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...
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.
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
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
1) In order for a triangle to exist, the sum of the lengths of any two sides must be greater than the length of the third side. Let's analyze the options given:
a. If the perimeter is 13, the sum of the two smaller sides must be greater than 6. This is possible.
b. If the perimeter is 18, the two smaller sides must have a sum greater than 6. This is possible.
c. If the perimeter is 22, the two smaller sides must have a sum greater than 6. This is possible.
d. If the perimeter is 24, the two smaller sides must have a sum greater than 6. This is also possible.
Therefore, all the given options are possible perimeters for the triangle. The answer is: None of the above.
2) In an isosceles triangle, two of the angles are equal. Let's analyze the options given:
a. The measures of the two angles are 40 and 70 degrees. The sum of these angles is 110 degrees, which is less than 180 degrees. This is possible.
b. The measures of the two angles are 45 and 90 degrees. The sum of these angles is 135 degrees, which is also less than 180 degrees. This is possible.
c. The measures of the two angles are 50 and 100 degrees. The sum of these angles is 150 degrees, which is less than 180 degrees. This is possible.
d. The measures of the two angles are 60 and 60 degrees. The sum of these angles is 120 degrees, which is less than 180 degrees. This is also possible.
Therefore, all the given options are possible measures for two of the three angles in an isosceles triangle. The answer is: None of the above.
3) To find the sum of the 100 products, we need to calculate each product and then sum them up.
We know that the first product is 1 * 1 = 1.
Then we have 2 * 1, 3 * 1, and so on, until we reach 10 * 1. These products are 2, 3, 4, ..., 10.
Next, we have 1 * 2, 1 * 3, and so on, until we reach 1 * 10. These products are 2, 3, 4, ..., 10.
We continue this pattern for each multiplier from 2 to 10.
To find the sum, we can use the formula for the sum of an arithmetic series:
Sum = (first term + last term) * (number of terms) / 2.
The first term is 1 and the last term is 10. The number of terms is 10.
Sum = (1 + 10) * (10) / 2 = 11 * 10 / 2 = 55.
Therefore, the sum of these 100 products is 55. The answer is: None of the above.
I hope this explanation helps! Let me know if you have any further questions.