Answer:
We are effectively looking for a and b such that 5, a, b, 135 is a geometric sequence.
This sequence has common ratio <span><span>3<span>√<span>1355</span></span></span>=3</span>, hence <span>a=15</span> and <span>b=45</span>
Explanation:
In a geometric sequence, each intermediate term is the geometric mean of the term before it and the term after it.
So we want to find a and b such that 5, a, b, 135 is a geometric sequence.
If the common ratio is r then:
<span><span>a=5r</span><span>b=ar=5<span>r2</span></span><span>135=br=5<span>r3</span></span></span>
Hence <span><span>r3</span>=<span>1355</span>=27</span>, so <span>r=<span>3<span>√27</span></span>=3</span>
Then <span>a=5r=15</span> and <span>b=ar=15⋅3=45</span>
Steps to solve:
3x + 2 = 8
~Subtract 2 to both sides
3x = 6
~Divide 3 to both sides
x = 2
Best of Luck!
Answer:
a=-7.275*967
Step-by-step explanation:
Answer:
- 2(L +W) ≤ 600
- W ≤ 200
- L ≥ 2W
Step-by-step explanation:
We assume the problem wording means the length is to be at least 2 times <em>as long as</em> the width. (<em>Longer than</em> usually refers to a difference, not a scale factor.)
If we let "W" and "L" represent the width and length, respectively, then we can translate the problem statement to ...
2(L + W) ≤ 600 . . . . . . the perimeter is twice the sum of length and width
W ≤ 200 . . . . . . . . . . . . the width is at most 200 inches
L ≥ 2W . . . . . . . . . . . . . the length is at least twice the width
