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serious [3.7K]
3 years ago
9

Triangle Midsegment Theorem What are the coordinates of the endpoints of the midsegment for △LMN that is parallel to LN¯¯¯¯¯ ?

Mathematics
2 answers:
egoroff_w [7]3 years ago
6 0
<span>A midsegment of a triangle is a segment that connects the midpoints of two sides of a <span>triangle. The midsegment of the triangle parallel to side LN is the midsegment connecting the midpoint of side LM and the midpoint of side MN.

From the given figure, the midpoint of side LM is given by \left( \frac{0+6}{2} ,\,  \frac{5+4}{2} \right)=(3,\, 4.5)

Also, the midpoint of the side MN is given by </span></span>\left( \frac{0+6}{2} ,\,  \frac{2+4}{2} \right)=(3,\, 3)

Therefore, <span>the coordinates of the endpoints of the midsegment for △LMN that is parallel to side LN are: (3, 4.5) and (3, 3).</span>
lesya [120]3 years ago
3 0
All you have to do is use the Midpoint formula  Plug in Rs and RT points to get (3,3) and (3,4.5). just took the test this is the answer <3 
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<span>x+y=30.....(1) and x-y=2......(2) so(1)+(2)......x+y+x-y=30+2 so 2x=32...x=16 now apply x=16 in (1) so we get......16+y=30 so .......y=14</span>
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Step-by-step explanation:

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Describe the process of creating a linear equation using two points and the point-slope form.
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8 0
3 years ago
HELPP!!! 30 POINTS!!! WILL MARK BRAINLYIST!! Step 1: Written Response (30 points) Using complete sentences, compare the fat cont
yan [13]

Answer:

median a= 65 b= 75 c=68

This shows the median of each fat content of each restaurant showing that restaurant c is closer to that of 'restaurant a ' than a is to 'restaurant b' and near to 1/7 lower than that of restaurant b which shows they are all highly skewed with each other upon their graph.

The measures of spread start with 55 to 72 for 'restaurant a' where the upper and lower quartiles  are 70-60 showing a distribution 10-17 into the whisker plot.

Compared to restaurant b which shows a measure of spread start with 65 to 90 for 'restaurant b' where the upper and lower quartiles are 85 - 68 into the whisker plot and show a spread of 3 in the left exterior and 5 spread in the right exterior to the upper quartile. So in comparison to restaurant a that restaurant b has a greater spread in the interquartile range = 15  where restaurant a =10.

For the outer quartile the comparison to restaurant a is twice as small meaning restaurant a is greater by over double and shown as   53 <a < 60 which when showing both outer quartiles we can use the amounts being 7 and 70<a<72 = 2

and show closed dots on equality line number line separately showing a<-7  and a>2 for each outer exterior quartile. For restaurant b this shows opposite similar equalities 65<b< 68 = b>-3 and   82<b<89 = b>7

So the differences are smaller for b for left side by 4 and larger for b by 5 up on the right side.

We compare both to restaurant c and find c =   60<c<61 = c>-1 and 70<c<71 = c>1 so the differences are that the outer quartile for c through the other quartiles = -2> c < -6 . This shows how smaller c is compared to a and b outer quartiles). We can also prove that while the outer quartiles are much more smaller for c than a and b we cna prove that c actually has an inner quartile more similar to c and closer distribution of b as the median is more closer for a and c where b has a greater output for median as restaurant b has the higher fat content and greater distribution within the inner quartiles over all.

Summary findings Restaurant c outer quartile is moderately skewed as they show -1 -0.5  on the left side and 1-0.5 on the right side. Restaurant a has a closer median inner quartile to c and closer distribution of inner distribution of c. It's output outer quartile distribiution is a distribution that is higher than b but a smaller inner quartile compared to b, when this happens then the distribution spread shows less range and fewer products to account for.

So i think restaurant b has the higher fat content to its menu.

Where restaurant a must be the healthiest as it holds the lowest range and larger gap is such range for healthier food.ie) when compared to the others. 

Meanings of what we are asked.

In a box and whisker plot: the ends of the box are the upper and lower quartiles, so the box spans the interquartile range. the median is marked by a vertical line inside the box. the whiskers are the two lines outside the box that extend to the highest and lowest observations.

Skewness refers to distortion or asymmetry in a symmetrical bell curve, or normal distribution, in a set of data. If the curve is shifted to the left or to the right, it is said to be skewed. Skewness can be quantified as a representation of the extent to which a given distribution varies from a normal distribution.

As a general rule of thumb:

If skewness is less than -1 or greater than 1, the distribution is highly skewed.

If skewness is between -1 and -0.5 or between 0.5 and 1, the distribution is moderately skewed.

If skewness is between -0.5 and 0.5, the distribution is approximately symmetric.

4 0
4 years ago
Holly made 12 pair of earrings for each of her girlfriends. To wrap the pairs individually, she creates a simple cardboard gift
Mekhanik [1.2K]

Answer:

3280cm²

Step-by-step explanation:

We are told the gift box is folded into a Pyramid with a square base.

Hence, we are to find the Surface Area of a square pyramid

From the question, we are given:

Length of the base of the square = 10cm

Triangle

Side length = 10cm

Height = 8 2/3 cm

A square Pyramid has 5 faces: 4 triangles and 1 square.

Step 1

Find the Area of the square

Formula = Side length² this is because all the side of a square is equal.

Length of the base = 10cm

Area of the square = (10cm)²

= 100cm²

Step 2

Find the Area of the triangle. We are told the 4 triangles are equilateral, which means their side are equal.

Hence, for one triangle.

Area of a triangle = 1/2 × base × height

Base = side length = 10cm

Height = 8 2/3 = 8.6666666667cm

Area of a triangle = 1/2 × 10 × 8.6666666667

Area of a triangle = 43.333333333cm²

Since we have 4 triangles

We would have : 4 × 43.333333333cm²

= 173.33333333cm²

The Area of the 4 equilateral triangles =

173.33333333cm²

Step 3

The Surface area of the Square Pyramid

= Sum of the areas of all the faces.

= 100cm² + 173.33333333cm²

= 273.33333333cm²

Therefore, the minimum amount of cardboard needed to make one gift box = 273.33333333cm²

Step 4

The minimum amount of cardboard Holly needs to create 12 gift boxes to wrap each pair of earrings individually is calculated as: 12 × 273.33333333cm²

= 3280cm²

7 0
3 years ago
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