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3 years ago

Which is equivalent to 4 radical 9^1/2x

Mathematics

1 answer:

jek_recluse [69]3 years ago

8 0

Answer: 9 1/8x

Step-by-step explanation:

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3 years ago

A rectangular table is fivetimes as long as it is wide. if the area is 80ftsquared,find the length and the width of the table.

SOVA2 [1]

A = L * W
A = 80
L = 5W

80 = W(5W)
80 = 5W^2
80/5 = W^2
16 = W^2
sqrt 16 = W
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3 years ago

What is the equation, in slope-intercept form, of the line that is parallel to the given line and passes through the point (0,

Mamont248 [21]

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Step-by-step explanation:

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7 0

3 years ago

Use the diagram to complete the statement. Triangle J K L is shown. Angle K J L is a right angle. Angle J K L is 52 degrees and

zzz [600]

Answer:

$\bold{sin(38^\circ)=cos(52^\circ)}$

Step-by-step explanation:

Given that $\triangle KJL$ is a right angled triangle.

$\angle JKL = 52^\circ\\\angle KLJ = 38^\circ$

and

$\angle KJL = 90^\circ$

Kindly refer to the attached image of $\triangle KJL$ in which all the given angles are shown.

To find:

sin(38°) = ?

a) cos(38°)

b) cos(52°)

c) tan(38°)

d) tan(52°)

Solution:

Let us use the trigonometric identities in the given $\triangle KJL$ .

We have to find the value of sin(38°).

We know that sine trigonometric identity is given as:

$sin\theta =\dfrac{Perpendicular}{Hypotenuse}$

$sin(\angle JLK) = \dfrac{JK}{KL}\\OR\\sin(38^\circ) = \dfrac{JK}{KL}$ ....... (1)

Now, let us find out the values of trigonometric functions given in options one by one:

$cos\theta =\dfrac{Base}{Hypotenuse}$

$cos(\angle JLK) = \dfrac{JL}{KL}\\OR\\cos(38^\circ) = \dfrac{JL}{KL}$ ....... (2)

By (1) and (2):

sin(38°) $\neq$ cos(38°).

$cos(\angle JKL) = \dfrac{JK}{KL}\\OR\\cos(52^\circ) = \dfrac{JK}{KL}$ ...... (3)

Comparing equations (1) and (3):

we get the both are same.

$\therefore \bold{sin(38^\circ)=cos(52^\circ)}$

6 0

3 years ago