The measure of angle ∠EGF is 65°. And the measure of the angle ∠CGE is 115°.
<h3>What is the triangle?</h3>
A triangle is a three-sided polygon with three angles. The angles of the triangle add up to 180 degrees.
Triangle GEF is shown with its exterior angles.
Line GF extends through point B.
Line FE extends through point A.
Line EG extends through point C.
Angles ∠FEG and ∠EGF are congruent.
∠FEG = ∠EGF = x
Sides EF and GF are congruent.
Angle ∠EFG is 50° degrees.
∠EFG + ∠FGE + ∠GEF = 180°
50° + x + x = 180°
2x = 130°
x = 65°
∠FEG = ∠EGF = 65°
Then angle ∠CGF will be
∠CGF + ∠FGE = 180°
∠CGF + 65° = 180°
∠CGF = 115°
More about the triangle link is given below.
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Well, if you just divide 4000 by 50, you can see that the answer will come out as:
<em><u>80
</u></em>~TO check: 80 multiplied by 50 would come out to the answer of 4,000.
<em>~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~</em><em><u>
</u>Hope this helped! :D</em>
Answer:
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Answer:
s(x) is t(x) ...
- horizontally compressed by a factor of 2,
- reflected across the y-axis, and
- translated downward 5 units.
Domain and Range
- t(x) has a domain of x ≤ 0, and a range of y ≥ 0.
- s(x) has a domain of x ≥ 0, and a range of y ≥ -5.
Step-by-step explanation:
t(x) is the square root function reflected across the y-axis and compressed horizontally by a factor of 2. That is, in f(x) = √x, the x has been replaced by -2x.
s(x) has the function t(x) <em>reflected back across the y-axis</em> and compressed horizontally by another factor of 2. It is also <em>translated downward by 5 units</em>, so that its origin (vertex) is at (0, -5).
_____
The graph shows you the domain and range of s(x). The domain is all numbers to the right of x=0, including x=0. That is ...
domain: x ≥ 0
The range is all numbers -5 or above:
range: y ≥ -5
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For t(x), the argument of the square root function must not be negative, which means the value of x cannot be positive.
domain: x ≤ 0
For non-negative values of radicand, the t(x) function will have non-negative values. So, the range is ...
range: y ≥ 0
_____
<em>Comment on solving problems like this</em>
Your graphing calculator can be your friend.