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>
Answer:
Consider f: N → N defined by f(0)=0 and f(n)=n-1 for all n>0.
Step-by-step explanation:
First we will prove that f is surjective. Let y∈N be any natural number. Define x as the number x=y+1. Then x∈N, and f(x)=x-1=(y+1)-1=y. We conclude that f is surjective.
However, f is not injective. Take x1=0 and x2=1. Then x1≠x2 but f(x1)=0 and f(x2)=x2-1=1-1=0. We have shown that there are two natural numbers x1,x2 such that x1≠x2 but f(x1)=f(x2), that is, f is not injective.
Note:
If 0∉N in your definition of natural numbers, the same reasoning works with the function f: N → N defined by f(1)=1 and f(n)=n-1 for all n>1. The only difference is that you consider x1=1, x2=2 for the injectivity.
X=8 because add 11 to both sides and it becomes 28=3.5x and divide both sides by 3.5 so x=8
Answer:
y = 3/5x + 6
Step-by-step explanation:
When given two points, it is convenient to start with the 2-point form of the equation of a line, then rearrange it to the form needed for the problem. For points (x1, y1) and (x2, y2), the equation is ...
y = (y2 -y1)/(x2 -x1)·(x -x1) +y1
Filling in the given point coordinates, this is ...
y = (3 -0)/(-5 -(-10))·(x -(-10)) + 0
y = 3/5(x +10)
y = (3/5)x + 6
