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natima [27]
3 years ago
13

Find the missing side round to the nearest tenth can someone please help me this is my last question

Mathematics
1 answer:
jeka57 [31]3 years ago
8 0

The question is related to trigonometric ratio rule .

In the given diagram, there is a right angle triangle, with adjacent =57, and the unknown side is y. Given angle measurement is 39 .

To find the measurement of y, we will use tan rule which is

tan \Theta = \frac{opposite}{adjacent}

tan 39=\frac{y}{57}

y = 57 tan 39 = 46.2

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4 bracelets on Thursday and 3 on Friday she uses 5/6 yard for each bracelet
qwelly [4]

Answer:

35/6 yards

Step-by-step explanation:

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Should I have plugged in values of theta rather than x to have received the points to graph the equations?
Artyom0805 [142]

Answer:

No, you did a good job!

4 0
3 years ago
Triangle GHK has an area of 117cm2. Write and solve an equation to find the height of the triangle GHK
stellarik [79]

Answer:

h=58.5/b

Step-by-step explanation:

This isn't necessarily possible, unless you know the base, but I can solve it with what we have, understanding the answer will contain variables:

The formula for the area of a triangle is A=1/2bh

117=(1/2)bh

58.5=bh

h=58.5/b

I'm not sure if this is the answer you were looking for, but I hope it helps

6 0
3 years ago
The restrooms for the former Estabrook Hall were known to each be 117 cubic feet in volume. We are tasked with creating a 1:4 sc
agasfer [191]

Answer:

29.25

Step-by-step explanation:

Make an equation:

\frac{1}{4} =\frac{x}{117}

117 = 4x

x = 29.25

7 0
3 years ago
Consider the following. Cube roots of −343 (a) Use the formula zk = n r cos θ + 2πk n + i sin θ + 2πk n to find the indicated ro
Lapatulllka [165]

Answer:

Z0 = 7 ( cos 60° + isin60°)

Z1 = 7( cos180° + isin180° )

Z2 = 7 ( cos300° + isin300°)

Step-by-step explanation:

Given that;

cube root of  -343

so, z = -343 + 0i

∴ r = √ (( -343)² + (0)²)  = 343

so tan∅ = y/x ⇒ tan∅ = 0/-343

∅ = tan⁻¹ (0/-343)

= 0 or 180°

but we are going yo make use of 180° since -343 is negative x-axis

Zk = ∛343 ( cos 180/3 + 360K/3) + isin(180/3 + 360k/3)

here k = 0, 1, 2, 3 .........

SO z0 = z1 = z2 = ???

k=0

Z0 =  ∛343 ( cos 180/3 + isin180/3)

=  ∛343 ( cos 60° + isin60°)

Z0 = 7 ( cos 60° + isin60°)

K=1

Z1 = ∛343 ( cos 180/3 + 360×1 / 3 + isin180/3 + 360×1 / 3 )

= ∛343 ( cos180° + isin180°)

Z1 = 7( cos180° + isin180° )

K=2

Z2 = ∛343 ( cos 180/3 + 360×2 / 3 + isin180/3 + 360×2 / 3 )

= ∛343 ( cos300° + isiN300°)

Z2 = 7 ( cos300° + isin300°)

6 0
3 years ago
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