The sum of two numbers is 196. One number is 64 more than the other. Find the numbers.
2 answers:
Answer:
Step-by-step explanation:
Hi there!
Let one of the number be "x" then the next number is x+64.
According to the question,
The sum of two numbers is 196.
i.e x + (x+64) = 196
or, 2x +64 = 196
Subtract 64 from both sides,
2x +64-64 = 196 - 64
2x = 132
Divide both sides by 2,
=
x = 66
And x+64 = 66+64 = 130
Therefore, the one number was 66 and another was 130.
Hope it helps!
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Answer:
(1) Option A is correct , The measure of angle A = 26°
(2) Option D is correct, The measure of angle B =70°
Step-by-step explanation:
(1)
In a right angle triangle figure-1 ,
* Hypotenuse(H) is always opposite the right angle and it is always the longest side of the triangle.
* Adjacent side (B) is the non-hypotenuse side that is next to a given angle A.
* Opposite side(P) is across from a given angle A.
To find the Angle A we use the Cosine ratio:
Cosine ratio states that the ratio of the adjacent side of a right triangle to the hypotenuse.
By definition of cosine ratio we have;
....[1]
In the figure - 1 we have B = 18 in and H = 20 in
substitute these in [1] to find angle A
or
⇒
therefore, the measure of angle A is,
(2)P
In figure -2
Find the measure of angle B.
First find the side BC, using Pythagoras theorem;
Pythagoras theorem states that in the right angle triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Apply Pythagoras theorem in rt. angle ABC;
Substitute the values of AB = 5.1 cm and AC =4.8 cm to solve for BC
Simplify:
or
Now, find the angle B, using tangent ratio;
Tangent ratio states that the ratio of the Opposite side of a right triangle to the adjacent side.
Therefore, the measure of angle B is; .
Answer:
A batch of 20 semiconductor chips is inspected by choosing a sample of 3 chips. Assume 10 of the chips do not conform to customer requirements.
a) Number of different samples = 20C3 = =1140
b) 2 good and one bad chip
Number of samples = 10C2 * 10C1 = 45 * 10 =450
c) 2 good 1 bad + 1 good 2 bad + 3 bad
Number of samples = 10C2 * 10C1 + 10C1 * 10C2 + 10C3
=