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3 years ago

5

If 1 is a root of the equation

mula1" title="5x^{2} +kx - 3 = 0 " alt="5x^{2} +kx - 3 = 0 " align="absmiddle" class="latex-formula">
find the other root

1 answer:

Mazyrski [523]3 years ago

6 0

Step-by-step explanation:

5x² + kx - 3 = 0 has a root x = 1.

=> 5(1)² + (1)k - 3 = 0

=> 5 + k - 3 = 0

=> k = -2.

Therefore we have 5x² - 2x - 3 = 0.

=> (x - 1)(5x + 3) = 0

=> x = 1 or x = -3/5.

Hence the other root is -3/5.

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heba ate 1/12 of a box of cereal now the box is 3/4 full what fraction of a full box was there before heba ate it

abruzzese [7]

Now box is 3/4 full

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3 years ago

Read 2 more answers

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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Step-by-step explanation:

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