Answer:
Step-by-step explanation:
log8 62 = x1
8^x1= 62
8^2 =64 >62 so <u>x1<2</u>
log7 50=x2
7^x2=50
7^2=49<50 so <u>x2>2</u>
we have, x1<2 and 2<x2
x1<2<x2
x1<x2
log8 62<log7 50
Answer:
Gimme some time
Step-by-step explanation:
9514 1404 393
Answer:
1.63 cm (across the centerline from release)
Step-by-step explanation:
If we assume time starts counting when we release the weight from its fully-extended downward position, then the position at 1.15 seconds can be found from ...
h(t) = -7cos(2πt/4)
h(1.15) = -7cos(π·1.15/2) = -7(-0.233445) ≈ 1.63412 . . . cm
That is, 1.15 seconds after the weight is released from below the resting position, it will be 1.63 cm above the resting position.
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If it is released from <em>above</em> the resting position, it will be 1.63 cm <em>below</em> the resting position at t=1.15 seconds.
You can say
x+y=5
x=5-y
You know the value of x, put it into the next equation.
3(5-y)-7y=19
15-3y-7y=19
15-19 = 10y
-4=10y
-4/10=y
-2/5=y
Put y value back into either equation to find value of x.
x+y=5
x-2/5=5
x=5+2/5
x=27/5
Therefore, x,y = (27/5, -2/5)
Hope I helped :)
The standard form of a parabola is

So, if you expand the square, you get

which is the standard form