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3 years ago

The value of x that makes lines a and b parallel is

Mathematics

2 answers:

ra1l [238]3 years ago

5 0

The value of x that makes lines a and b parallel is 76°.

This is because when two parallel lines are cut by a transversal, the alternate angles formed by the transversal are equal. In the figure below, the angle x is an alternative interior angle to the angle measured 76°, so they are equal. That is, x = 76° (provided that line a is parallel to line b)

murzikaleks [220]3 years ago

5 0

Answer:

76°

Step-by-step explanation:

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The Slow Fix auto shop charges $28 for parts and $48 per hour of labor. The We Work Cheaper auto shop charges $59 for parts and

Sidana [21]

Answer: 10 hours

Step-by-step explanation:

Given

Slow fix auto charge $28 for parts and $48 per hour of labor

We work charge $59 for parts and $44.90 per hour of labor

Suppose both works for x hours

Cost of Slow fix= $\$(28+48x)$

Cost of We work= $\$(59+44.90x)$

When these costs are the same

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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: $F=0.055 L\s$

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