#10 On a blueprint, 1 inch = 13 feet. If the dimensions of the playroom on the blueprint are 0.35 inches x 1.25 inches, what are its actual measurements? *
*Show all your work for full credit*
#11 The coordinate of rectangle ABCD are given. Find the coordinates of its image after a dilation with the scale factor of 2.
A (2,4) B (-3, -1) C (0,9) D (6,-4)
#12 Your friend has a mini train set modeled after a real train. The model of the train is 6 feet long and 3 feet wide. The actual train is 70 yards long. What is the train’s actual width? If necessary, round to the nearest tenth of a mile. *
*Show all your work for full credit*
CAN YOU PLEASE HELP ME I HAVE NO IDEA HOW TO DO THIS PLEASE PLEASE PLEASE!!!!
10. Multiply the dimensions by 13.
11. I don't know.
12. There's an error in this problem.
actually i got #12 its 6/3 = 70/h
6 x h = 3 x 70
6h = 210
6 ÷ 6 = 210 ÷ 6
h = 35
Ms. Sue? Is Savannah's answer correct?
No.
The actual train is 70 yards long -- but you're supposed to state the width in miles???? That doesn't make sense.
ok, put aside that question, what i 11????? i cannot figure it out
Cut a piece of ribbon 6.18 m long into 6 equal pieces
Of course, I'll be happy to help you. Let's go through each question step by step:
#10: To find the actual measurements of the playroom, we need to use the conversion rate given on the blueprint. It states that 1 inch on the blueprint represents 13 feet in real life.
To find the actual length of the playroom, you multiply the length on the blueprint (0.35 inches) by the conversion factor (13 feet/inch):
Length = 0.35 inches * 13 feet/inch = 4.55 feet
Similarly, to find the actual width of the playroom, you multiply the width on the blueprint (1.25 inches) by the conversion factor (13 feet/inch):
Width = 1.25 inches * 13 feet/inch = 16.25 feet
Therefore, the actual measurements of the playroom are approximately 4.55 feet by 16.25 feet.
#11: To find the coordinates of the image (after dilation with a scale factor of 2) of rectangle ABCD, we need to multiply each coordinate of ABCD by the scale factor.
For point A:
New x-coordinate = 2 (original x-coordinate) * 2 (scale factor) = 4
New y-coordinate = 4 (original y-coordinate) * 2 (scale factor) = 8
So the new coordinate of point A is (4, 8).
Similarly, you can apply the same process to find the new coordinates for points B, C, and D.
For point B:
New x-coordinate = -3 (original x-coordinate) * 2 (scale factor) = -6
New y-coordinate = -1 (original y-coordinate) * 2 (scale factor) = -2
So the new coordinate of point B is (-6, -2).
For point C:
New x-coordinate = 0 (original x-coordinate) * 2 (scale factor) = 0
New y-coordinate = 9 (original y-coordinate) * 2 (scale factor) = 18
So the new coordinate of point C is (0, 18).
For point D:
New x-coordinate = 6 (original x-coordinate) * 2 (scale factor) = 12
New y-coordinate = -4 (original y-coordinate) * 2 (scale factor) = -8
So the new coordinate of point D is (12, -8).
Therefore, the coordinates of the image of rectangle ABCD after dilation with a scale factor of 2 are: A (4, 8), B (-6, -2), C (0, 18), D (12, -8).
#12: To find the actual width of the train, we'll use the scale of the model train compared to the actual train.
Given that the model train is 6 feet long and 3 feet wide, and the actual train is 70 yards long, we need to find the width of the actual train.
First, we convert the width of the model train from feet to yards:
Model train width = 3 feet = 1 yard (since 3 feet = 1 yard)
Next, we set up a proportion using the model train and the actual train:
Model train width / Actual train length = Model train length / Actual train width
Substituting the values we have:
1 yard / 70 yards = 6 feet / Actual train width
Simplifying the proportion:
1/70 = 6 / Actual train width
To isolate Actual train width, we cross-multiply and divide:
(1/70) * Actual train width = 6
Actual train width = 6 * 70 = 420 yards
Finally, if necessary, we can round the actual train width to the nearest tenth of a mile:
420 yards ≈ 0.2386 miles ≈ 0.2 miles (rounded to the nearest tenth)
Therefore, the actual width of the train is approximately 0.2 miles.