2 pages back to back.....x and x + 1
x + x + 1 = 203
2x + 1 = 203
2x = 203 - 1
2x = 202
x = 202/2
x = 101
x + 1 = 101 + 1 = 102
the 2 page numbers are : 101 and 102
Its 40 you basically add it kus if it gets colder than the temperature must of been higher so you add
J/8 is the answer youre looking for
Answer:
Step-by-step explanation:
Given ,
A right angle triangle ΔABC with right angle at C and ∠A=30° and AC=5√5 units .
Let ∠A=A,∠B=B∠C=C and AC=b,BC=a,CA=b .
Implies perimeter of triangle = a+b+c .
now
![tanA=\frac{a}{b} \\\a=b*tanA\\a= 5\sqrt{5} *tan30^0\\a=\frac{5\sqrt{5}}{\sqrt{3}}](https://tex.z-dn.net/?f=tanA%3D%5Cfrac%7Ba%7D%7Bb%7D%20%5C%5C%5Ca%3Db%2AtanA%5C%5Ca%3D%205%5Csqrt%7B5%7D%20%2Atan30%5E0%5C%5Ca%3D%5Cfrac%7B5%5Csqrt%7B5%7D%7D%7B%5Csqrt%7B3%7D%7D)
and
![cosA=\frac{b}{c} \\\c=\frac{b}{cosA} \\c=\frac{10\sqrt{5}}{\sqrt{3} }](https://tex.z-dn.net/?f=cosA%3D%5Cfrac%7Bb%7D%7Bc%7D%20%5C%5C%5Cc%3D%5Cfrac%7Bb%7D%7BcosA%7D%20%5C%5Cc%3D%5Cfrac%7B10%5Csqrt%7B5%7D%7D%7B%5Csqrt%7B3%7D%20%7D)
implies ,
![perimeter = \frac{5\sqrt{5}}{\sqrt{3}} + 5\sqrt{5} +\frac{10\sqrt{5}}{\sqrt{3} }\\perimeter= 5\sqrt{15} + 5\sqrt{5}](https://tex.z-dn.net/?f=perimeter%20%3D%20%5Cfrac%7B5%5Csqrt%7B5%7D%7D%7B%5Csqrt%7B3%7D%7D%20%2B%205%5Csqrt%7B5%7D%20%2B%5Cfrac%7B10%5Csqrt%7B5%7D%7D%7B%5Csqrt%7B3%7D%20%7D%5C%5Cperimeter%3D%205%5Csqrt%7B15%7D%20%2B%205%5Csqrt%7B5%7D)
Answer: -3
Step-by-step explanation:8-5=3 take the negative so it’s -3