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3 years ago

What is the value of the fourth term in a geometric sequence for which a=15 and r= 1/3

Mathematics

1 answer:

Andreyy893 years ago

6 0

Answer:

Step-by-step explanation:

hello :

the n-ieme term is : An=A1×r^(n-1)

A1 the first term r : the common ratio

in this exercice : A1 =15 r = 1/3 n = 4

A4=15×(1/3)^(4) =15×4^4 =3840

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The sum of three consecutive even integers is -78. What is the smallest integer? Show all working outs (5)

svet-max [94.6K]

Answer:

The smallest integer is -27 and the integers are -27, -26, and -25

Step-by-step explanation:

We can represent this with:

x + x - 1 + x - 2 = -78

Simplify.

3x - 3 = -78

Add 3 to both sides.

3x = -75

Divide both sides by 3.

x = -25

So, the first integer is -25.

Subtract two to get the smallest integer.

-25 - 2 = - 27

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3 years ago

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3 years ago

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poizon [28]

Answer:

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Step-by-step explanation:

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3 years ago

−1/3−(−1/2)

Radda [10]

Hi ;-)

$-\frac{1}{3}-(-\frac{1}{2})=-\frac{1}{3}+\frac{1}{2}=-\frac{2}{6}+\frac{3}{6}=\frac{3}{6}-\frac{2}{6}=\frac{1}{6}\\\\3\frac{1}{3}-5=-(5-3\frac{1}{3})=-(4\frac{3}{3}-3\frac{1}{3})=-1\frac{2}{3}\\\\-1\frac{4}{5}-(-2\frac{7}{8})=-1\frac{4}{5}+2\frac{7}{8}=-1\frac{32}{40}+2\frac{35}{40}=2\frac{35}{40}-1\frac{32}{40}=1\frac{3}{40}\\\\-3\frac{3}{8}-\frac{7}{8}=-(3\frac{3}{8}+\frac{7}{8})=-3\frac{10}{8}=-3\frac{5}{4}=-4\frac{1}{4}\\\\4\frac{3}{4}-(-1\frac{1}{12})=4\frac{9}{12}+1\frac{1}{12}=5\frac{10}{12}$

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2 years ago

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The answer is B.

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3 years ago

Read 2 more answers