Solve the simultaneous equations 2p- 3q= 43p + 2q= 9

Mathematics

1 answer:

Rzqust [24]4 years ago

8 0

Answer:

q = 6/13, p = 35/13

Step-by-step explanation:

3p + 2q = 9

2p - 3q = 4

_________ (subtract the two equation from each other.)

p + 5q = 5

p = 5 - 5q

2(5 - 5q) - 3q = 4 (substitute value of p into equation)

10 - 10q - 3q = 4

10 - 13q = 4

-13q = -6

q = 6/13

3p + 2(6/13) = 9 (substitute value of q into equation)

3p + 12/13 = 9

3p = 105/13

p = 35/13

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Use right triangle ABC, with right angle C, as shown to answer the question.

kompoz [17]

The ratios are: sin B = 4/5 , cos B = 3/5 and tan B = 4/3

Step-by-step explanation:

We can see in the diagram that the values of all sides are given. In case of trigonometric ratios, the side in front of angle under consideration is perpendicular and the side adjacent to angle is base.

In the given case, we have to find trigonometric ratios for angle B; So,

Base = a = 3

Perpendicular = b = 4

Hypotenuse = c = 5

Hence,

The ratios will be:

$sin\ B = \frac{perpendicular}{hypotenuse} = \frac{b}{c} =\frac{4}{5}\\cos\ B = \frac{base}{hypotenuse} = \frac{a}{c} =\frac{3}{5}\\tan\ B = \frac{perpendicular}{base} = \frac{b}{a} = \frac{4}{3}$