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2 years ago

prove that cos theta by 1 minus tan theta + sin theta by 1 minus cot theta equal to sin theta + cos theta

1 answer:

3 0

Here is the answer for prove that cos theta by 1 minus tan theta + sin theta by 1 minus cot theta equal to sin theta + cos theta

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4cm 10cm Base= Height= Area=

Anton [14]

Answer:

Step-by-step explanation:

If you're talking about a square, then the area would be 40.

If you're talking about a triangle, then the area would be 20.

8 0

3 years ago

Please help me answer the question in the picture

coldgirl [10]

Answer:

D. y-5 = 3(x+1)

Step-by-step explanation:

Slope of the perpendicular line is 3.

Slope intercept form is,

y-5 = 3 (x-(-1))

y-5 = 3(x+1)

4 0

3 years ago

What is the ratio of one hour to 600 seconds?

Marat540 [252]

Answer:

6:1

Step-by-step explanation:

1 hr= 60 minutes

60 minutes= 3600 seconds

3600:600

36:6

5 0

3 years ago

Read 2 more answers

Graph y+6 = 4/5 (x+3 using the point and slope given in the equation.

egoroff_w [7]

Answer:

Slope: 4/5

Point: (-3|-6)

4 0

3 years ago

What is the expression in radical form? (4x3y2)310

sertanlavr [38]

Given:

Consider the given expression is

$(4x^3y^2)^{\frac{3}{10}}$

To find:

The radical form of given expression.

Solution:

We have,

$(4x^3y^2)^{\frac{3}{10}}=(2^2)^{\frac{3}{10}}(x^3)^{\frac{3}{10}}(y^2)^{\frac{3}{10}}$

$(4x^3y^2)^{\frac{3}{10}}=(2)^{\frac{6}{10}}(x)^{\frac{9}{10}}(y)^{\frac{6}{10}}$

$(4x^3y^2)^{\frac{3}{10}}=(2)^{\frac{3}{5}}(x)^{\frac{9}{10}}(y)^{\frac{3}{5}}$

$(4x^3y^2)^{\frac{3}{10}}=\sqrt[5]{2^3}\sqrt[10]{x^9}\sqrt[5]{y^3}$ $[\because x^{\frac{1}{n}}=\sqrt[n]{x}]$

$(4x^3y^2)^{\frac{3}{10}}=\sqrt[5]{8y^3}\sqrt[10]{x^9}$ $[\because x^{\frac{1}{n}}=\sqrt[n]{x}]$

Therefore, the required radical form is $\sqrt[5]{8y^3}\sqrt[10]{x^9}$ .

8 0

3 years ago

prove that cos theta by 1 minus tan theta + sin theta by 1 minus cot theta equal to sin theta + cos theta​

prove that cos theta by 1 minus tan theta + sin theta by 1 minus cot theta equal to sin theta + cos theta