Answer:
222 fence posts
Step-by-step explanation:
Since the log is 20 m long, correct to the nearest metre and has to be cut into fence posts which must be 90 cm long, correct to the nearest 10 centimetres, there can be x fence posts 90 cm long.
So, we have x × 90 cm = 20 m
x × 0.9 m = 20 m
x = 20 m/0.9 m
x = 222.22
x ≅ 222 fence posts
So, the largest number of fence posts that can possibly be cut from the log is 222 fence posts.
59
47 - 11 = 36
23 + 36 = 59
I hope this helps. Brainliest appreciated.
3(p + 2) = 18
3p + 6 = 18
3p = 12
p = 4
Check:
3(p + 2) = 18
3(4 + 2) = 18
3(6) = 18
18 = 18
This proves that p = 4.
⭐ Answered by Hyperrspace (Ace) ⭐
⭐ Brainliest would be appreciated, I'm trying to reach genius! ⭐
⭐ If you have questions, leave a comment, I'm happy to help! ⭐
1
Simplify n\times \frac{6}{9}n×96 to \frac{2n}{3}32n
\frac{2n}{3}=\frac{3}{12}32n=123
2
Simplify \frac{3}{12}123 to \frac{1}{4}41
\frac{2n}{3}=\frac{1}{4}32n=41
3
Multiply both sides by 33
2n=\frac{1}{4}\times 32n=41×3
4
Simplify \frac{1}{4}\times 341×3 to \frac{3}{4}43
2n=\frac{3}{4}2n=43
5
Divide both sides by 22
n=\frac{\frac{3}{4}}{2}n=243
6
Simplify \frac{\frac{3}{4}}{2}243 to \frac{3}{4\times 2}4×23
n=\frac{3}{4\times 2}n=4×23
7
Simplify 4\times 24×2 to 88
n=\frac{3}{8}n=83
The cosecant is the reciprocal of the sine function. That means you invert all the fractions and you divide 1 by the non-fractions.
csc(0°) = 1/0
csc(30°) = 2/1 = 2
csc(45°) = (√2)/1 = √2
csc(60°) = 2/√3
csc(90°) = 1/1 = 1
