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2 years ago

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In right triangle XYZ, the right angle is located at vertex Y. The length of line segment XY is 12.4 cm. The length of line segm

ent YZ is 15.1 cm. Which is the approximate measure of angle YZX? 34.8° 39.4° 50.6° 55.2°

2 answers:

saw5 [17]2 years ago

7 0

Answer:

∠ YZX ≈ 39.4°

Step-by-step explanation:

Using the tangent ratio in the right triangle

tan XYZ = $\frac{opposite}{adjacent}$ = $\frac{XY}{YZ}$ = $\frac{12.4}{15.1}$ , thus

∠ XYZ = $tan^{-1}$ ( $\frac{12.4}{15.1}$ ) ≈ 39.4°

BabaBlast [244]2 years ago

5 0

Answer:

39.4

Step-by-step explanation:

tan(YZX) = (12.4/15.1)

tan-1(12.4/15.1) = YZX

tan-1(39.4) = YZX

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Frank had hip replacement surgery and was given a prescription with instructions to take a 200 milligram (mg) tablet three times

IgorC [24]

Answer:

12,600mg per 21 days

Step-by-step explanation:

200mg three times a day → (200mg)(3) = 600mg per a day

21 days → (21)(600mg) = 12,600mg per 21 days

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2 years ago

Triangle A has a height of 2.5\text{ cm}2.5 cm2, point, 5, start text, space, c, m, end text and a base of 1.6\text{ cm}1.6 cm1,

konstantin123 [22]

Answer:

Option A

Option D

Option E

Step-by-step explanation:

we know that

If the height and base of triangle B are proportional to the height and base of triangle A

then

Triangle A and Triangle B are similar

Remember that

If two triangles are similar then the ratio of its corresponding sides is proportional and its corresponding angles are congruent

so

$\frac{h_A}{h_B} =\frac{b_A}{b_B}$

where

h_A and h_B are the height of triangle A and triangle B

b_A and b_B are the base of triangle A and triangle B

In his problem we have

$h_A=2.5\ cm\\b_A=1.6\ cm$

substitute

$\frac{2.5}{h_B} =\frac{1.6}{b_B}$

Rewrite

$\frac{2.5}{1.6} =\frac{h_B}{b_B}$

$\frac{h_B}{b_B}=1.5625$

<u><em>Verify all the options</em></u>

A) we have

$h_B=2.75\ cm\\b_B=1.76\ cm$

Find the ratio of the height to the base of triangle B and compare the result with the ratio of height to the base of triangle A (the value is 1.5625)

substitute the values in the proportion

$\frac{2.75}{1.76}=1.5625$

The ratios are the same

That means that are proportional

therefore

These values could be the height and base of triangle B

B) we have

$h_B=9.25\ cm\\b_B=9.16\ cm$

Find the ratio of the height to the base of triangle B and compare the result with the ratio of height to the base of triangle A (the value is 1.5625)

substitute the values in the proportion

$\frac{9.25}{9.16}=1.0098$

The ratios are not equal

That means that are not proportional

therefore

These values could not be the height and base of triangle B

C) we have

$h_B=3.2\ cm\\b_B=5\ cm$

Find the ratio of the height to the base of triangle B and compare the result with the ratio of height to the base of triangle A (the value is 1.5625)

substitute the values in the proportion

$\frac{3.2}{5}=0.64$

The ratios are not the same

That means that are not proportional

therefore

These values could not be the height and base of triangle B

D) we have

$h_B=1.25\ cm\\b_B=0.8\ cm$

Find the ratio of the height to the base of triangle B and compare the result with the ratio of height to the base of triangle A (the value is 1.5625)

substitute the values in the proportion

$\frac{1.25}{0.8}=1.5625$

The ratios are the same

That means that are proportional

therefore

These values could be the height and base of triangle B

E) we have

$h_B=2\ cm\\b_B=1.28\ cm$

Find the ratio of the height to the base of triangle B and compare the result with the ratio of height to the base of triangle A (the value is 1.5625)

substitute the values in the proportion

$\frac{2}{1.28}=1.5625$

The ratios are the same

That means that are proportional

therefore

These values could be the height and base of triangle B

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mote1985 [20]

Answer:

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Step-by-step explanation:

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$\dfrac{b+c}{2}$

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SpyIntel [72]

Answer:

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Step-by-step explanation:

The given coordinates of the vertices of the quadrilateral TRAM are;

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By a rotation of -90 degrees = Rotation of 90 degrees clockwise, we get;

The coordinates of the preimage before rotation = (x, y)

The coordinates of the image after rotation = (y, -x)

Therefore, we get for the the quadrilateral T'R'A'M', by rotating TRAM -90 degrees as follows;

T(-5, 1) → T'(1, 5)

R(-7, 7) → R'(7, 7)

A(-1, 7) → A'(7, 1)

M(-5, 4) → M'(4, 5)

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