Answer:
The intensity of the first earthquake is 6.31 times the second one
Step-by-step explanation:
The given formulae is M=log(
)
Here for both the earthquakes the value of S is same.
Let the Intensity of the first earthquake be I1 and the second be I2.
Given M1=7.5
M2=6.7
Now M1=log(
)------1
M2=log(
)-----2
Now subtract 1 and 2 we get
M1-M2=log(
)
I1/I2=
I1=I2(
I1=I2*6.31
Answer:
B. Multiply the divisor and dividend by 100.
Step-by-step explanation:
Multiply by 100 to get rid of the decimal
The area of the triangle is
A = (xy)/2
Also,
sqrt(x^2 + y^2) = 19
We solve this for y.
x^2 + y^2 = 361
y^2 = 361 - x^2
y = sqrt(361 - x^2)
Now we substitute this expression for y in the area equation.
A = (1/2)(x)(sqrt(361 - x^2))
A = (1/2)(x)(361 - x^2)^(1/2)
We take the derivative of A with respect to x.
dA/dx = (1/2)[(x) * d/dx(361 - x^2)^(1/2) + (361 - x^2)^(1/2)]
dA/dx = (1/2)[(x) * (1/2)(361 - x^2)^(-1/2)(-2x) + (361 - x^2)^(1/2)]
dA/dx = (1/2)[(361 - x^2)^(-1/2)(-x^2) + (361 - x^2)^(1/2)]
dA/dx = (1/2)[(-x^2)/(361 - x^2)^(1/2) + (361 - x^2)/(361 - x^2)^(1/2)]
dA/dx = (1/2)[(-x^2 - x^2 + 361)/(361 - x^2)^(1/2)]
dA/dx = (-2x^2 + 361)/[2(361 - x^2)^(1/2)]
Now we set the derivative equal to zero.
(-2x^2 + 361)/[2(361 - x^2)^(1/2)] = 0
-2x^2 + 361 = 0
-2x^2 = -361
2x^2 = 361
x^2 = 361/2
x = 19/sqrt(2)
x^2 + y^2 = 361
(19/sqrt(2))^2 + y^2 = 361
361/2 + y^2 = 361
y^2 = 361/2
y = 19/sqrt(2)
We have maximum area at x = 19/sqrt(2) and y = 19/sqrt(2), or when x = y.
