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8090 [49]
2 years ago
5

PLEASE HELP!!!!!

Mathematics
2 answers:
Debora [2.8K]2 years ago
8 0

The correct answer is; "△DEF is not a right triangle because no two sides are perpendicular." - I took the test.

Alenkasestr [34]2 years ago
7 0
<span>â–łDEF is not a right triangle because no two sides are perpendicular. There are two ways to determine if DEF is a right triangle. You can use the pythagorean theorem to see of a^2 + b^2 = c^2, or you can calculate the slope of each pair of lines to see if their product is -1. I'll demonstrate both methods, but given the available options, I suspect that the product of the slopes is the one you need. Slope method. Slope of DE = (1-(-1))/(-4-3) = 2/-7 = -2/7 Slope of EF = (-1-(-4))/(3-(-1)) = 3/4 Slope of DF = (1-(-4))/(-4-(-1)) = 5/-3 = -5/3 Since we're looking for -1 as a product and we have 2 negative slopes and 1 positive slope, we only need to check DE against EF and EF against DF. So -2/7 * 3/4 = -6/28 = -3/14. This isn't -1, so not perpendicular. 3/4 * -5/3 = -15/12 = -5/4. Still not -1, so not perpendicular. Since no sides of perpendicular to any other sides, DEF is not a right triangle. Now for the pythagorean theorem method. First, let's take a look at the length of the 3 sides of DEF. Actually, we just need to look at the square of the lengths of each side. So square of length DE = (-4 - 3)^2 + (1 - (-1))^2 = -7^2 + 2^2 = 49 + 4 = 53. square of length EF = (3-(-1))^2 + (-1-(-4))^2 = 4^2 + 3^2 = 16+9 = 25 square of length DF = (-4-(-1))^2 + (1-(-4))^2 = -3^2 + 5^2 = 9 + 25 = 34 The longest side is DE, so lets add up the squares of the other 2 sides to see if they equal the square of DE. 25 + 34 = 59. And 59 is not equal to 53. So we know that triangle DEF is NOT a right triangle. Now let's look at the options and see what's correct. â–łDEF is not a right triangle because no two sides are perpendicular. * This is a true statement. So it's the correct answer. The remaining 3 other options all claim that one side is perpendicular to another and as such are incorrect.</span>
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In J, K is the midpoint. The coordinates of ) are (2, 2), and the coordinates of Kare (10, 11). What are the coordinates of L?
DiKsa [7]

Answer:

L(18, 20)

Step-by-step explanation:

In JL, K is the midpoint. The coordinates of J are (2, 2), and the

coordinates of K are (10, 11). What are the coordinates of L?

Solution:

If O(x, y) is the midpoint between two points A(x_1,y_1) and B(x_2,y_2). The equation to determine the location of O is given by:

x=\frac{x_1+x_2}{2} \\\\y=\frac{y_1+y_2}{2}

Since JL is a line segment and K is the midpoint. Given the location of J as (2, 2) and K as (10, 11). Let (x_2,y_2) be the coordinate of L. Therefore:

10=\frac{2+x_2}{2} \\\\20=2+x_2\\\\x_2=18

11=\frac{2+y_2}{2} \\\\22=2+y_2\\\\y_2=20

Therefore L = (18, 20)

4 0
2 years ago
ABCD diagonals intersect at E. If m∠BAC=4x+5 and m∠CAD=5x-14, then find m∠CAD.
Travka [436]

Answer: m∠CAD = 81°

Step-by-step explanation: <u>Diagonal</u> is a line that unites opposite sides.

ABCD is a prallelogram. One property of diagonal in a parallelogram is it separates the parallelogram in 2 congruent triangles.

The figure below shows ABCD with its diagonals.

Since diagonal divides a parallelogram in 2 congruent triangles, it means the internal angles are also congruent. So

m∠BAC = m∠CAD

4x + 5 = 5x - 14

x = 19

Then, m∠CAD is

m∠CAD = 5(19) - 14

m∠CAD = 81

The angle m∠CAD is 81°.

7 0
2 years ago
What is the percentage of "85% of 78.00?" (Also needs calculation)
Tatiana [17]

-21.85.999

I found the ansewr by dividing both of  the numbers to equal tthat

6 0
2 years ago
A new bank account is opened on week 1 with a $200 deposit. After that first week, weekly deposits of $55 are made in the accoun
inysia [295]
Y=55x+200, you multiply the number of weeks by the amount of money, and add 200 because they started with that.   
7 0
3 years ago
Read 2 more answers
for any positive integer n, the sum of the first n positive integers equals n(n 1)/2. what is the sum of all the even integers b
marusya05 [52]

The sum of all the even integers between 99 and 301 is 20200

To find the sum of even integers between 99 and 301, we will use the arithmetic progressions(AP). The even numbers can be considered as an AP with common difference 2.

In this case, the first even integer will be 100 and the last even integer will be 300.

nth term of the AP = first term + (n-1) x common difference

               ⇒    300 = 100 + (n-1) x 2

Therefore, n = (200 + 2 )/2 = 101

That is, there are 101 even integers between 99 and 301.

Sum of the 'n' terms in an AP = n/2 ( first term + last term)

                                                = 101/2 (300+100)

                                                = 20200

Thus sum of all the even integers between 99 and 301 = 20200

Learn more about arithmetic progressions at brainly.com/question/24592110

#SPJ4        

7 0
1 year ago
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