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Eva8 [605]
3 years ago
15

Find the length of side a in the right triangle if b=3 and c=5a=?

Mathematics
1 answer:
Assoli18 [71]3 years ago
4 0
To solve this you will you the Pythagorean Theorem. 
the equation for the Pythagorean Theorem is a²+b²=c².
fill in the information you have:
a²+3²=5²
now solve for a:
a²+9=25
    -9   -9
a²=16
√a²=√16
a=4
The length of the side a of the right triangle is 4.
Hope I helped...
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6x + 2x - 1 = 5x + 11
sammy [17]
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        3x - 1 = 11
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             3x = 12
              3      3
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4 0
3 years ago
A+5b-c=-20<br> 4a-5b+4c=19<br> -a-5b-5c=2 <br> Solve for a b and c
ludmilkaskok [199]

Answer:a = - 2

b = - 3

c = 3

Step-by-step explanation:

a + 5b - c = - 20 - - - - - - - - - 1

4a - 5b + 4c = 19 - - - - - - - - - 2

-a - 5b - 5c = 2 - - - - - - - - - - - 3

Adding equation 1 and equation 2, it becomes

- 6c = - 18

c = - 18/ - 6 = 3

Multiplying equation 2 by 1 and equation 3 by 4, it becomes

4a - 5b + 4c = 19

-4a - 20b - 20c = 8

Adding both equations, it becomes

- 25b - 16c = 27

- 25b = 27 + 16c = 27 + 16 × 3

- 25b = 75

b = 75/- 25 = - 3

Substituting b = - 3 and c = 3 into equation 1, it becomes

a + 5 × - 3 - 3 = - 20

a - 15 - 3 = - 20

a - 18 = - 20

a = - 20 + 18 = - 2

5 0
3 years ago
How to make 27/102 into a simplest form and 16/32 into simplest form?
Reil [10]
My caculator got 9/34
7 0
3 years ago
Read 2 more answers
Cos s=-2/5 and sin t=4/5, s and t are in quadrant II<br> find cos(s+t) and cos(s-t)
mojhsa [17]

Answer:

•cos(s+t) = cos(s)cos(t) - sin(s)sin(t) = (-⅖).(-⅗) - (√21 /5).(⅘) = +6/25 - 4√21 /25 = (6-4√21)/25

•cos(s-t) = cos(s)cos(t) + sin(s)sin(t) = (-⅖).(-⅗) + (√21 /5).(⅘) = +6/25 + 4√21 /25 = (6+4√21)/25

cos(t) = ±√(1 - sin²(t)) → -√(1 - sin²(t)) = -√(1 - (⅘)²) = -⅗

sin(s) = ±√(1 - cos²(s)) → +√(1- cos²(s)) = +√(1 - (-⅖)²) = √21 /5

5 0
3 years ago
If two lines are perpendicular, which statement must be true?
Kryger [21]

Answer:

Their slopes are opposites

8 0
2 years ago
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