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Eddi Din [679]
4 years ago
15

What is the value of w? inscribed angles (Image down below)

Mathematics
1 answer:
Cloud [144]4 years ago
5 0

Answer:

w = 100°

Step-by-step explanation:

Opposite angles in an inscribed quadrilateral in a circle are supplementary.

Therefore, w + 80 = 180

Subtract 80 from both sides

w + 80 - 80 = 180 - 80

w = 100

The value of w = 100°

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How do you know when to use the subtraction property of equality as a reason as opposed to subtraction of real numbers?
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Subtraction property of equality is used when u have 2 sides with an equal sign in the middle and u want to subtract from both sides.
Example :
2x + 3 = 7 ....subtract 3 from both sides
2x + 3 - 3 = 7 - 3
2x = 4
x = 4/2
x = 2
When subtracting real numbers..if the numbers are on the same side of the equal sign...u can just subtract on that one side
Example :
2x = 7 - 3
2x = 4
x = 4/2
x = 2

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3 years ago
Apply the distributive property to factor out the greatest common factor.
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I’m pretty sure it’s 86, because you need to subtract 2 from the first side then you have 86 = X
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286,489 is an odd number.how many times greater is the ten thousand place than the 8 in the tens place ?
jonny [76]

We're comparing 80,000 with 80.

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80,000 is one thousand times greater than 80.

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4 years ago
Read 2 more answers
The angle 0 lies in Quadrant II . <br>Cos 0 = <br><img src="https://tex.z-dn.net/?f=%20-%20%20%5Cfrac%7B2%7D%7B3%7D%20" id="TexF
mel-nik [20]

Answer:

\tan(\theta)=\frac{-\sqrt{5}}{2}

Step-by-step explanation:

Since we are in quadrant 2, sine is positive.  Since sine is positive and cosine is negative, then tangent is negative.

Now I'm going to find the sine value of this angle given using one of the Pythagorean Identities, namely \sin^2(\theta)+\cos^2(\theta)=1.

If given \cos(\theta)=\frac{-2}{3}, then we have \sin^2(\theta)+(\frac{-2}{3})^2=1 by substitution of \cos(\theta)=\frac{-2}{3}.

Let's solve:

\sin^2(\theta)+(\frac{-2}{3})^2=1 for \sin(\theta).

\sin^2(\theta)+\frac{4}{9}=1

Subtract 4/9 on both sides:

\sin^2(\theta)=1-\frac{4}{9}

Simplify:

\sin^2(\theta)=\frac{5}{9}

Square root both sides:

\sin(\theta)=\sqrt{\frac{5}{9}}

\sin(\theta)=\frac{\sqrt{5}}{\sqrt{9}}

\sin(\theta)=\frac{\sqrt{5}}{3}

===========

\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}=\frac{\frac{\sqrt{5}}{3}}{\frac{-2}{3}}

Multiplying top and bottom by 3 gives:

\tan(\theta)=\frac{\sqrt{5}}{-2}

I'm going to move the factor of -1 to the top:

\tan(\theta)=\frac{-\sqrt{5}}{2}

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3 years ago
What is the length of a leg of an isosceles right triangle
Doss [256]

Answer: 3 square root of 2

Step-by-step explanation:

4 0
3 years ago
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