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Black_prince [1.1K]
4 years ago
14

Alex is drawing rectangles with different areas on a centimetre grid he can draw 3 different rectangles with an area of 12cmsqua

red a) How many different rectangles could alex draw with an area of 11cmsquared? B)Give a reason for ur anwser
Mathematics
1 answer:
Naily [24]4 years ago
7 0

Answer:

Alex can only draw a rectangle with an area of 11 cm²

The reason for the answer is that 11 is a prime number

Step-by-step explanation:

We are told that Alex is drawing rectangles with different areas on a centimeter grid, and he can draw 3 different rectangles with an area of 12 cm²

We have the minimum grid length of centimeters = 1 cm, since we assume that only integers can be used.

That is to say:

12 = 1 x 12

12 = 2 x 6

12 = 3 x 4

These are the 3 different rectangles with an area of 12 cm²

Now we have to find how many rectangles Alex could draw with an area of 11 cm²

11 = 1 x 11

Therefore, only one factorization is possible. Which means that you can only draw a rectangle with an area of 11 cm²

The reason for the answer is that 11 is a prime number, which means only the common factor of 1 and the number itself, that is, 11.

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Murrr4er [49]
<h3>You have the correct answer. It is choice B. Nice work.</h3>

K is the midpoint, so \overline{IK} \cong \overline{GK} and \overline{JK} \cong \overline{HK}, and along with the congruent vertical angles (\angle GKH \cong \angle IKJ and \angle GKJ \cong \angle IKH), you would use the SAS (side angle side) congruence theorem to prove the inner pairs of triangles to be congruent.

So, \triangle HKG \cong \triangle JKI (top and bottom triangles) and \triangle GKJ \cong \triangle IKH (left and right triangles).

Then through CPCTC, we can show the corresponding pieces are congruent leading to \overline{GH} \cong \overline{JI} and \overline{GJ} \cong \overline{HI} showing the opposite sides of the quadrilateral are congruent. Therefore we do have a parallelogram and enough information to prove it as such.

Side note: CPCTC stands for "corresponding parts of congruent triangles are congruent".

3 0
4 years ago
Read 2 more answers
Ms. Lee wants to know what kinds of things her students do prepare for her tests. Ms. Lee placed the names of her students into
Darina [25.2K]

Answers:

  • a) The sample is the set of students Ms. Lee selects from the box.
  • b) The population is the set of all students in Ms. Lee's classroom.

=============================================

Explanation:

The first sentence tells us what the population is: it's the set of all her students. She's not concerned with any other students in any other classroom. So her "universe", so to speak, is solely focused on this classroom only. Once the population is set up, a sample of it would be a subset of the population.

If set A is a subset of set B, then everything in A is also in B, but not vice versa. For example, the set of humans is a subset of the set of mammals because all humans are mammals. However, a dog is a mammal but not a human. This shows that A is a subset of B, but not the other way around. In this example, A = humans and B = mammals.

Going back to the classroom problem, we have A = sample and B = population. If Ms. Lee has 30 students, and she randomly selects 5 of them, then those 30 students make up set B and the 5 selected make up set A. Selecting the names randomly should generate an unbiased sample. This sample should represent the population overall. If the population is small enough, the teacher could do a census and not need a sample. Though there may be scenarios that it's still effective to draw a sample.

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3 years ago
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Step-by-step explanation:

Answer

a - 8

------------+

2

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Answer:

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