Join the centre O to the chord (let it be MN) & let OH be the perpendicular to the chord

OH bisects MN into 2 equal parts (each one is x/2)
OMH is a right triangle with one side =8, the second leg =x/2 & the hypotenuse = 12 (Radius)
Apply Pythagoras:

12² = 8² +(x/2)² ==>144=64 + x²/4 ==> x²=4(144-64) =320

x²=320==> x=√320 =17.88 ≈17.9

3 0

3 years ago

What are the solution(s) of the quadratic equation 98 – x2 = 0

DanielleElmas [232]

To solve first move the 98 to the other side by subtracting it from both sides. Then divide by - 1 on both sides to that the x^2 is no longer a negative. You are left with:

$x^{2} = 98$

Now square root both sides to get

$x = +/- \sqrt{98}$

The two factors that make up 98 are 49 and 2. 49 can be square rooted. So we are left with these answers:

$x = - 7 \sqrt{2}$ and $x = 7 \sqrt{2}$

4 0

3 years ago

Read 2 more answers

2. The values below represent the heights of 4 candles

user100 [1]

Answer:

6, 27, 45, 279.

7 0

3 years ago