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kirza4 [7]
3 years ago
9

I only need to answer the first one, but I need you to tell me if the others are correct or incorrect, if they are incorrect ple

ase help me.
Rotation of 90 degrees counter clockwise of point A (7, 4).
Quadrant of given point: 1
Coordinates after the switch: 4, 7
Quadrant after rotation:
New coordinates after rotation:

Rotation of 180 degrees of point B at (-8, -2).
Quadrant of given point: 3
Quadrant after rotation: 1
New coordinates after rotation: 8, 2

Rotation of 270 degrees clockwise around the origin of point C at (9, -5).
Quadrant of given point: 4
Coordinates after the switch: 5, 9
Quadrant after rotation: 1
New coordinates after rotation: 5. 9

Rotation of 90 degrees clockwise around the origin of point D at (-6, -11).
Quadrant of given point: 3
Coordinates after the switch: 11, 6
Quadrant after the rotation: 2
New coordinates after rotation: -11, 6
Mathematics
2 answers:
Zepler [3.9K]3 years ago
7 0
The quadrant would be second
zavuch27 [327]3 years ago
4 0
Answer 2 I think sry if it’s wrong
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m = 3/2

Step-by-step explanation:

Use the slope formula: (y2 - y1) / (x2 - x1)

4 0
3 years ago
HELP!!!!! what 2 numbers can you multiply to get -6 and then add to get 5
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8 0
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

4 0
2 years ago
A container holds 13 quarts of water. How much is this in gallons?
patriot [66]

Answer:

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Step-by-step explanation:

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What is 11.625 rounded to the nearest whole number
Andru [333]

Answer:

your answer is 12

Step-by-step explanation:

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