Answer:
√5.
Step-by-step explanation:
Tan A = 1/2 means that the right triangle containing angle A has legs of length 1 and 2 units. So the hypotenuse = √(1^2 + 2^2) = √5 (using the Pythagoras theorem). The side opposite to < A = 1 unit and the adjacent side = 2 (as tan = opposite / adjacent).
so cos A = adjacent / hypotenuse = 2/√5.
and sin A = opposite / hypotenuse = 1 / √5
cos A / sin A = 2/√5 / 1/ √5 = 2.
sin A / (1 + cos A) = 1/√5 (1 + 2/ √5)
= 1 / √5 ( (√5 + 2) /√5)
= 1 / (√5 + 2)
So the answer is:
2 + 1 /(√5 + 2).
We can simplify it further by multiplying top and bottom of the fraction by the complement of √5 + 2 which is √5 - 2.
2 + 1 / (√5 + 2)
= 2(√5 + 2) + 1 / (√5 + 2 )
= { 2(√5 + 2) + 1 } / (√5 + 2)
Multiplying this by √5 - 2 / √5 - 2 we get:
(2(5 - 4) + √5 - 2) / (5 -4)
= 2 + √5 - 2 / 1
= √5.
1)we calculate the volume of the right rectangular prism
volume=lenght x width x heigth
volume=12 ft * 8.5 ft * 4 ft=408 ft³
density= mass / volume ⇒ mass=volume * density.
density=0.25 pound /ft³
2) we calculate the mass of the right rectangular prism.
mass=volume * density
mass=408 ft³(0.25 pound / ft³)=102 pounds.
answer: c. 102
Answer:
Step-by-step explanation:
1) we have to find the area of the cut out portion
Since both circles in one side makes up the side 28cm
The diameter of each circle is 14cm
radius = diameter / 2
r = 7cm
Area of a circle = πr²
A = 22/7 × 7²
= 22/7 × 7 × 7
= 22 × 7 = 154cm²
So the four circles will make
= 154 + 154 + 154 + 154
= 616 cm²
The Area of the square will be
= L × L
= 28 × 28
= 784 cm²
Therefore the area of the remaining portion will be
Area of the square - total area of the circles
= 784 cm² - 616 cm²
= 168 cm²
The area of the remaining part is 168 cm²
Answer : A
The word "increased" indicates a plus sign. The word "is equal" helps show us show that something is = to 15
You should must check for an extraneous solution when the variable appears both inside and outside the absolute value expression
<h3>
What is an extraneous solution?</h3>
An extraneous solution is a solution that in obtained after completely solving an equation but it does not work in the original given equation.
You should must check for an extraneous solution when the variable appears both inside and outside the absolute value expression (Option D).
Learn more about extraneous solution: brainly.com/question/14054707
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