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True [87]
3 years ago
11

Solve the triangle. B = 36°, a = 38, c = 18

Mathematics
2 answers:
vovikov84 [41]3 years ago
7 0
Using cosine law.

b= sqrt(38^2 + 18^2 - 2(38)(18) Cos 36)
using calculator solve.
b = sqrt (1444 + 324 - 1106.75)
b = 25.75
nignag [31]3 years ago
6 0

Finishing the answer above:

B=36

a=38

c=18

b=25.7

a-b-c (SSS) {a^2=b^2+c^2-2ab(CosA) [Law of Cosines]

38^2=25.7^2+18^2-2(25.7)(18)(CosA)

1444=660.49+324-925.2(CosA)

1444=984.49-935.2(CosA)

459.51=-935.2(CosA)

-0.491349443969204=CosA

A=119.43

b-B-C (SAA) {(SinC)/c=(SinB/b)} [Law of Sines]

(SinC)/18=(Sin36)/25.7

(SinC)/18=0.022871021489979

SinC=0.411678386819631

C=24.31

A=119.43

B=36

C=24.31

a=38

b=25.7

c=18

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Solve the equation.-21=m/3m<br> m=7<br> m=-7<br> m=-63<br> m=63
Jlenok [28]
-21


I hope this helps and have a wonderful day filled with joy!!


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6 0
3 years ago
I need help on this question please
Fynjy0 [20]

Answer: NEITHER

Step-by-step explanation:

From the equation of lines given above, to know if relationships exist between the two lines, we must  first determine that  from their gradients or slopes.

From first equation

2x - y = 5,

      y = 2x - 5

Therefore, m₁ = 2

From the second equation

3x - y = 5

       y = 3x - 5

Therefore m₂ = 3

Recall, For the two lines to be parallel, m₁ = m₂ , and for the two line to be perpendicular, m₁m₂ = -1

since none fulfilled the conditions stated above, the answer then is

NEITHER

5 0
3 years ago
Ryan flew from Wiley Post to Ponca City and back. Ryan maintained an average rate of 450 mph going to Ponca City and an average
Reil [10]
The answer is B, and here's why.  Set up a table for "there" and "back" and use the distance = rate * time formula, like this:
               d             r            t
there       d           450         t
back        d           400        1-t

Let me explain this table to you.  The distance is d, we don't know what it is, that's what we are actually looking for.  We only know that if we go somewhere from point A to point B, then back again to point A, the distance there is the same as the distance back.  Hence, the d in both spaces.  There he flew 450 mph, back he flew 400 mph.  If the total distance was 1 hour, he flew an unknown time there and one hour minus that unknown time back.  For example, if he flew for 20 minutes there, one hour minus 20 minutes means that he flew 60 minutes - 20 minutes = 40 minutes back.  See? Now, because the distance there = the distance back, we can set the rt in both equal to each other.  If d = rt there and d = rt back and the d's are the same, then we can set the rt's equal to each other.  450t = 400(1-t) and
450t = 400 - 400t and 850t = 400.  Solve for t to get t = .47058.  Now, t is time, not the distance and we are looking for distance. So multiply that t value by the rate (cuz d = r*t) to get that the distance one way is
d = 450(.470580 and d = 211. 76 or, rounded like you need, 212.
4 0
2 years ago
Convert 15 km per hour to meters per second
ipn [44]
1km = 1000m
1hr = 3600sec

Do proportions:
1km / 1000m = 15km / x
15000m = x

1hr = 3600sec

15000/3600 = 4.2m per second
Choice 4


6 0
2 years ago
Look at this shape:
worty [1.4K]

Answer:

In my opinion is B

4 0
2 years ago
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