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3 years ago

HELP ME PLSSSS

Mathematics

2 answers:

gogolik [260]3 years ago

8 0

The right answer would be C

Alchen [17]3 years ago

7 0

Answer:

the correct answer is option c. y= 192/x

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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

Advocard [28]

Answer:

The measure of angle YZX is 39.4°.

Step-by-step explanation:

Given information: ΔXYZ is a right angled triangle, ∠Y=90°, XY=12.4 cm, YZ=15.1 cm.

In a right angled triangle,

$\tan \theta = \frac{opposite}{adjacent}$

In triangle XYZ,

$\tan (\angle YZX) = \frac{XY}{YZ}$

$\tan (\angle YZX) = \frac{12.4}{15.1}$

Taking tan⁻¹ on both sides.

$\angle YZX =\tan^{-1} (\frac{12.4}{15.1})$

$\angle YZX =39.3925680393$

$\angle YZX \approx 39.4^{\circ}$

Therefore the measure of angle YZX is 39.4°.

8 0

4 years ago

A. -7/8x-3/2<br> B.y=7/8x+3/2<br> C.y=-8/7x+3/2<br> D.y=-3/2x+7/8<br> Hi I need help

uranmaximum [27]

Answer:

Hi! your answer is A. -7/8x-3/2

Step-by-step explanation:

to find the slope of the line (-7/8) you count the x and y difference between the 2 points. It goes down 7 and right 8 units.

to find the y-intercept you just take it from the graph. The line crosses the y-axis at y=-3/2.

8 0

3 years ago

I need help taking quiz !

Firlakuza [10]

Answer:

3,4,5,6,7

Step-by-step explanation:

thats the anwser

4 0

3 years ago

What is 3 times the power of 10

Paha777 [63]

I believe its 27 cause u have to multiply 3 times 3 which is 9, then 3 times 9 which is 27

8 0

4 years ago

Read 2 more answers

PLS HELP ME!!!!!!!! 99 POINTS

Amiraneli [1.4K]

Answer:

$\angle LKN=\arctan1.8\sqrt{97}\approx 86.77^{\circ}$

Step-by-step explanation:

Draw the second altitude MB (see attached diagram).

Quadrilateral HLMB is a rectangle, then LM = HB = 3 units.

Trapezoid KLMN is isosceles trapezoid (because KL=MN), thus

$KH = BN = \dfrac{KN-HB}{2}=\dfrac{13-3}{2}=5\ units\\ \\HN=HB+BN=3+5=8\ units$

Triangle LHN is right triangle, then by Pythagorean theorem,

$LN^2=LH^2+HN^2\\ \\89^2=LH^2+8^2\\ \\LH^2=89^2-8^2\\ \\LH^2=(89-8)(89+8)\\ \\LH^2=81\cdot 97\\ \\LH=9\sqrt{97}\ units$

Consider right triangle KLH. In this triangle,

$\tan\angle LKH=\{\tan\angle LKH\}=\dfrac{\text{opposite leg}}{\text{adjacent leg}}=\dfrac{LH}{KH}=\dfrac{9\sqrt{97}}{5}=1.8\sqrt{97}\ units$

So,

$\angle LKN=\arctan1.8\sqrt{97}\approx 86.77^{\circ}$

5 0

3 years ago