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stiv31 [10]
3 years ago
7

HOW CAN ARITHMETIC AND GEOMETRIC SEQUENCES BECOME LINEAR AND EXPONENTIAL FUNCTIONS?

Mathematics
1 answer:
vladimir1956 [14]3 years ago
3 0
A sequence is a set of numbers, called terms, arranged in some particular order. An arithmetic sequence is a sequence with the difference between two consecutive terms constant. The difference is called the common difference. A geometric sequence is a sequence with the ratio between two consecutive terms constant.
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In a competitive examination negative marks of 2 are awarded for each wrong answers and 4 marks for every correct answer. If nee
Alex777 [14]

9514 1404 393

Answer:

  Rajan gets 6 marks more

Step-by-step explanation:

Neerav's marks are (15)(4) +(5)(-2) = 60 -10 = 50.

Rajan's marks are (14)(4) = 56.

Rajan gets 6 more marks than Neerav.

3 0
3 years ago
Find exact value of the expression:<br><br> tan -1 square root 3/3
adell [148]
Hello,

 if tan x=√3/3 then x=30°+k*180° k: integer

4 0
3 years ago
(#5) A carpenter has two different-sized circular saws. The smaller saw has the radius shown. I couldn't show it, so it is 1.25
grin007 [14]

Answer:

4.71 in. difference

Step-by-step explanation:

so, the radius of the smaller saw is 1.25 and the diameter is double the radius. so it is 2.5 in diameter. the larger saw is 1.5 in greater which is 4 in.

to find circumference, you have to use the solution C = πd or circumference = pi x diameter.

3.14 x 2.5 = 7.85

4 x 3.14 = 12.56

subtract 7.85 from 12.56 to get your answer! i hope this helps!

5 0
3 years ago
Factor the following expression. Simplify your answer.<br> 3s(s - 1)^1/3 + 2(s - 1)^4/3
denis23 [38]

Answer:

Step-by-step explanation:

3s\sqrt[3]{s-1} + 2 \sqrt[4/3]{s-1} =\\3s\sqrt[3]{s-1} + 2 \sqrt[1/3]{(s-1)^4} =\\3s\sqrt[3]{s-1} + 2 (s-1)\sqrt[1/3]{s-1} =\\\sqrt[3]{s-1}*(3s + 2 (s-1)) =\\\sqrt[3]{s-1}*(3s + 2s-2)) =\\\sqrt[3]{s-1}*(5s -2) \\

6 0
2 years ago
Read 2 more answers
Mr. Smith is flying his single-engine plane at an altitude of 2400 feet. He sees a cornfield at an angle of depression of 30º. W
Svet_ta [14]

Answer:

Height of the fighter plane =1.5km=1500 m

Speed of the fighter plane, v=720km/h=200 m/s

Let be the angle with the vertical so that the shell hits the plane. The situation is shown in the given figure.

Muzzle velocity of the gun, u=600 m/s

Time taken by the shell to hit the plane =t

Horizontal distance travelled by the shell =u

x

t

Distance travelled by the plane =vt

The shell hits the plane. Hence, these two distances must be equal.

u

x

t=vt

u Sin θ=v

Sin θ=v/u

=200/600=1/3=0.33

θ=Sin

−1

(0.33)=19.50

In order to avoid being hit by the shell, the pilot must fly the plane at an altitude (H) higher than the maximum height achieved by the shell for any angle of launch.

H

max

=u

2

sin

2

(90−θ)/2g=600

2

/(2×10)=16km

4 0
3 years ago
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