Answer:
x = 1
Step-by-step explanation:
Given in the question the equation
y = -x² + 2x - 1
To find the x-intercept, substitute in 0 for y
0 = -x² + 2x - 1
To find value of x use quadratic equation
x = -b ± √b²-4ac / 2a
here a = -1
b = 2
c = -1
x1 = -2 + √2²-4(-1)(-1) / 2(-1)
= -2 + 0 / -2
= -2 / -2
= 1
x2 = -2 - √2²-4(-1)(-1) / 2(-1)
= -2 - 0 / -2
= -2 / -2
= 1
Check whether the two expressions 2x+3y2x+3y and 2y+3x2y+3x equivalent.
The first expression is the sum of 2x2x 's and 3y3y 's whereas the second one is the sum of 3x3x 's and 2y2y 's.
Let us evaluate the expressions for some values of xx and yy . Take x=0x=0 and y=1y=1 .
2(0)+3(1)=0+3=32(1)+3(0)=2+0=22(0)+3(1)=0+3=32(1)+3(0)=2+0=2
So, there is at least one pair of values of the variables for which the two expressions are not the same.
Answer:
Part c: Contained within the explanation
Part b: gcd(1200,560)=80
Part a: q=-6 r=1
Step-by-step explanation:
I will start with c and work my way up:
Part c:
Proof:
We want to shoe that bL=a+c for some integer L given:
bM=a for some integer M and bK=c for some integer K.
If a=bM and c=bK,
then a+c=bM+bK.
a+c=bM+bK
a+c=b(M+K) by factoring using distributive property
Now we have what we wanted to prove since integers are closed under addition. M+K is an integer since M and K are integers.
So L=M+K in bL=a+c.
We have shown b|(a+c) given b|a and b|c.
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Part b:
We are going to use Euclidean's Algorithm.
Start with bigger number and see how much smaller number goes into it:
1200=2(560)+80
560=80(7)
This implies the remainder before the remainder is 0 is the greatest common factor of 1200 and 560. So the greatest common factor of 1200 and 560 is 80.
Part a:
Find q and r such that:
-65=q(11)+r
We want to find q and r such that they satisfy the division algorithm.
r is suppose to be a positive integer less than 11.
So q=-6 gives:
-65=(-6)(11)+r
-65=-66+r
So r=1 since r=-65+66.
So q=-6 while r=1.
