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galina1969 [7]
3 years ago
10

In each figure a || b. Determine whether the third figure is a translation image of the first figure. Write yes or no. Explain y

our answer.

Mathematics
1 answer:
harina [27]3 years ago
6 0

Answer:

  No

Step-by-step explanation:

The second and third figures are translations of each other. The first figure appears to be a reflection of the second, not a translation. Hence the first and third figures are not translations of each other.

__

If an image is translated, all of the pre-image line segments are parallel to the image line segments. That is not the case for the first and third figures, in which the top line segments go in different directions.

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Use technology to approximate the solution(s) to the system of equations to the nearest tenth of a unit.
kherson [118]

Answer:

  • A.)  (3.6, 0.6)
  • D.)  (2.6, 0.4)

Step-by-step explanation:

See below for a graph.

___

Choices B, C, E can be eliminated on the basis that neither x nor g(x) can be negative. The domain of f(x) is x>0; the range of g(x) is x≥0.

7 0
3 years ago
Please help!!! Don’t mind the one I picked, I accidentally clicked one.
natima [27]

Answer:

y<2/3x-1

Step-by-step explanation:

plotted point is on -1

slope of the line is rise 2, run 3

so its the last option

8 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
-2 1/4 divided by -1 1/2
xeze [42]

Answer: 1 1/2

Step-by-step explanation:

5 0
3 years ago
Zane bought a pair of jeans that originally cost $56. He used a coupon for 25% off and paid 8% in sales tax. How much did he pay
sattari [20]

Answer:

$38.64

Step-by-step explanation:

so the equation needed to solve is $56*0.25 and that number is 14. Since it is a coupon you subtract 14 from 56 and end up with 42. Now multiply 0.08*42 and you got your sales tax, $3.36. now subtract that from 42 and you have your answer! Don't forget the dollar sign!

hope this helped! : )

5 0
3 years ago
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