Answer:
b+6
Problem:
If the average of b and c is 8, and d=3b-4, what is the average of c and d in terms of b?
Step-by-step explanation:
We are given (b+c)/2=8 and d=3b-4.
We are asked to find (c+d)/2 in terms of variable, b.
We need to first solve (b+c)/2=8 for c.
Multiply both sides by 2: b+c=16.
Subtract b on both sides: c=16-b
Now let's plug in c=16-b and d=3b-4 into (c+d)/2:
([16-b]+[3b-4])/2
Combine like terms:
(12+2b)/2
Divide top and bottom by 2:
(6+1b)/1
Multiplicative identity property applied:
(6+b)/1
Anything divided by 1 is that anything:
(6+b)
6+b
b+6
The answer is AAAAAAAAAAAAAAAA
Step-by-step explanation:
We know that the length of the diagonals will be more than the side.
Lets suppose the quadrilateral is a rectangle, with sides 6 and 4.
The length of a diagonal is:
√(6*6)+(4*4)
√36+16
√52
7.21(rounded off to 3 sig. fig.)
So, 6+4+6+4<2(7.21+7.21)
20< 2*14.42
20<28.84, which is true.
:)
Answer:
y= 163 because of vertical angle
The generic equation of a third degree polynomial is given by:
y = ax ^ 3 + bx ^ 2 + cx + d
We must make a system of equations to find the values of a, b, c, d.
We have then:
For (1, 3):
3 = a (1) ^ 3 + b (1) ^ 2 + c (1) + d
3 = a + b + c + d
For (2, -2):
-2 = a (2) ^ 3 + b (2) ^ 2 + c (2) + d
-2 = 8a + 4b + 2c + d
For (3, -5):
-5 = a (3) ^ 3 + b (3) ^ 2 + c (3) + d
-5 = 27a + 9b + 3c + d
For (4.0):
0 = a (4) ^ 3 + b (4) ^ 2 + c (4) + d
0 = 64a + 16b + 4c + d
We obtain the following system of equations:
3 = a + b + c + d
-2 = 8a + 4b + 2c + d
-5 = 27a + 9b + 3c + d
0 = 64a + 16b + 4c + d
Whose solution is:
a = 1
b = -5
c = 3
d = 4
The polynomial will then be:
y = x ^ 3 - 5x ^ 2 + 3x + 4
Answer:
y = x ^ 3 - 5x ^ 2 + 3x + 4
