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SCORPION-xisa [38]
3 years ago
13

We have seen that isosceles triangles have two sides of equal length. The angles opposite these sides have the same measure. Use

the information to the right to help find the measure of angles 1 through 5​

Mathematics
1 answer:
dedylja [7]3 years ago
6 0

angle 1 = 180° - 115° (Linear Pair)

<u>angle 1 = 65°</u>

<u>angle 2 = angle 1 = 65°</u> ( Angles on the equal sides of an isosceles triangle )

angle 3 = 180° - (65° + 65°) (Angle Sum Property)

angle 3 = 180° - 130°

<u>angle</u><u> </u><u>3</u><u> </u><u>=</u><u> </u><u>5</u><u>0</u><u>°</u>

angle 4 = 180° - 65° (Linear Pair)

<u>angle</u><u> </u><u>4</u><u> </u><u>=</u><u> </u><u>1</u><u>1</u><u>5</u><u>°</u>

<u>angle</u><u> </u><u>5</u><u> </u><u>=</u><u> </u><u>1</u><u>1</u><u>5</u><u>°</u> (Vertically Opposite Angles)

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Answer:

The statement is missing. The statement is -- "A ray can be part of a line."

The answer is : The converse is not true, so Jahmiah is correct.

Step-by-step explanation:

A conditional statement is represented by showing p → q. It means if p is correct or true,  then q is also correct or true.

And the converse of p → q can be shown as  q → p.

But we know that the converse of a statement is not always true, it may be true and may not be true.

In the context, the statement is " a ray can be a part of a line." And so the converse would be  "A line can be a part of the ray".

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Thus ray is part of line but line is not a part of the ray. So the converse of the statement is not correct.

Hence, Jahmiah is correct.

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Select the correct expressions. 1 Identify each expression that represents the slope of a tangent to the curve y= * +1 at any po
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The expressions which represents the slope of a tangent to the curve y=\frac{1}{x+1} at any point (x, y) are:

                                ​f'(x) =limh \rightarrow 0\frac{-h}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{-h}{x^2h+2xh+xh^2+h^2 +1}

<h3>The slope of a tangent to the curve.</h3>

Mathematically, the slope of a tangent line to the curve is given by this equation:

f'(x) =limh \rightarrow 0\frac{f(x+h)-f(x)}{h}

Given the function:

f(x)=y=\frac{1}{x+1}

When (x + h), we have:

f(x+h)=y=\frac{1}{x+h+1}

Next, we would find the derivative of f(x):

f'(x) =limh \rightarrow 0\frac{\frac{1}{x+h+1} -\frac{1}{x+1}}{h}\\\\f'(x) =limh \rightarrow 0\frac{x+1 -(x+h+1)}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{x+1 -x-h-1}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{-h}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{-h}{x^2h+2xh+xh^2+h^2 +1}\\\\f'(x) = \frac{-1}{(x+1)(x+1)} \\\\f'(x) = \frac{-1}{(x+1)^2}

Read more on slope of a tangent here: brainly.com/question/26015157

#SPJ1

5 0
2 years ago
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