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eduard
3 years ago
6

Please help! What is the answer to this question?​

Mathematics
2 answers:
Crank3 years ago
8 0

Answer:

The answer is e= 8,9-6=2y

Step-by-step explanation:

e=+2=59

59y-27x=2y

bezimeni [28]3 years ago
8 0

Since k and j are parallel, angle 4 is equal to the angle to the left of 60 degrees. Since j is a line, the angle to the left of 60 added on to 60 is 180.

180 -60 = 120. The angle is 120 degrees.

Since angle 4 is equal to that angle, angle 4 is also 120 degrees.

Hope this helps

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-3/2 and 0 rational numbers find​ only 3 rational numbers
slamgirl [31]

Step-by-step explanation:

The given nos. are , 0 and -3/2 . We can write -3/2 as -1.5. Therefore three rational numbers between them will be ,

  • -1
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  • -1.3
6 0
3 years ago
What is the value of the fourth term in a geometric sequence for which a1 = 30 and r = 1/2?.
KIM [24]
For a Geometric sequence:
a n = a ( n-1 ) * r
In this case: a 1 = 30
a 2 = 30 * 1/2 = 15
a 3 = 15 * 1/2 = 7.5
a 4 = 7.5 * 1/2 = 3.75 = 3 3/4
Answer: The value of the fourth term is 3.75 or 3 3/4.
8 0
3 years ago
Pls help me do these problems. I need them by today.
valina [46]

Answer:

see below

Step-by-step explanation:

Question 1: D

Question 2: A

Question 3: B

Question 4: B

Question 5: C

Question 6: D

Question 7: A

Question 8: D

5 0
3 years ago
If sine ⁡theta equals negative 8/17 and the terminal side of theta lies in quadrant IV, find cosine ⁡ theta .
Lelechka [254]
\sin^2\theta+\cos^2\theta=1\implies \cos\theta=\pm\sqrt{1-\sin^2\theta}

Since \sin\theta=-\dfrac8{17}, it follows that

\cos\theta=\pm\sqrt{1-\dfrac{64}{289}}=\pm\sqrt{\dfrac{225}{289}}=\pm\dfrac{15}{17}

where the sign depends on the location of the terminal side of the angle. Since the terminal side lies in the fourth quadrant, that means the cosine of the angle is positive, so

\cos\theta=\dfrac{15}{17}
3 0
3 years ago
Which polygons can be mapped onto each other similarity transmations
soldi70 [24.7K]
Which polygons can be mapped onto each other similarity transformations?
If you are talking about this image, the answer is; polygons 1 and 3.
Correct me if I'm wrong.

3 0
3 years ago
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