9514 1404 393
Answer:
(a) x^2/16 +y^2/9 = 1
Step-by-step explanation:
The form for the equation of an ellipse centered at the origin is ...
(x/(semi-x-axis))^2 +(y/(semi-y-axis))^2 = 1
The vertex values tell you the semi-x-axis is 4 units, and the semi-y-axis is 3 units. Then you have ...
(x/4)^2 +(y/3)^2 = 1
x^2/16 +y^2/9 = 1
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In case you don't remember that form, you can try any of the points in the equations. The equation that works will quickly become apparent.
If it is a right angled triangle the hypotenuse is inocorrect , it should be 12 not 13
Answer:
2(x^2+9)
Step-by-step explanation:
2x^2 +18
Factor out a 2
2(x^2+9)
You just need to equate the expressions to find the values for a,b and c. First, let's settle that 2^5/4=8, and 8 is 2^3. Also, an important tool here is the indicie rule:

1st expression: 2^5/4=8
2nd expression: It tells us that 2^5/2^a=2^3 (I know that this equals to 8 as all the expressions are equal therefore is the same as our first expression). Using the indicie law, this means that a=2
3rd expression: 2^b must equal to 8 as well, so b=3
4th expression: c=8 as all of the other expressions we have figured out equal to 8
So a=2, b=3 and c=8
Answer:
its c
Step-by-step explanation: