Answer:
Step-by-step explanation:
Consider a quadratic equation given by
.... (1)
The quadratic formula then is
The quadratic equation provided is
This can be expressed as
.... (2)
By comparing (1) and (2), we derive:
Plug these values into the quadratic formula.
Factoring out common elements.
The two solutions are
and 
We recognize that
Thus,
Consequently, root is positive while is negative.
The area calculation for the shaded section, as seen in the attached diagram, involves subtracting the area of the kite from that of the rectangle. The area of the rectangle is calculated as (3x+x)*(x+x), which simplifies to 4x*2x, equating to 8x². The area of the kite is determined using the formula (1/2)*[d1*d2], where d1 and d2 represent the diagonals, specifically d1=4x and d2=2x. Therefore, the area of the kite becomes (1/2)*[4x*2x], leading to 4x². Consequently, the area of the shaded region can be computed as 8x²-4x², resulting in 4x². Thus, the solution is 4x².
Answer:
Step-by-step explanation:
I concluded the answer is.25 because it makes sense and I was correct.
Utilizing the normal distribution and the central limit theorem, there's a 0.0284 or 2.84% chance of observing a sample mean mass of 695g or less.
The diagrams for parts A and C are included here. For part B, we have circle O. We begin by drawing two radii OA and OC, connecting points A and C to create chord AC. The radius intersects chord AC at point B, bisecting AC into equal segments AB and BC. This gives us two triangles, ΔOBA and ΔOBC, where OA equals OC (since they're radii), OB equals OB (by the reflexive property), and AB is equal to BC (as stated in the question). By applying the SSS triangle congruence criterion, we conclude that ΔOBA is congruent to ΔOBC, allowing us to deduce that ∡OBA equals ∡OBC, both measuring 90°. Thus, OB is perpendicular to AC. Moving on to part D, we again work with circle O and draw the two radii OA and OC, joining points A and C to create chord AC. The radius intersects AC at point B, where AB is perpendicular to AC, meaning ∡B equals 90°. We then consider the right triangles ΔOBA and ΔOBC, and given OA equals OC (the radii), and OB equals OB (reflexive property), we conclude through the HL triangle congruence that ΔOBA is congruent to ΔOBC. Consequently, we find BA equal to BC, thus OB bisects AC.
