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Grace [21]
3 years ago
8

Find the common ratio of the geometric sequence: −320,80,−20,5,…−320,80,−20,5,… A. −14−14 B. 1414 C. -4 D. 4

Mathematics
1 answer:
vredina [299]3 years ago
3 0
I'm assuming the sequence is: -320, 80, -20, 5

Divide each term by the previous term to get the common ratio r

r = (term 2)/(term 1) = 80/(-320) = -1/4
r = (term 3)/(term 2) = -20/80 = -1/4
r = (term 4)/(term 3) = 5/(-20) = -1/4

No matter which term you pick (as long as it's not the first one) dividing it by the previous term leads to -1/4 or -0.25 as the common ratio

Answer: -1/4

It seems like none of the answer choices have -1/4 but it could be choice A was supposed to show "-1/4" instead of "-14"? Perhaps the division sign didn't show up for some reason?

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Find the area of a circle with a diameter of 26 cm. Give your answer in exact form.
ELEN [110]
The area of a circle is calculated as pi*(r^2). Given a diameter of 26 cm, the radius is half the diameter, which is 26/2 = 13 cm.
Therefore the area is pi*(13^2) = 169 pi.

Another alternative formula for the area is pi*(d^2 / 4). Substituting d = 26 would give pi*(26^2 / 4) = pi*(676/4) = 169 pi, which is the same answer.

So the correct answer is C.
6 0
4 years ago
Volume and surface area plz!!!
atroni [7]

volume is 180

surface area (area) is 105.97


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Oh no! The mathematical magicians only finished 20% of the calculators they need to make this year. If they needed to make 300 c
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4 0
3 years ago
Me ayudan a resolverlo por favor (3x+8y+8)+(4x+6y-6)
Anna007 [38]

Answer:

7x + 14y + 2

Step-by-step explanation:

(3x+8y+8)+(4x+6y-6)

3x + 8y + 8 + 4x + 6y - 6

3x + 4x + 8y + 6y + 8 - 6

7x + 14y + 2

5 0
3 years ago
Two automobiles leave the same city simultaneously and both head towards another city. The speed of one is 10 km/hour greater th
Veronika [31]

Answer:

5091 Km/hr and 505 km/hr

Step-by-step explanation:

Speed = Distance / Time

Let the speed of first automobile be 'x' and that of the second be 'y'

Since speed of one is 10 times greater than the other. therefore;

⇒ x = 10 y

also let time for faster automobile be 'T' and time for slower auto mobile be 't'

Since first arrive one hour earlier than second, therefore;

⇒ t = T + 1

⇒ For first automobile; x X T = 560 ; substituting for 'x' and 'T'. Therefore;

⇒ 10y (t+1) = 560

⇒ For Second automobile; y X t = 560

⇒ y = \frac{560}{t}

⇒ 10(\frac{560}{t})(t + 1) = 560

⇒ 5600 + \frac{5600}{t} = 560

⇒ 5600 - 560 =  -  \frac{5600}{t}

⇒ t = 1.11 hr

also ; T = 1.11 - 1 = 0.11 hr

Speed of 1st auto  = 560/0.11 = 5091 km /hr

Speed of 2nd auto = 560/1.11 = 505 km/hr

8 0
3 years ago
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