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Ira Lisetskai [31]
3 years ago
5

If ΔMNL is rotated 180° about point N, which additional transformation could determine if ΔONP and ΔMNL are similar by the AA si

milarity postulate?
Segments OM and LP intersect at point N; triangles are formed by points LNM and ONP; line k intersects with both triangles at point N.

Reflect ONP over line k.
Reflect M'N'L' over line k.
Dilate ONP from point N by a scale factor of segment NP over segment NL.
Dilate M'N'L' from point N by a scale factor of segment NP over segment NL
Mathematics
2 answers:
Oksi-84 [34.3K]3 years ago
6 0

The answer is D because  you need to make it smaller by dilation. This leaves the last two options. D makes the most sense.

lesya [120]3 years ago
4 0

Answer:

  Dilate M'N'L' from point N by a scale factor of segment NP over segment NL

Step-by-step explanation:

Multiplying the length of N'L' by the factor NP/NL will give it the length of NP, making the dilated version of ΔM'N'L' congruent to ΔONP. This is apparently your goal.

__

Reflection over line k doesn't seem to do anything useful, and the other offered dilation is by the wrong factor. You want to ...

dilate M'N'L' from point N by a scale factor of segment NP over segment NL.

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Hello I need help on this question!
Sloan [31]

Answer:

5

Step-by-step explanation:

Refer to attachment for marking of sides.

In the given figure , ∆ABC , ∆ABD and ∆ADC are right angled triangles . Therefore here we can use the Pythagoras theorem , as ,

\longrightarrow base² + perpendicular² = hypotenuse ² .

<u>•</u><u> </u><u>In </u><u>∆</u><u>A</u><u>B</u><u>D</u><u> </u><u>,</u><u> </u><u>we </u><u>have</u><u> </u><u>;</u>

\longrightarrow AB² + BD² = AD²

\longrightarrow AB² + x² = 10²

\longrightarrow AB² = 10² - x²

\longrightarrow AB² = 100 - x²

\rule{200}2

<u>•</u><u> </u><u>Again</u><u> </u><u>in </u><u>∆</u><u>A</u><u>D</u><u>C</u><u> </u><u>,</u><u> </u><u>we </u><u>have</u><u> </u><u>;</u>

\longrightarrow AC² + AD² = CD²

\longrightarrow AC² = 20² - 10²

\longrightarrow AC² = 400 - 100

\longrightarrow AC² = 300

\rule{200}2

<u>Again</u><u> </u><u>in </u><u>∆</u><u>A</u><u>B</u><u>C</u><u> </u><u>,</u><u> </u><u>we </u><u>have</u><u> </u><u>,</u>

\longrightarrow AC² = AB² + BC²

Substituting the values from above ,

\longrightarrow 300 = 100-x² + (20-x)²

\longrightarrow 300 = 100 - x² + 400 + x² - 40x

\longrightarrow 40x = 500 - 300

\longrightarrow 40x = 200

\longrightarrow x = 200/40

\longrightarrow x = 5

<h3>Hence the required answer is 5 .</h3>

8 0
2 years ago
How do you solve these problems without numbers
insens350 [35]

i dont know if you want me to answer them or not, but you have to COMBINE LIKE TERMS if you dont then your problems would be messed up example if you have a+a it would be 2a make sense? so you want everything combined then you have to solve. theres no possible way of solving these without numbers number would have to be added although youll probably end up getting rid of the numbers in the end. after you combine like terms do the rest of solving


hope that helps if you need more help pm me, :)

7 0
3 years ago
1. Is it true, on your scale, that 8 fluid ounces of cornflakes weigh only 1 ounce?
mariarad [96]

Answer:  Yes, it is true, on the scale that 8 fluid ounces of cornflakes weighs only 1 ounce

2. Peanut butter Conversion Chart Near 2.2 US fluid ounces.

3. 1 US cup (236.59mL) = 8 US fluid ounces

Step-by-step explanation:

3 0
2 years ago
Pls write the steps u did​
frez [133]

When two parallel lines are intersected by a transversal, the same-side exterior angles are supplementary. That means that their sum is 180.

Using that logic, if the two roads were parallel, then the sum of their same-side exterior angles will add up to 180. Yet their same-side exterior angles add up to 170 (130 + 40 = 170), hence they can't be parallel.

See the drawing attached below.

Using supplmenatry angles (two angles whose sum of measures add up to 180 or a straight line), we can say that:

m<DIE + m<HID = 18

40 + m<HID = 180

m<HID = 140

Similarly:

m<BHC + m<CHI = 180

130 + m<CHI = 180

m<CHI = 50

Using verticle angles therome, (when two lines intersect, the angles opposite to eachother are congruent, or have the same measure), we can say that:

m<DIE = m<GIH = 40

m<GIE = m<HID = 140

m<CHI = m<AHB = 50

m<BHC = m<AHI = 130

5 0
2 years ago
in june kayla spent all of the $80 she had earned. the difference between her earnings in june and her money remaining from may
docker41 [41]

Answer:

She earned $60

Step-by-step explanation:

Since she had 20 dollars remaining from may, she earned the rest of the money in june so:

80-20=60

7 0
3 years ago
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