I must assume that you meant <span>X^2-10x =7. If that is indeed the case, then:
</span>X^2-10x =7 => x^2 -10x + 25 - 25 = 7, so that (x-5)^2 = 32 (Add 25, then
subtract 25)
Then x-5 = plus or minus sqrt(16)sqrt(2) = 4sqrt(2).
Then x = 5 plus or minus 4 sqrt(2) (answers)
The equation of a circle with radius r and center at point (a, b) is given by
(x - a)^2 + (y - b)^2 = r^2
1.) For the circle given by x^2 + y^2 = 24
(x - 0)^2 + (y - 0)^2 = (√24)^2
Therefore, center = (0, 0) and radius = √24 = 4.9 units
3.) x^2 + y^2 -10x -2y = -10
x^2 - 10x + (-5)^2 + y^2 - 2y + (-1)^2 = -10 + (-5)^2 + (-1)^2
x^2 - 10x + 25 + y^2 - 2y + 1 = -10 + 25 + 1
(x - 5)^2 + (y - 1)^2 = 16
(x - 5)^2 + (y - 1)^2 = 4^2
D is the answer multiply width times length to find area for quadrilaterals for a triangle use base times height divided by two.
Answer:
wrong
Step-by-step explanation:
it is 5/16 do KCF and 1*5 is 5 and 2*8 is 16 so you get 5/16
Hope this helps!
Answer:
B) ∠B ≅ ∠K
Step-by-step explanation:
The Angle-Angle-Side Postulate (AAS) states that if two angles and a non-included side of one triangle are congruent to two angles and the corresponding non-included side of another triangle, then the two triangles are congruent.
Given that ∠A ≅ ∠J and AC ≅ JL, then we need that ∠B ≅ ∠K to satisfy postulate AAS (the two angles are ∠A and ∠B in one triangle and ∠K and ∠J in the other one, and the non-included side is AC in one triangle and JL in the other one).
