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GalinKa [24]
3 years ago
6

Which are vertical angles?

Mathematics
2 answers:
vichka [17]3 years ago
7 0

Answer:

You should republish this question but with a picture because this doesnt really make sense

Arisa [49]3 years ago
7 0

Answer:

Its A.

Step-by-step explanation:

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SOMEONE PLEASE HELP ME I WILL GIVE BRAINLY
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Answer:

answer is B

Step-by-step explanation:

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Katie could already play five songs on her harp. She decided to learn to new songs each week. If S = songs and W = weeks, which
denis-greek [22]

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she would have to record a the amount of songs She did every week

5 0
3 years ago
Which statements are true the ordered pair (1, 2) and the system of equations?
galina1969 [7]

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when (1.2) is substituted into the second equation the equation is true

Step-by-step explanation:

further you substitute x and the then solve

4 0
3 years ago
Which is true about the degree of the sum and difference of the polynomials 3x5y – 2x3y4 – 7xy3 and –8x5y + 2x3y4 + xy3?
Alekssandra [29.7K]
<span>First we have to find the sum and the difference of those polynomials- The sum is: ( 3 x^5y - 2 x^3y^4 - 7 xy^3 ) + ( - 8 x^5y + 2 x^3y^4 + xy^3 ) = 3 x^5 - 2 x^3y^4 - 7xy^3 - 8 x^5y + 2 x^3y^4 + xy^3 = - 5 x^5y - 6 xy^3. And the difference: ( 3 x^5y - 2 x^3y^4 - 7 xy^3 ) - ( - 8 x^5y + 2 x^3y^4 + xy^3 ) = 3 x^5y - 2 x^3y^4 - 7 xy^3 + 8 xy^5 - 2 x^3y^4 - xy^3 = 11 xy^5 - 4 x^3y^4 - 8xy^3. The highest exponent in both polynomials is 5. Answer: The degree of the polynomials is 5.</span>
7 0
4 years ago
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The limit as h approaches 0 of (e^(2+h)-e^2)/h is ?
Whitepunk [10]
<span>If you plug in 0, you get the indeterminate form 0/0. You can, therefore, apply L'Hopital's Rule to get the limit as h approaches 0 of e^(2+h),
which is just e^2.
</span><span><span><span>[e^(<span>2+h) </span></span>− <span>e^2]/</span></span>h </span>= [<span><span><span>e^2</span>(<span>e^h</span>−1)]/</span>h

</span><span>so in the limit, as h goes to 0, you'll notice that the numerator and denominator each go to zero (e^h goes to 1, and so e^h-1 goes to zero). This means the form is 'indeterminate' (here, 0/0), so we may use L'Hoptial's rule:
</span><span>
=<span>e^2</span></span>
3 0
3 years ago
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