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Travka [436]
3 years ago
13

The length of the hypotenuse of a right triangle is 157 units. The length of one leg of the triangle is 132. Lara wrote the foll

owing step to find the length of the unknown leg: Length of the unknown leg = 1572 – 1322 = 24,649 – 17,424 = 7,225 units Which statement best explains whether Lara's step is correct or incorrect? It is correct because the length of the unknown side is the difference of the lengths of the sides. It is correct because the length of the unknown side is the difference of the squares of the sides. It is incorrect because the length of the unknown side is the square root of 42,073. It is incorrect because the length of the unknown side is the square root of 7,225.
Mathematics
2 answers:
Softa [21]3 years ago
6 0

Answer:

the correct answer for this is the last choice:

It is incorrect because the length of the unknown side is the square root of 7,225.

ASHA 777 [7]3 years ago
4 0

the correct answer for this is the last choice:

It is incorrect because the length of the unknown side is the square root of 7,225.

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7 0
3 years ago
The dye dilution method is used to measure cardiac output with 3 mg of dye. The dye concentrations, in mg/L, are modeled by c(t)
Lemur [1.5K]

Answer:

Cardiac output:F=0.055 L\s

Step-by-step explanation:

Given : The dye dilution method is used to measure cardiac output with 3 mg of dye.

To Find : Find the cardiac output.

Solution:

Formula of cardiac output:F=\frac{A}{\int\limits^T_0 {c(t)} \, dt} ---1

A = 3 mg

\int\limits^T_0 {c(t)} \, dt =\int\limits^{10}_0 {20te^{-0.06t}} \, dt

Do, integration by parts

[\int{20te^{-0.6t}} \, dt]^{10}_0=[20t\int{e^{-0.6t} \,dt}-\int[\frac{d[20t]}{dt}\int {e^{-0.6t} \, dt]dt]^{10}_0

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20}{0.6}\int {e^{-0.6t} \,dt]^{10}_0

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20e^{-0.6t}}{(0.6)^2}]^{10}_{0}

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-200e^{-6}}{0.6}+\frac{20e^{-6}}{(0.6)^2}]+\frac{20}{(0.60^2}

[\int{20te^{-0.6t}} \, dt]^{10}_0=\frac{20(1-e^{-6}}{(0.6)^2}-\frac{200e^{-6}}{0.6}

[\int{20te^{-0.6t}} \, dt]^{10}_0\sim {54.49}

Substitute the value in 1

Cardiac output:F=\frac{3}{54.49}

Cardiac output:F=0.055 L\s

Hence Cardiac output:F=0.055 L\s

4 0
3 years ago
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