...........(2)

substitute r= 4 in equation(2)

now we get a(4)=8

dividing "4" on both sides , we get a = 2

Geometric series:-

$a+ar+ar^{2}+ar^{3}+$ .....is an infinite geometric series.

sum of this infinite series will be the upper limit of the fungal spores

that is we have to find sum of infinite series

$S_{n} = \frac{a(r^{n}-1) }{r-1} if r>1$

$S_{n} = \frac{2(4^{n}-1) }{4-1} if r>1$

$S_{n} = \frac{2(4^{n}-1) }3} if r>1$

Final answer:-

sum of this infinite series will be the upper limit of the fungal spores

$S_{n} = \frac{2(4^{n}-1) }3} if r>1$

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3 years ago

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worty [1.4K]

Answer:

Step-by-step explanation:

angles A & L are not equal

sides are not proportional 4 ≠ 5.6

6 8

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3 years ago

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The answer to the question

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3 years ago