Answer:
It can't be C or D. I think it could be A.
Answer:
- as written, c ≈ 0.000979 or c = 4
- alternate interpretation: c = 0
Step-by-step explanation:
<em>As written</em>, you have an equation that cannot be solved algebraically.
(32^2)c = 8^c
1024c = 8^c
1024c -8^c = 0 . . . . . . rewrite as an expression compared to zero
A graphical solution shows two values for c: {0.000978551672551, 4}. We presume you're interested in c = 4.
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If you mean ...
32^(2c) = 8^c
(2^5)^(2c) = (2^3)^c . . . . rewriting as powers of 2
2^(10c) = 2^(3c) . . . . . . . simplify
10c = 3c . . . . . . . . . . . . . .log base 2
7c = 0 . . . . . . . . . . . . . . . subtract 3c
c = 0 . . . . . . . . . . . . . . . . divide by 7
Just sub x in with 5.
So 6(5)^3+8(5) = 6 (125) + 40 = 750 + 40 = 790.
So 790
Y = 3x - 5
y = 6x - 8
y = y
3x - 5 = 6x - 8
3x - 6x = -8 + 5
-3x = -3
x = -3/-3
x = 1
y = 3x - 5 ⇒ 3(1) - 5 = 3 - 5 ⇒ y = -2
y = 6x - 8 ⇒ 6(1) - 8 = 6 - 8 ⇒ y = -2
x = 1 ; y = -2 or (1,-2) is the point of intersection.
Answer:
see explanation
Step-by-step explanation:
If the triangle is right then by Pythagoras' identity, the longest side ( hypotenuse) will be equal to the sum of the squares on the other 2 sides.
Given sides 3, 5 and 6 ( the longest side ), then
6² = 36
3² + 5² = 9 + 25 = 34 ≠ 36
Thus the given sides do not make a right triangle