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mestny [16]
2 years ago
9

By the definition of complementary angles, since

Mathematics
2 answers:
svlad2 [7]2 years ago
8 0

The blank spaces in the task content can be filled with; Substitution property, 90° and subtraction property.

<h3>What are the missing statements in the proof?</h3>

Since it follows from the vertical angles theorem that angle 1 = angle 4 as they are congruent.

Additionally, angles 1 and 2 are said to be complementary and hence, the sum of their measures is equal to 90°.

Finally, by virtue of the subtraction property of equality, the measure of angle 2 is; 90° - 40° = 50°.

Read more on complementary angles;

brainly.com/question/16281260

#SPJ1

Mariulka [41]2 years ago
5 0

Blank 1: Substitution property of equality

Blank 2: 50

Blank 3: Subtraction property of equality

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S + 2.5s + 2.5s - 3.3 =
Vinvika [58]

Answer: 6s-3.3

Step-by-step explanation:

4 0
3 years ago
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Please help!
Vlada [557]

Option 3: 2^{\frac{5}{6}} is the right answer

Step-by-step explanation:

Given expression is:

(\sqrt{2})(\sqrt[3]{2} )

In order to simplify the expression we have to convert the radicals in exponents

2^{\frac{1}{2}} . 2^{\frac{1}{3}}

As he base is same, the powers can be added

=2^{\frac{3+2}{6}}\\=2^{\frac{5}{6}}

Hence,

Option 3: 2^{\frac{5}{6}} is the right answer

Keywords: Exponents, radicals

Learn more about exponents at:

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6 0
3 years ago
A^-1/[a^-1 - b^-1] + a^-1/[a^-1 + b^-1]
o-na [289]

\dfrac{a^{-1}}{a^{-1}-b^{-1}}+\dfrac{a^{-1}}{a^{-1}+b^{-1}}=a^{-1}\left(\dfrac{a^{-1}+b^{-1}}{(a^{-1}-b^{-1})(a^{-1}+b^{-1})}+\dfrac{a^{-1}-b^{-1}}{(a^{-1}+b^{-1})(a^{-1}-b^{-1})}\right)

=a^{-1}\left(\dfrac{(a^{-1}+b^{-1})+(a^{-1}-b^{-1})}{a^{-2}-b^{-2}}\right)

=\dfrac{2a^{-2}}{a^{-2}-b^{-2}}

=\dfrac{2b^2}{b^2-a^2}

5 0
3 years ago
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Please help me ergent ):
Temka [501]
What do you need help with

6 0
3 years ago
What is the range of f?
Pachacha [2.7K]

Answer:

Range(R)= (5,1,-4,-3)

Step-by-step explanation:

f={ (2,5) , (-6,1) , (0,-4), (4,-3), (7,-3) }

Same range is not repeated.

3 0
3 years ago
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