Answer:
Let the no. of marbles john had be x
Then the no. of marbles jivanti had = (45-x)
The no. of marbles left with john, when he lost 5 marbles = 45-x-5=40 - x
Therefore their product = (x-5)(40-x)
=40x - x2 -200+5x
=x2+45x-200
So,-x2 + 45x-200=124. (given , product =124)
so, x2-45x+324=0
Therefore, the no. of marbles john had , satisfies the quadratic eqn ,
x2-45x+324.
When u will find the value x of this eqn that no. will be the no. of marbles they started with!
Hope it helps
Step-by-step explanation:
The answer to ten times ten is 100
Answer:
72
Step-by-step explanation:
The amplitude is 2, so A = 2 (see note 1 below)
The period is T = pi/2 which means
B = 2pi/T
B = 2pi/(pi/2)
B = 2pi*(2/pi)
B = 4
The phase shift is C = pi/4
The midline is y = -3 indicating that D = -3
Put this all together and we go from this (see note 2)
y = A*sin(B(x-C)) + D
to this
y = 2*sin(4(x-pi/4)) + (-3)
and that simplifies to
y = 2sin(4x-pi) - 3
which is one way to write the final simplified answer (see note 3)
-------------------------------------------------------------
Extra Stuff:
Note 1: The value of A can be A = -2. Recall that the amplitude is equal to |A|. So If A = -2, then |A| = |-2| = 2. The amplitude is the vertical distance from the midline to the peak or valley. Distance can't be negative. To make things simple, I went with A = 2.
Note 2: The general formula involving sine can be replaced with cosine instead. The cosine function is basically sine but a phase shift of it (it has the same shape and pattern, but has been moved over). You cannot use tangent as that is a completely different class of function.
Note 3: I distributed the 4 through to the (x-pi/4) to get 4x-pi. Your teacher or book may want you to keep things factored and not distribute. If so, then the answer would be y = 2sin(4(x-pi/4)) - 3. That is of course you went with using sine instead of cosine. The benefit of not distributing is that we can more easily see what the phase shift is.
Answer:
178.3 mm²
Explanation:
The surface area of the regular pyramid is equal to the sum of the base and lateral areas:()