A function assigns the values. The graph can be plotted as shown below.
<h3>What is a Function?</h3>
A function assigns the value of each element of one set to the other specific element of another set.
Given the function, g(x) = -3(1/2)ˣ, therefore, the value of g(x) can be written as,
g(x) = -3(1/2)ˣ
g(-2) = -3(1/2)² = -0.75
g(-1) = -3(1/2)¹ = -1.5
g(0) = -3(1/2)⁰ = -3
g(1) = -3(1/2)¹ = -1.5
Hence, the graph can be plotted as shown below.
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The length of side walk is 500 feet
<em><u>Solution:</u></em>
Given that, A rectangle park measures 300 ft by 400 ft
Length = 300 feet
Width = 400 feet
A sidewalk runs diagonally from one comer to the opposite corner
We have to find the length of side walk
Which means, we have to find the length of diagonal of rectangle
<em><u>The diagonal of rectangle is given by formula:</u></em>
Where,
d is the length of diagonal
w is the width and l is the length of rectangle
<em><u>Substituting the values in formula, we get</u></em>
Thus length of side walk is 500 feet
1. I complete some of my cyber: 7:00am to 10:00am
2. I spend time doing assignments in votech: 12:00pm to 2:30pm
Answers:
x = -8/5 or x = 8/5
Sum of the first ten terms where all terms are positive = 4092
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Explanation:
r = common ratio
- first term = 4
- second term = (first term)*(common ratio) = 4r
- third term = (second term)*(common ratio) = (4r)*r = 4r^2
The first three terms are: 4, 4r, 4r^2
We're given that the sequence is: 4, 5x, 16
Therefore, we have these two equations
Solve the second equation for r and you should find that r = -2 or r = 2 are the only possible solutions. If r = -2, then 5x = 4r solves to x = -8/5. If r = 2, then 5x = 4r solves to x = 8/5.
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To find the sum of the first n terms, we use this geometric series formula
Sn = a*(1 - r^n)/(1 - r)
We have
- a = 4 = first term
- r = 2, since we want all the terms to be positive
- n = 10 = number of terms to sum up
So,
Sn = a*(1 - r^n)/(1 - r)
S10 = 4*(1 - 2^10)/(1 - 2)
S10 = 4*(1 - 1024)/(-1)
S10 = 4*(-1023)/(-1)
S10 = 4092
Answer:
(-3,8) would NOT fit the table.
