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mel-nik [20]
2 years ago
7

A certain right triangle with integer side lengths has perimeter $72$. What is its area?

Mathematics
1 answer:
JulijaS [17]2 years ago
7 0

Answer: 216 square units

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

Explanation:

A common pythagorean triple you may be familiar with is the 3-4-5 right triangle. This has two legs of 3 and 4, and a hypotenuse of 5. The perimeter is 3+4+5 = 7+5 = 12. Note how this is a factor of 72.

If we multiply the perimeter (12) by 6, then 12*6 = 72. So we have scaled the triangle by a factor of 6. Each length is 6 times longer

the side length 3 becomes 3*6 = 18

the side length 4 becomes 4*6 = 24

the side length 5 becomes 5*6 = 30

The new perimeter is 18+24+30 = 42+30 = 72

The last step is to find the area. The two legs of this triangle are the base and height

area = 0.5*base*height

area = 0.5*18*24

area = 9*24

area = 216

-----

Or you could find the area of the 3-4-5 right triangle to get

area = 0.5*base*height = 0.5*3*4 = 6

then multiply by 36 to get 6*36 = 216. The 36 is the square of the scale factor 6 we applied above. The new lengths are 6 times longer, so the new area is 6^2 = 36 times larger.

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Gary used landscape timbers to create a border around a garden shaped like a right triangle. The longest two timbers he used are
Cloud [144]

Answer:

9 feet

Step-by-step explanation:

Given:

The border of the garden is a right angled triangle.

Two lengths are given as 12 ft and 15 ft.

Let the length of the shortest timber be 'x' feet.

Now, in a right angled triangle, the longest length is called the hypotenuse.

As 15 feet is the largest length, it is the hypotenuse of the triangle. Now, applying Pythagoras theorem, we get:

(Leg1)^2+(Leg2)^2=(Hypotenuse)^2\\x^2+12^2=15^2\\x^2+144=225\\x^2=225-144\\x^2=81\\x=\pm \sqrt{81}=\pm 9

The negative value is neglected as length can never be negative.

Therefore, the length of the shortest timber is 9 feet.

8 0
2 years ago
600 can be written as 2a x b x cd where a,b,c and d are all prime numbers find the values of a, b, c and d
mihalych1998 [28]
The prime factorisation of 600 is given by

600 = 2 \times 2 \times 2 \times 3 \times 5 \times 5 = 2^3 \times 3 \times 5^2

Therefore, a = 3, b = 3, c = 5 and d = 2.
4 0
2 years ago
There are 10 pots exposed in the shop, 2 of which have hidden defects. the customer buys two pieces. what is the probability tha
Tpy6a [65]
Check the tree diagram, of all the possible scenarios and the probabilities.

All the possible scenarios are: 

1st.non defective,  2nd.non defective
1st.non defective,  2nd. defective
1st. defective,  2nd. non defective
1st. defective,  2nd. defective

consider the case 1st.non defective,  2nd.non defective

the probability that the first one is non defective is 8/10, since 8 out of 10 are non defective.

In this scenario, the second one is non defective as well. The probability for this to happen is 7/9, since now we have 7 non defective pots, and 9 in total.

This means that the probability of the first to be non defective, and the second non-defective is (8/10) * (7/9)=0.622

similarly we calculate the other cases, as shown in the picture.

P(at least one is defective)=
P(1st.non defective,  2nd. defective)+P(1st. defective,  2nd. non defective)+P(1st. defective,  2nd. defective)=
0.178+0.178+0.022=0.378


Answer: 0.378

8 0
2 years ago
Which of the following statements are true concerning the mean of the differences between two dependent samples​ (matched pairs)
elena-s [515]

Answer:

A and C

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations we can use it.  

And in order to conduct it we need to have some assumptions:

1) The dependent variable must be continuous (interval/ratio).

2) The observations are independent of one another.

3) The dependent variable should be approximately normally distributed.

4) The dependent variable should not contain any outliers.

Let's analyze one by one the options on this case:

A. The methods used to evaluate the mean of the differences between two dependent variables apply if one has 86 IQ scores of taxpayers from Texas and 86 IQ scores of taxpayers from Ohio

We can assume that is true since the two variables are dependent and we are assuming that all the other conditions are satisfied to use the t paired test.

B. The requirement of a simple random sample is satisfied if we have matched pairs of voluntary response data.

That's not neccesary true since is not a requirement in order to use the t pairedtest.

C. If one has more than 23 matched pairs of sample​ data, one can consider the sample to be large and there is no need to check for normality.

We can assume that is true since we need to ensure normality in order to apply the test and if the sample size is large enough large we can apply the test.

D. If one has twenty matched pairs of sample​ data, there is a loose requirement that the twenty differences appear to be from a normally distributed population.

False the requirement of normality is important to apply the test not a loose requirement.

E. If one wants to use a confidence interval to test the claim that mu Subscript d Baseline greater than 0 with a 0.01 significance​ level, the confidence interval should have a confidence level of 98​%.

That's not correct the significance level is \alpha=1-0.98=0.02 and that not correspond to the significance level given on this case.

7 0
2 years ago
What is the simplest form of 2sqrt2/sqrt3-sqrt2?
Alchen [17]

Answer:

2 sqr 6+4 or A

Step-by-step explanation:

7 0
2 years ago
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