5(c+8) or 5c+40
B = 5A C = 4A - 8 The sum of the angles of a triangle = 180° A + B + C = 180 A + 5A + (4A - 8) = 180 10A = 188 A = 18.8° (and therefore B = 94° and C = 67.2°)
To find the number of combinations of five numbers from a set of eight, you can use the combination formula ( C(n, k) = \frac{n!}{k!(n-k)!} ). In this case, ( n = 8 ) and ( k = 5 ). Thus, ( C(8, 5) = \frac{8!}{5!(8-5)!} = \frac{8!}{5!3!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 ). Therefore, there are 56 possible combinations.
To determine the number of ways to pick a set of 5 crayons from a box of 8 crayons, we can use the combination formula, which is given by ( C(n, k) = \frac{n!}{k!(n-k)!} ). Here, ( n = 8 ) and ( k = 5 ). Therefore, the number of combinations is ( C(8, 5) = \frac{8!}{5!(8-5)!} = \frac{8!}{5!3!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 ). Thus, there are 56 different ways to pick a set of 5 crayons from the box.
To multiply a sum by a number,multiply each addend of the sum by the number outside the parentheses.For any numbers a, b, and c, a(b+c) = ab + ac and a(b-c) = ab - ac.Example: 2(5-3)= (2 times* 5) - (2 times* 3). Times*_multiply
what is the lcm of 5 and 8
There is no sequence of adds or subtracts of 5, 6, 8, 9, and 10 that sum to 1. Check it with this C++ code... for (int i=0; i<32; ++i) { int sum = 0; if (i&1) sum += 5; else sum -= 5; if (i&2) sum += 6; else sum -= 6; if (i&4) sum += 8; else sum -= 8; if (i&8) sum += 9; else sum -= 9; if (i&16) sum += 10; else sum -= 10; cout << i << " " << sum << endl; }
B = 5A C = 4A - 8 The sum of the angles of a triangle = 180° A + B + C = 180 A + 5A + (4A - 8) = 180 10A = 188 A = 18.8° (and therefore B = 94° and C = 67.2°)
Um, x2+3x-5=0? This is ax2+bx+c where a=1, b=3, and c=-5. The sum of the roots is -b/a so that means the sum of the roots is -3. Also, product of the roots is c/a. That means the product of the roots is -5. -3+(-5)= -8. There you have it.
To find the number of combinations of five numbers from a set of eight, you can use the combination formula ( C(n, k) = \frac{n!}{k!(n-k)!} ). In this case, ( n = 8 ) and ( k = 5 ). Thus, ( C(8, 5) = \frac{8!}{5!(8-5)!} = \frac{8!}{5!3!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 ). Therefore, there are 56 possible combinations.
There is not enough information to determine the answer.For example, 6 + 2 = 8 = 5 + 3But 6 - 2 = 4 while 5 - 3 = 2.
-3
To determine the number of ways to pick a set of 5 crayons from a box of 8 crayons, we can use the combination formula, which is given by ( C(n, k) = \frac{n!}{k!(n-k)!} ). Here, ( n = 8 ) and ( k = 5 ). Therefore, the number of combinations is ( C(8, 5) = \frac{8!}{5!(8-5)!} = \frac{8!}{5!3!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 ). Thus, there are 56 different ways to pick a set of 5 crayons from the box.
28 1 2 3 4 -------------------| 5 | 6 | 7 -------------------- 8 1 clicks 7 times 2 clicks 6 3 clicks 5 4 c 4 5 c 3 6 c 2 7 c 1 8 c 0 ______ 28
(c + 8)(c - 8)= c^2 - 64
c + j = 42; c + 5 = 3(j + 5) c = 42 - j so 42 - j + 5 = 3j + 15 ie 47 - j = 3j + 15 ie 4j = 32 so Jared is 8 and Chim 34 and in 5 years Jared will be 13 and Chim 39. QED
To multiply a sum by a number,multiply each addend of the sum by the number outside the parentheses.For any numbers a, b, and c, a(b+c) = ab + ac and a(b-c) = ab - ac.Example: 2(5-3)= (2 times* 5) - (2 times* 3). Times*_multiply
-3