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3 years ago

Find the equation of the linear relationship.

Mathematics

1 answer:

rosijanka [135]3 years ago

3 0

Answer:

y = -3x + 150

Step-by-step explanation:

2 points on the graph:

(30, 60) and (40, 30)

Slope:

m=(y2-y1)/(x2-x1)

m=(30 - 60)/(40 - 30)

m= (-30)/10

m= -3

Slope-intercept:

y - y1 = m(x - x1)

y - 60 = -3(x - 30)

y - 60 = -3x + 90

y = -3x + 150

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3 years ago

Solve the above que no. 55

aleksandr82 [10.1K]

Answer:

Let $\left(1+\frac{1}{\tan^{2}A} \right)\cdot \left(1+\frac{1}{\cot^{2}A} \right)$ , we proceed to prove the trigonometric expression by trigonometric identity:

1) $\left(1+\frac{1}{\tan^{2}A} \right)\cdot \left(1+\frac{1}{\cot^{2}A} \right)$ Given

2) $\left(1+\frac{\cos^{2}A}{\sin^{2}A} \right)\cdot \left(1+\frac{\sin^{2}A}{\cos^{2}A} \right)$ $\tan A = \frac{1}{\cot A} = \frac{\sin A}{\cos A}$

3) $\left(\frac{\sin^{2}A+\cos^{2}A}{\sin^{2}A} \right)\cdot \left(\frac{\cos^{2}A+\sin^{2}A}{\cos^{2}A} \right)$

4) $\left(\frac{1}{\sin^{2}A} \right)\cdot \left(\frac{1}{\cos^{2}A} \right)$ $\sin^{2}A+\cos^{2}A = 1$

5) $\frac{1}{\sin^{2}A\cdot \cos^{2}A}$

6) $\frac{1}{\sin^{2}A\cdot (1-\sin^{2}A)}$ $\sin^{2}A+\cos^{2}A = 1$

7) $\frac{1}{\sin^{2}A-\sin^{4}A}$ Result

Step-by-step explanation:

Let $\left(1+\frac{1}{\tan^{2}A} \right)\cdot \left(1+\frac{1}{\cot^{2}A} \right)$ , we proceed to prove the trigonometric expression by trigonometric identity:

1) $\left(1+\frac{1}{\tan^{2}A} \right)\cdot \left(1+\frac{1}{\cot^{2}A} \right)$ Given

2) $\left(1+\frac{\cos^{2}A}{\sin^{2}A} \right)\cdot \left(1+\frac{\sin^{2}A}{\cos^{2}A} \right)$ $\tan A = \frac{1}{\cot A} = \frac{\sin A}{\cos A}$

3) $\left(\frac{\sin^{2}A+\cos^{2}A}{\sin^{2}A} \right)\cdot \left(\frac{\cos^{2}A+\sin^{2}A}{\cos^{2}A} \right)$

4) $\left(\frac{1}{\sin^{2}A} \right)\cdot \left(\frac{1}{\cos^{2}A} \right)$ $\sin^{2}A+\cos^{2}A = 1$

5) $\frac{1}{\sin^{2}A\cdot \cos^{2}A}$

6) $\frac{1}{\sin^{2}A\cdot (1-\sin^{2}A)}$ $\sin^{2}A+\cos^{2}A = 1$

7) $\frac{1}{\sin^{2}A-\sin^{4}A}$ Result

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