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irakobra [83]
3 years ago
5

Which of the following are necessary when proving that the opposite angles of a parallelogram are congruent? A. Opposite sides a

re perpendicular B. Opposite sides are congruent C. Angle addition postulate D. Segment Addition postulate
Mathematics
2 answers:
Misha Larkins [42]3 years ago
8 0

Answer:

C and B

Step-by-step explanation:

The correct option is option B and C. The necessary condition to prove that the opposite angles of a parallelogram are congruent:

C. Angle Addition Postulate.

B. Opposite sides are congruent

yanalaym [24]3 years ago
3 0

Answer:

The statements that are necessary when proving that the opposite angles of a parallelogram are congruent are:

B. Opposite sides are congruent - This has to be true as opposite sides should be similar.

C. Angle Addition Postulate.  It states that if a point lies on the interior of an angle, that angle is the sum of two smaller angles with legs that go through the given point.

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Amiraneli [1.4K]

Answer:

I think it would be c 160

Step-by-step explanation:

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3 years ago
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9. An item is priced al $13.62. If the sales tax is 6%, what does the
zheka24 [161]

Answer:

A (14.44)

Step-by-step explanation:

13.62 multiplied by 6%= 0.8172

Add 13.62 and 0.8172= 14.4372

Round to the hundredths=14.44

5 0
3 years ago
Meghan has $1.10 I'm dimes nickels and pennies. She has three times as many nickels as pennies, and the number of dimes is two l
Dmitriy789 [7]
N=3p and d=p-2  we are also told that:

p+n+d=110 using n and d from above we get:

p+3p+p-2=110

5p-2=110

5p=112

p=112/5

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There are 3 times as many nickels as pennies

There are pennies-2 dimes 

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7 0
3 years ago
Evaluate the following limit, if it exists : limx→0 (12xe^x−12x) / (cos(5x)−1)
Amanda [17]

Answer:

\lim_{x \to 0} \frac{12xe^x-12x}{cos(5x)-1}=-\frac{24}{25}

Step-by-step explanation:

Notice that \lim_{x \to 0} \frac{12xe^x-12x}{cos(5x)-1}=\frac{12(0)e^{0}-12(0)}{cos(5(0))-1}=\frac{0}{0}, which is in indeterminate form, so we must use L'Hôpital's rule which states that \lim_{x \to c} \frac{f(x)}{g(x)}=\lim_{x \to c} \frac{f'(x)}{g'(x)}. Basically, we keep differentiating the numerator and denominator until we can plug the limit in without having any discontinuities:

\frac{12xe^x-12x}{cos(5x)-1}\\\\\frac{12xe^x+12e^x-12}{-5sin(5x)}\\ \\\frac{12xe^x+12e^x+12e^x}{-25cos(5x)}

Now, plug in the limit and evaluate:

\frac{12(0)e^{0}+12e^{0}+12e^{0}}{-25cos(5(0))}\\ \\\frac{12+12}{-25cos(0)}\\ \\\frac{24}{-25}\\ \\-\frac{24}{25}

Thus, \lim_{x \to 0} \frac{12xe^x-12x}{cos(5x)-1}=-\frac{24}{25}

3 0
2 years ago
ANSWER ASAP! 1 question, explanation would be appreciated but not required!
IrinaK [193]

Answer:

choice 3) 50 in

Step-by-step explanation:

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radius = 25

d = 2(radius)

d = 50 in

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3 years ago
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