ax² + bx + c = 0
x = (-b ± √(b² - 4ac))/2a
First, rewrite the first equation so that the first coefficient is 1. Divide everything by a.
(ax² + bx + c = 0)/a =
x² + (b/a)x + (c/a) = 0
Isolate (c/a) by subtracting (c/a) from both sides
x² + (b/a)x + (c/a) (-(c/a) = 0 (- (c/a)
x² + (b/a)x = 0 - (c/a)
Add spaces
x² + (b/a)x = -c/a
Take 1/2 of the middle term's coefficient and square it. Remember that what you add to one side, you add to the other.
x² + (b/a)x + (b/2a)² = -c/a + (b/2a)²
Simplify the left side of the equation.
x² + (b/a)x + (b/2a)² = (x + (b/2a))²
(x + b/2a))² = ((b²/4a²) - (4ac/4a²)) -> ((b² - 4ac)/(4a²))
Take the square root of both sides of the equation
√(x + b/2a))² = √((b²/4a²) - (4ac/4a²))
x + b/(2a) = (±√(b² - 4ac)/2a
Simplify. Isolate the x.
x = -(b/2a) ± (∛b² - 4ac)/2a = (-b ± √(b² - 4ac))/2a
~
I + 6 = r
4r = s
i + r + s = 126
4i + 24 = 4r
s= 4r
i = r-6
4i+24=i+6
-3i=18
i = 6
Solve the rest
The picture is very blurry I can’t see
Answer:
the areas of these triangles are 83.2cm² and 46.8cm²
Step-by-step explanation:
1. If the triangles are similar and the ratio of the perimeter ois 4:3, then the areas are in the following ratio:
4²:3²
16:9
2. The sum of their areas is 65 cm², then, you can calculate the area of the larger triangle as following:
130(16/16+9)
130(0.64)
=83.2cm²
3. The area of the smaller triangle is:
130(9/16+9)
130(0.36)
46.8cm²
<u>Hope this help</u>s
