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ASHA 777 [7]
3 years ago
9

Can anyone help me with this question on edmentum?

Mathematics
1 answer:
quester [9]3 years ago
5 0

Answer:

1.

-  \infty  < x <  \infty

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X^2/81 + y^2/64 = 1 The major intercepts are at (±9, 0) (±8, 0) (0, ±9)
Inessa [10]

Answer:

Step-by-step explanation:

Your question is unclear. What are "major intercepts"?

x²/81 + y²/64 = 1

The major axis is horizontal.

0²/81 + y²/64 = 1

y-intercepts = ±8

x²/81 + 0/64 = 1

x-intercepts = ±9

4 0
3 years ago
In between what 2 integers is the square root of 48
Murljashka [212]

Answer:

7,6

Step-by-step explanation:

the square root of 49 is 7 and

the square root of 36 is 6

so it must be between 7 and 6

8 0
3 years ago
Write the ratio as a fraction in lowest terms.<br> 15 minutes to 3 hours
mel-nik [20]

Answer:

Part A is 5/6 of the whole

Part B is 1/6 of the whole

Solution:

Assuming there are no other parts,

the Whole = A + B is the denominator:

Whole = 15 + 3 = 18

Part A = 15 and Part B = 3 are numerators for each fraction.

The fractions are then:

15/18 and 3/18

Meaning:

Part A is 15/18 of the whole

Part B is 3/18 of the whole

Reducing the fractions, it is also true that:

Part A = 5/6 of the whole

Part B = 1/6 of the whole

7 0
3 years ago
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! URGENT EMERGENCY !<br> what is the value of x and y
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The pic is black I can’t see
5 0
3 years ago
Suppose you have 40 meters of fencing what is the greatest rectangular area that you can enclose
Gelneren [198K]
It'd help if you could sketch this situation.  Note that the area of a rectangle is equal to the product of its width and length:  A = L W.

Consider the perimeter of this rectangular area.  It's P = 2 L + 2 W.  Note that P = 40 meters in this problem.

Thus, if we choose to use W as our independent variable, then P = 40 meters = 2 L + 2 W.  Let's express L in terms of W.  Divide both sides of the following equation by 2:  40 = 2 L + 2 W.  We get 20 = L + W.  Thus, L = 20 - w.

Then the area of the rectangle is  A = ( 20 - W)*W.

Multiply this out.  Your result will be a quadratic equation.  Graph this quadratic equation (in other words, graph the function that represents the area of the rectangle).  For which W value is the area at its maximum?  

Alternatively, find the vertex of this graph:  remember that the x- (or W-) coordinate  of the vertex is given by

W = -b/(2a), where a is the coefficient of W^2 an b is the coefficient of W in your quadratic equation.

Finally, substitute this value of W into your quadratic equation, to calculate the maximum area.
8 0
4 years ago
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