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3 years ago

Show that sec^2x+csc^2x=sec^2xcsc^2x

2 answers:

5 0

Sec^2x+csc^2x=sec^2xcsc^2x
Divide both sides by csc^2x
sec^2x+(csc^2x/csc^2x)=sec^2xcsc^2x/csc^2x
sec^2x=sec^2x
Divide both sides by sec^2x
0=0 which is a true statement.

Hope this helps :)

Arte-miy333 [17]3 years ago

5 0

You might know that sec²(x) - tan²(x) = 1. If we solve for sec²(x), then we obtain:

sec²(x) = 1 + tan²(x) 

Then, 

csc²(x)(1 + tan²(x)) 
==> csc²(x) + csc²(x)tan²(x) 
==> csc²(x) + 1/sin²(x) * sin²(x)/cos²(x) 
==> csc²(x) + 1/cos²(x) 
==> csc²(x) + sec²(x) 
==> RHS 

I hope this helps!

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Find the equation for the circle with center (4,3) and passing through (2,-4)

notsponge [240]

Answer:

$(x - 4)^{2} + (y - 3)^{2} = 53$

Step-by-step explanation:

The general equation of a circle is as follows:

$(x - x_{c})^{2} + (y - y_{c})^{2} = r^{2}$

In which the center is $(x_{c}, y_{c})$ , and r is the radius.

In this problem, we have that:

$x_{c} = 4, y_{c} = 3$

$(x - 4)^{2} + (y - 3)^{2} = r^{2}$

Passing through (2,-4)

We replace into the equation to find the radius.

$(2 - 4)^{2} + (-4 - 3)^{2} = r^{2}$

$4 + 49 = r^{2}$

$r^{2} = 53$

The equation of the circle is:

$(x - 4)^{2} + (y - 3)^{2} = 53$

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3 years ago

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irga5000 [103]

It is $12.35, because this number is 5% of 247.

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4 years ago

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Part A: Explain why x = 6 makes 4x − 5 ≤ 19 true but not 4x − 5 < 19. (5 points)

vovikov84 [41]

This solution shows that 6 is included in the solution due to the equal sign and not true for the inequality 4x − 5 < 19 since 6 is excluded

<h3>Inequality expression</h3>

Inequalities are equation not separated by an equal sign

Given the inequality expression

4x − 5 ≤ 19

4x ≤ 19 + 5

4x ≤ 24

x≤ 6

This solution shows that 6 is included in the solution due to the equal sign and not true for the inequality 4x − 5 < 19 since 6 is excluded

Given the equation

4x + 3 = 23

4x = 23 - 3

4x =20

x = 5

Hence 5 is the solution from the given set

Learn more on inequality here: brainly.com/question/24372553

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3 years ago

A 98% confidence interval for a proportion is found to be (0.62, 0.68).

Vilka [71]

Answer:

I think the answer is C ,The sample proportion is the fraction of samples which were successes, so. (1) For large , has an approximately normal distribution.

Step-by-step explanation:

If the queston true please votes 5 stars and mark as brainliest for me.Thank you so much

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