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Salsk061 [2.6K]
4 years ago
15

Determine which relation is a function.

Mathematics
2 answers:
slamgirl [31]4 years ago
5 0
The last one is the correct answer
kolbaska11 [484]4 years ago
5 0
(-3,2) (-2.3) (-1,2) (0,4) (1,1)
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Find the equation of the parabola with vertex (1,3) and focus (1/2,3).
daser333 [38]

Answer:

A (y-3)^2 = -2(x-1) is the correct choice :)


4 0
3 years ago
In this triangle, which of the following is true?​
Svetradugi [14.3K]

We know that this is a right triangle, so one of the sides has to be 90 (indicated by the square on the corner)  The angles of all triangles must add to exactly 180.  Subtract 90 and 35 from 180

If we do that, we get 55.  We know that the unknown angle is 55 now.  Now, it asks us for b.  We can pythagoras theorem which states that (a^2 + b^2 = c^2)  We already know c (c is 20)  It does not tell us 2 of the sides but we do know that the answer is either B or D.  

The side opposite of 90* should be the largest.  The size opposite of 55 would be the second largest one.  The side opposite of 35 would be the smallest one.  Using pythagoras theorem again, I can try plugging in both 11.47 and 16.38 with c^2 - b^2 = a^2  

11.47:

400 - 132 =268  (sq root now = 16.37) this would mean that A is larger than B.  B should be larger than A.  Try the other one

16.37:

400 -268 = 132 (sq root = 16.37) this would mean that B is larger than A which is what we want.

Thus, 16.37  = B and the last option would be correct (D)

Hope this helps, give an honest rating for me please

7 0
3 years ago
(12a^(2) - 3)/(2)*(2a + 1)^(-2)*((6)/((2a + 1)))^(-1)
Elena L [17]

Answer:

\frac{144a^{4}+144a^{3} - 36 - 12a} {2a+1}\\

Step-by-step explanation:

\frac{(12a^2-3)}{2*(2a+1)^{-2}}*\frac{6}{(2a+1)^{-1} }\\ = \frac{(12a^2-3)}{2 * \frac{1}{(2a+1)^{2} }} * \frac{6}{\frac{1}{2a+1}  }\\ =\frac{(12a^2-3)(2a+1)^{2}}{2} * \frac{6}{2a+1}\\=\frac{(12a^2-3)(2a+1)^{2}}{2} * \frac{6}{2a+1}

= \frac{(12a^2-3)(4a^{2} + 1 + 2 (2a)(1))}{2} * \frac{6}{2a+1}\\= \frac{(12a^2-3)(4a^{2} + 1 + 4a)}{2} * \frac{6}{2a+1}\\= \frac{(12a^2-3)(4a^{2} + 1 + 4a)}{1} * \frac{3}{2a+1}\\= \frac{3(12a^2-3)(4a^{2} + 1 + 4a)} {2a+1}\\\\= \frac{3(12a^2(4a^{2} + 1 + 4a) - 3 (4a^{2}+1+4a)} {2a+1}\\= \frac{3(48a^{4}+12a^{2}+48a^{3}  - 12a^{2} -12 - 4a)} {2a+1}\\= \frac{144a^{4}+36a^{2}+144a^{3}  - 36a^{2} - 36 - 12a)} {2a+1}

= \frac{144a^{4}+36a^{2}+144a^{3}  - 36a^{2} - 36 - 12a} {2a+1}\\= \frac{144a^{4}+144a^{3} - 36 - 12a} {2a+1}\\

6 0
4 years ago
If (10,3) and (6,31) are anchor point on a trend line, find the equation of the line
Readme [11.4K]

Answer:5.6

Step-by-step explanation:

4 0
3 years ago
Please answer ASAP <br>Will mark BRAINLYEST and 40 points​
Nana76 [90]

Answer:

a. a + 10 is the correct option.

8 0
3 years ago
Read 2 more answers
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