The answer to your problem is 10x-y=0
9514 1404 393
Answer:
18 square units
Step-by-step explanation:
Referring to the figure, we see that the base AB has a slope of 1, and the altitude CD has a slope of -1. The number of unit squares crossed by these segments are, respectively 6 and 3, so the length of each is ...
AB = 6√2
CD =3√2
The area is half the product of the base (AB) and height (CD) so is ...
A = 1/2bh = (1/2)(6√2)(3√2) = 18
The area of ΔABC is 18 square units.
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<em>Additional comment</em>
It is useful to remember that the diagonal of a unit square is √2. We used that fact here. If you need to figure it using the Pythagorean theorem, you find ...
c² = a² +b²
c = 1² +1² = 2
c = √2
To find the answer. you have to use the pythagorean theorem.
a² + b² = c² where "a" and "b" are the legs and "c" is the hypotenuse.
24² + b² = 26²
576 + b² = 676
b² = 676 - 576
b² = 100
√b² = √100
b = 10 so 10 is the number of the second leg.
hope this helps, God bless!
The function, as presented here, is ambiguous in terms of what's being deivded by what. For the sake of example, I will assume that you meant
3x+5a
<span> f(x)= ------------
</span> x^2-a^2
You are saying that the derivative of this function is 0 when x=12. Let's differentiate f(x) with respect to x and then let x = 12:
(x^2-a^2)(3) -(3x+5a)(2x)
f '(x) = ------------------------------------- = 0 when x = 12
[x^2-a^2]^2
(144-a^2)(3) - (36+5a)(24)
------------------------------------ = 0
[ ]^2
Simplifying,
(144-a^2) - 8(36+5a) = 0
144 - a^2 - 288 - 40a = 0
This can be rewritten as a quadratic in standard form:
-a^2 - 40a - 144 = 0, or a^2 + 40a + 144 = 0.
Solve for a by completing the square:
a^2 + 40a + 20^2 - 20^2 + 144 = 0
(a+20)^2 = 400 - 144 = 156
Then a+20 = sqrt[6(26)] = sqrt[6(2)(13)] = 4(3)(13)= 2sqrt(39)
Finally, a = -20 plus or minus 2sqrt(39)
You must check both answers by subst. into the original equation. Only if the result(s) is(are) true is your solution (value of a) correct.
