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
Answer:
3 + 4 = 7 which is true.
Step-by-step explanation:
Whenever you have an equation with a plain variable (that is, no exponent included), there is only one number that will work when substituted for x.
To solve it, you have to "undo" what is done to the variable. You also go in the reverse order of operations, so you do the addition/subtract first, then multiplication/division.
You also have to do the same to both sides, kind of like keeping a balance scale in balance.
In this case, we subtract 4 from both sides first:
3x + 4 -4 = 7 - 4
The + 4 - 4 cancel each other out, so you get:
3x = 3
3x means "3 times x" so you divide by 3 to undo it. I will use the / to indicate division:
3x / 3 = 3 /3
so 1x = 1.
Since 1x is "1 times x" it is the same as x by itself, so:
x=1
AND, if we substitute 1 back into the original equation (the asterisk stands for multiply):
3 * 1 + 4 = 7
3 * 1 is 3, so:
3 + 4 = 7 which is true.
1 is the only number that works.
Hope this helped.
Answer:
The line AB, with A(Ax, Ay), B(Bx, By) and midpoint M(Mx, My) satisfying:
Ax + Bx = 2Mx
Ay + By = 2My
=>
2 + Bx = 2*1
5 + By = 2*2
=> Bx = 0
=> By = -1
=> B(0, -1)
Hope this helps!
:)
<span> As the x-values go to negative infinity the functions values go to positive infinity</span>
