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3 years ago

PLEASE HELP!!!!! WORTH 50 POINTS! PLEASE ANSWER ALL QUESTIONS (6,7,8,9)

Mathematics

1 answer:

iogann1982 [59]3 years ago

7 0

Answer:

6. Additional information is required on the two triangles

7. The pair of triangles are similar by SAS similarity postulate

8. Yes, by triangle proportionality theorem

9. DE = 9

Step-by-step explanation:

When two or more triangles are similar, the ratio of the corresponding sides of both triangles is constant

6. From the given figure, more information is required on the two triangles

However, whereby the side with length 10 is parallel to the corresponding side on the other triangle, then the alternate angles formed by the transversal to the two parallel lines are equal and the two triangles are similar by Angle Angle, AA, similarity postulate

7. From the given triangles, the ratio of the corresponding sides relative to the given included angles are;

18/45 and 10/25

18/45 = 2/5

10/25 = 2/5

Therefore, given that two pair of corresponding sides have the same included angle and that the two included corresponding sides of the of the two triangles are proportional, then the two triangles are similar by the Side Angle Side SAS similarity postulate

8. By triangle proportionality theorem, we have;

If $\overline{MN}$ ║ $\overline{KL}$ , then we have;

3/6 = 4/8 = 2

∴ $\overline{MN}$ ║ $\overline{KL}$

9. By triangle proportionality theorem, where $\overline{BE}$ ║ $\overline{CD}$ , we have;

12/20 = DE/15

∴ DE = 15 × 12/20 = 9

DE = 9

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Answer:

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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)

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Answer:

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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

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

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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