Answer:
The pairs are (13,15) and (-15,-13).
Step-by-step explanation:
If n is an odd integer, the very next odd integer will be n+2.
n+1 is even (so we aren't using this number)
The sum of the squares of (n) and (n+2) is 394.
This means
(n)^2+(n+2)^2=394
n^2+(n+2)(n+2)=394
n^2+n^2+4n+4=394 since (a+b)(a+b)=a^2+2ab+b^2
Combine like terms:
2n^2+4n+4=394
Subtract 394 on both sides:
2n^2+4n-390=0
Divide both sides by 2:
n^2+2n-195=0
Now we need to find two numbers that multiply to be -195 and add up to be 2.
15 and -13 since 15(-13)=-195 and 15+(-13)=2
So the factored form is
(n+15)(n-13)=0
This means we have n+15=0 and n-13=0 to solve.
n+15=0
Subtract 15 on both sides:
n=-15
n-13=0
Add 13 on both sides:
n=13
So if n=13 , then n+2=15.
If n=-15, then n+2=-13.
Let's check both results
(n,n+2)=(13,15)
13^2+15^2=169+225=394. So (13,15) looks good!
(n,n+2)=(-15,-13)
(-15)^2+(-13)^2=225+169=394. So (-15,-13) looks good!
A <span>counterclockwise rotation of 270º about the origin is equivalent to a </span><span>clockwise rotation of 90º about the origin.
Given a point (4, 5), the x-value, i.e. 4 and the y-value, i.e. 5 are positive, hence the point is in the 1st quadrant of the xy-plane.
A clockwise rotation of </span><span>90º about the origin of a point in the first quadrant of the xy-plane will have its image in the fourth quadrant of the xy-plane. Thus the x-value of the image remains positive but the y-value of the image changes to negative.
Also the x-value and the y-value of the original figure is interchanged.
For example, given a point (a, b) in the first quadrant of the xy-plane, </span><span>a counterclockwise rotation of 270º about the origin which is equivalent to a <span>clockwise rotation of 90º about the origin will result in an image with the coordinate of (b, -a)</span>
Therefore, a </span><span>counterclockwise rotation of 270º about the origin </span><span>of the point (4, 5) will result in an image with the coordinate of (5, -4)</span> (option C)
You have to square the binomial.
(x^3 + 4y) (x^3 + 4y)
Using FOIL:
x^9 + 4(x^3)y + 4(x^3)y + 16y^2
x^9 + 8(x^3)y + 16y^2
You could also use the formula: (a+b)^2 = (a^2 + 2ab + b^2).
Answer:
6*8=48 groups with elements of order 7
Step-by-step explanation:
For this case the first step is discompose the number 168 in factors like this:
And for this case we can use the Sylow theorems, given by:
Let G a group of order 
where p is a prime number, with 
and p not divide m then:
1) 
2) All sylow p subgroups are conjugate in G
3) Any p subgroup of G is contained in a Sylow p subgroup
4) n(G) =1 mod p
Using these theorems we can see that 7 = 1 (mod7)
By the theorem we can't have on one Sylow 7 subgroup so then we need to have 8 of them.
Every each 2 subgroups intersect in a subgroup with a order that divides 7. And analyzing the intersection we can see that we can have 6 of these subgroups.
So then based on the information we can have 6*8=48 groups with elements of order 7 in G of size 168
Answer:
134.189
Step-by-step explanation:
1+ 4x√7
√7 = 2.646
=4 x 2.646 = 10.584
1+10.584= 11.584
11.584 squared
=134.189
