Help with trigonometry

Mathematics

1 answer:

diamong [38]3 years ago

4 0

Answer:

cos ∠CBD = - 4√41 / 41

Step-by-step explanation:

AB = 4 ΔABC = (4 x AC) / 2 = 10

AC = 5

BC = √5² + 4² = √41

cos ∠CBD = cos (180° - ∠CBA) = cos 180° cos ∠CBA + sin 180° sin ∠CBA

(cos 180° = - 1 sin 180° = 0 cos ∠CBA = 4 / √41 )

cos ∠CBD = (-1) x cos ∠CBA = - 4 / √41 = - 4√41 / 41

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there are a total of 500 students in grades 1 through 5 in an elementary school. 17% of the total number of students are in 1st

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17% of the total number of students are in 1st grade
→0.17×500=85 students in 1st grade
19% of the total number of students are in 4th grade
→0.19×500=95 students in 4th grade
The number of 3rd-grade students is 9 less than the number of 4th-grade students.
→95-9=86 students in 3rd grade
The number of 2nd-grade students is 10 less than the number of 5th-grade students.
1st grade+4th grade+3rd grade=85+95+86=266
500-266=234=total number of students in 2nd grade and 5th grade
Let x=2nd grade students
Let y=5th grade students
x+y=234
2nd grade students is 10 less than the number of 5th grade students
→x=y-10
Since x=y-10 so we substitute y-10 for x
→y-10+y=234
→2y-10=234
+10 +10
→2y=244
→2y/2=244/2
→y=122 students in 5th grade
x+y=234
→x+122=234
→x=112 students in 2nd grade
ANSWERS:
1st: 85
2nd: 112
3rd: 86
4th: 95
5th: 122

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