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Sliva [168]
3 years ago
14

Given a term in a geometric sequence and the common ratio find the first five terms, the explicit formula, and the recursive for

mula. A1 = 4, r = 5 Question 4 options: First Five Terms: 5, 20, 80, 320, 1280; an=5*4n−1;a1=5 First Five Terms: 4, 9, 13, 18, 23; an=5*4n−1;a1=4 First Five Terms: 4, 20, 100, 500, 2500; an=4*5n−1;a1=4 First Five Terms: 20, 25, 30, 35, 40; an=4*5n−1;a1=4
Mathematics
1 answer:
Fed [463]3 years ago
3 0
<span>−2.5 ⋅ 4n − 1. 10) a n. = −4 ⋅ 3n − 1. Given the recursive formula for a geometric sequence find thecommon ratio, the first five terms, and the explicit formula. 11) a n ... Given a term in a geometric sequence and the common ratio find the first five terms, the explicit formula, and the recursive formula. 21) a. 4 = 25, r = −5. 22) a.</span><span>
</span>
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A. What are the coordinates of the vertices of the triangle below?
Viktor [21]

Answer:

Solution given:

<u>A.coordinate are</u>

A(-2,3)

B(0,-3)

C(4,5)

<u>B</u><u>.</u><u>Each</u><u> </u><u>length</u><u> </u><u>are</u><u> </u><u>:</u>

we have

length \sqrt{(x2-x1)²+(y2-y1)²}

now

AB:\sqrt{(-2-0)²+(3+3)²}=2\sqrt{10}units

BC:\sqrt{(0-4)²+(-3-5)²}=4\sqrt{5}units

AC:\sqrt{(-2-4)²+(3-5)²}=2\sqrt{10}units

<u>C.</u><u> the </u><u>figure</u><u>:</u>

<u>By</u><u> </u><u>using</u><u> </u><u>Pythagoras</u><u> </u><u>law</u>

base[b]=AB=perpendicular [p]=AC

hypotenuse [h]=BC

we have

h²=p²+b²

substituting value

(4\sqrt{5})²=2p²

16*5=2*(2\sqrt{10})²

80=2*4*10

80=80

<u>SO</u><u> </u><u>IT</u><u> </u><u>IS</u><u> </u><u>RIGHT</u><u> </u><u>ANGLED</u><u> </u><u>ISOSCELES</u><u> </u><u>TRIANGLE</u><u>.</u>

4 0
3 years ago
Does anyone know how to do this?
sleet_krkn [62]
1 because it is 1pound
4 0
3 years ago
What’s 27337 + 3939393
Lynna [10]

Answer:

3966730

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Jerald jumped from a bungee tower. If the equation that models his height, in feet, is h = –16t2 + 729, where t is the time in s
Snezhnost [94]

Answer. First option: t > 6.25


Solution:

Height (in feet): h=-16t^2+729

For which interval of time is h less than 104 feet above the ground?

h < 104

Replacing h for -16t^2+729

-16t^2+729 < 104

Solving for h: Subtracting 729 both sides of the inequality:

-16t^2+729-729 < 104-729

-16t^2 < -625

Multiplying the inequality by -1:

(-1)(-16t^2 < -625)

16t^2 > 625

Dividing both sides of the inequality by 16:

16t^2/16 > 625/16

t^2 > 39.0625

Replacing t^2 by [ Absolute value (t) ]^2:

[ Absolute value (t) ]^2 > 39.0625

Square root both sides of the inequality:

sqrt { [ Absolute value (t) ]^2 } > sqrt (39.0625)

Absolute value (t) > 6.25

t < -6.25 or t > 6.25, but t can not be negative, then the solution is:

t > 6.25



5 0
4 years ago
Read 2 more answers
The average of the terms, X, X+3, 2x+4, and 3x+8 is 100. If the term 4x is included,
oksian1 [2.3K]

Answer:

the new average of the term is 100x+4

4 0
3 years ago
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