General formula for n-th term of arithmetical progression is
a(n)=a(1)+d(n-1).
For 3d term we have
a(3)=a(1) +d(3-1), where a(3)=7
7=a(1)+2d
For 7th term we have
a(7)=a(1) +d(7-1)
a(7)=a(1) + 6d
Also, we have that the <span>seventh term is 2 more than 3 times the third term,
a(7)=3*a(3)+2= 3*7+2=21+2=23
So we have, </span>a(7)=a(1) + 6d and a(7)=23. We can write
23=a(1) + 6d.
Now we can write a system of equations
23=a(1) + 6d
<span> - (7=a(1)+2d)
</span>16 = 4d
d=4,
7=a(1)+2d
7=a(1)+2*4
a(1)=7-8=-1
a(1)= - 1
First term a(1)=-1, common difference d=4.
Sum of the 20 first terms is
S=20 * (a(1)+a(20))/2
a(1)=-1
a(n)=a(1)+d(n-1)
a(20) = -1+4(20-1)=-1+4*19=75
S=20 * (-1+75)/2=74*10=740
Sum of 20 first terms is 740.
The average of 6 quiz scores, where 
is the score of the last quiz, is
The student needs a minimum average of 86 to earn an A:
so the student needs a score of 91.
The picture shows the answer I got!
Also! For the future, I use a site called
math way and it is great for problems like these.
Congruence refers to the exact similarity in terms of the dimensions of a shape or polygon including as well as the angles of the shape. Rotating does not change the image, as well as reflecting and translating. Hence the answers that apply are A, D and E.
Answer:
X=20; pqr = 130°
Step-by-step explanation:
They key here is to notice that the instructions say that qs bisects pqr, meaning that it evenly cuts it into 2 pieces. So, to find x, you just solve for x in the equation 3x+5=2x+25. Then, you plug it back into either side of the equation, at which point, you should get 65. Since it is half of pqr, just double it to get your final answer of 130°
