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Mrrafil [7]
2 years ago
5

Find the slope on the line.

Mathematics
2 answers:
AysviL [449]2 years ago
5 0

Answer: -2 is the slope

(0,0) (1,4) (rise over run) *Y2 - Y1/X2 - X1*

X1= 0

Y1= 4

X2= 1

Y2= 2

2-4= -2

1-0= 1

-2/1= -2

the slope is -2

Lynna [10]2 years ago
4 0
Slope=4

Rise 4
Run 1

Rise over run - 4/1=4
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3 years ago
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What is the net force acting on a 3000kg sled that accelerated a 5 m/s2
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Answer:

15000 Newtons

Step-by-step explanation:

Force (in newtons)= weight * acceleration = 3000kg * 5 m/s^2 = 15000 N

5 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
Which does NOT have a value of 4 when
12345 [234]

Answer:

3 beause when you dividend it -5.6+4(3) .

7 0
3 years ago
What is the slope of the line graphed in the coordinate grid?
zvonat [6]

Answer:

3/4

Step-by-step explanation:

I drew a slope triangle and found the slope by finding the distance between two points on the line.

vertical distance/horizontal distance

Hope this helped!

5 0
3 years ago
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