1) Each side of a triangle is a different length. One side is 6, one side is more than 6, and one side is less than 6. The perimeter of the triangle could not be...

Question

a.13
b.18
c.22
d.24

2)Which of the following could not be the measures of two of the three angles in the same isosceles triangle?
a.40 and 70
b.45 and 90
c.50 and 100
d.60 and 60

3)I multiplied all ten integers from 1 to 10 by 1. Then I multiplied the same ten integers (from 1 to 10) by 2, 3, 4, 5, 6, 7, 8, 9, & 10 respectively. What is the sum of these 100 products?
A. 2475
B. 2500
C. 3025
D. 3575

Please explain how you get your answer. I don't understand.

Answer 1

1)
biggest side <12 because must stretch 6+6
so perimeter < 24

2) sum has to be 180
40+40 + 70 = 150 no
but 40 + 70 + 70 = 180 ok

45+45 + 90 = 180 ok

100 + 100 too big so C is no good

60+60 + 90 = 180 ok

3)
make a table 1 to 10 across
1 to 10 down
start to fill it in with products
then sum the first row
you will get 55 which is 10 (1 +10)/2
the sum of the next row will be 110
which is 10 (2+20)/2
in fact each row will be the number at the left times 55
so what we really have is
55 + 2*55 + 3*55 + 4*55 etc

which is of course an arithmetic series
a1 = 55 = a and d = 55
a2 = 55+55 = a1 + 1*55
a3 = a2 + 55 = a1 + 2*55
a4 = a3 + 55

a10 = a1 + 9*55

sum = (n/2)(2 a + (n-1)d )

a = 55 and d = 55 and n = 10

sum = 5 (110 + 9(55) ) = 5 (11*55)
= 3025 or C

Answer 2

or for #3 you have

1*(1+2+3+...+9+10) +
2*)1+2+3+...+9+10) +
...
10*(1+2+3+...+9+10)
add them:
(1+2+3+...+9+10)(1+2+3+...+9+10)
= 55*55
= 3025

Answer 3

For question 1, the sides do not have to be whole numbers. So it could be 6,5,2 to eliminate a. 6, 5,7 to eliminate b or 6, 5.9, 10.1 to eliminate c. But whatever you do it cannot give you d.