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koban [17]
2 years ago
15

Joan wants to reduce the area of her posters to be 2/3 of the original size. Match the original dimensions in the left column wi

th the correct new area in the right column.
Mathematics
1 answer:
svet-max [94.6K]2 years ago
3 0

Answer:

Actual                                      2/3

30 inch. by 12 inch               240 inch²

30 inch. by 18 inch               360 inch²

12 inch. by 15 inch                  120inch²

18 inch. by 15 inch                 180inch²

Step-by-step explanation:

The answers are given in the same line .

Actual                                                      1/3                          2/3

30 inch. by 12 inch                                120 inch²                 240 inch²

30 inch. by 18 inch                                180 inch²                 360 inch²

12 inch. by 15 inch                                  60 inch²                  120inch²

18 inch. by 15 inch                                90 inch²                    180inch²

We  find the areas and multiply by 2/3 to get the answers

30 * 12= 360 inch²

2/3 * 360 = 240 inch²

30 * 18= 540 inch²

2/3 * 540 = 360inch²

12 * 15= 180 inch²

2/3 * 180 = 120 inch²

18 * 15= 270 inch²

2/3 * 270 = 180inch²

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A surveyor leaves her base camp and drives 42km on a bearing of 032degree she then drives 28km on a bearing of 154degree,how far
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Answer:

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Step-by-step explanation:

The final position of the surveyor is represented by the following vectorial sum:

\vec r = \vec r_{1} + \vec r_{2} + \vec r_{3} (1)

And this formula is expanded by definition of vectors in rectangular and polar form:

(x,y) = r_{1}\cdot (\cos \theta_{1}, \sin \theta_{1}) + r_{2}\cdot (\cos \theta_{2}, \sin \theta_{2}) (1b)

Where:

x, y - Resulting coordinates of the final position of the surveyor with respect to origin, in kilometers.

r_{1}, r_{2} - Length of each vector, in kilometers.

\theta_{1}, \theta_{2} - Bearing of each vector in standard position, in sexagesimal degrees.

If we know that r_{1} = 42\,km, r_{2} = 28\,km, \theta_{1} = 32^{\circ} and \theta_{2} = 154^{\circ}, then the resulting coordinates of the final position of the surveyor is:

(x,y) = (42\,km)\cdot (\cos 32^{\circ}, \sin 32^{\circ}) + (28\,km)\cdot (\cos 154^{\circ}, \sin 154^{\circ})

(x,y) = (35.618, 22.257) + (-25.166, 12.274)\,[km]

(x,y) = (10.452, 34.531)\,[km]

According to this, the resulting vector is locating in the first quadrant. The bearing of the vector is determined by the following definition:

\theta = \tan^{-1} \frac{10.452\,km}{34.531\,km}

\theta \approx 16.840^{\circ}

And the distance from the camp is calculated by the Pythagorean Theorem:

r = \sqrt{(10.452\,km)^{2}+(34.531\,km)^{2}}

r = 36.078\,km

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

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Answer:

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Step-by-step explanation:

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