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aleksandrvk [35]
3 years ago
15

Help me asssopppppppppppp​

Mathematics
1 answer:
Mrrafil [7]3 years ago
7 0

52%4=13

13*3=39

Answer: A

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-3c-4=2<br> ---------------
Lilit [14]

Answer:

Solution given:

-3c-4=2

add 4 on both side

-3c-4+4=2+4

-3c=6

divide both side by -3

-3c/-3=6/-3

c=-2

7 0
3 years ago
Read 2 more answers
What is the midpoint of the segment shown below? For geometry worksheet!
Dafna1 [17]

Answer:

D

Step-by-step explanation:

The y-coordinate will be the average of 5 and -2 which is 3/2. This eliminates options A and B. C doesn't make sense because the x-coordinate (which is -7) of the midpoint can't be larger or smaller than the x-coordinates of the endpoints, so that means that the answer is D.

7 0
3 years ago
CHECK MY ANSWERS PLEASE
Papessa [141]

Step-by-step explanation:

For a geometric sequence,

\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}

1. The sequence is :

3, 13, 23, 33,...

\dfrac{13}{3}\ne \dfrac{23}{13}

It is not geometric. It is false

2. The sequence is :

5, -25, 125, -625

\dfrac{-25}{5}=\dfrac{125}{-25}\\\\-5=-5

So, the sequence is geometric as the common ratio is same. It is true.

4 0
3 years ago
Y= -4 is this relation linear?
QveST [7]
Yes, as there is only one variable, it makes the line straight
6 0
3 years ago
Someone help me do this question am giving brainliest
irina1246 [14]

Step-by-step explanation:

Given:

r^2(b^2\cos^2{\theta} + a^2\sin^2{\theta}) = a^2b^2

or

b^2(r^2\cos^2{\theta}) + a^2(r^2\sin^2{\theta}) = a^2b^2

Let x = r\cos{\theta} and y = \sin{\theta}. Substituting these into our given equation, we get

b^2x^2 + a^2y^2 = a^2b^2

Dividing both sides by a^2b^2, we get

\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1

which we recognize as an equation for an ellipse.

3 0
3 years ago
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