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Gre4nikov [31]
2 years ago
12

Marie heated water in a beaker. When she began, the water had a temperature of 40 degrees, and she heated it at a rate of 2.5 de

grees per minute. Kathy heated water in a different beaker. When she began, the water had a temperature of 50 degrees, and she heated it at a rate of 1.5 degrees per minute. Which equation could be used to determine how many minutes passed before the water temperature in both beakers was the same?
Mathematics
1 answer:
kykrilka [37]2 years ago
4 0

Answer:

2.5x + 40 = 1.5x + 50

Step-by-step explanation:

To find how many minutes passed before the water temperature in both beakers are the same, you need to set both of the expressions that show how hot the water in each beaker is to be equal to each other

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If the pressure inside a rubber balloon is 1500 mmHg, what is this pressure in pounds-force per square inch (psi)? Answer: 29.0
vivado [14]

Answer:

Pressure magnitude of 1500 mmHg equals 29.00 pounds per square inch.

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4 years ago
Consider the functions z = 4 e^x ln y, x = ln (u cos v), and y = u sin v.
Katen [24]

Answer:

remember the chain rule:

h(x) = f(g(x))

h'(x) = f'(g(x))*g'(x)

or:

dh/dx = (df/dg)*(dg/dx)

we know that:

z = 4*e^x*ln(y)

where:

y = u*sin(v)

x = ln(u*cos(v))

We want to find:

dz/du

because y and x are functions of u, we can write this as:

dz/du = (dz/dx)*(dx/du) + (dz/dy)*(dy/du)

where:

(dz/dx)  = 4*e^x*ln(y)

(dz/dy) = 4*e^x*(1/y)

(dx/du) = 1/(u*cos(v))*cos(v) = 1/u

(dy/du) = sin(v)

Replacing all of these we get:

dz/du = (4*e^x*ln(y))*( 1/u) + 4*e^x*(1/y)*sin(v)

          = 4*e^x*( ln(y)/u + sin(v)/y)

replacing x and y we get:

dz/du = 4*e^(ln (u cos v))*( ln(u sin v)/u + sin(v)/(u*sin(v))

dz/du = 4*(u*cos(v))*(ln(u*sin(v))/u + 1/u)

Now let's do the same for dz/dv

dz/dv = (dz/dx)*(dx/dv) + (dz/dy)*(dy/dv)

where:

(dz/dx)  = 4*e^x*ln(y)

(dz/dy) = 4*e^x*(1/y)

(dx/dv) = 1/(cos(v))*-sin(v) = -tan(v)

(dy/dv) = u*cos(v)

then:

dz/dv = 4*e^x*[ -ln(y)*tan(v) + u*cos(v)/y]

replacing the values of x and y we get:

dz/dv = 4*e^(ln(u*cos(v)))*[ -ln(u*sin(v))*tan(v) + u*cos(v)/(u*sin(v))]

dz/dv = 4*(u*cos(v))*[ -ln(u*sin(v))*tan(v) + 1/tan(v)]

5 0
3 years ago
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