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4 years ago

Find a counterexample to disprove the statement "The Associative Property is true for division."

Mathematics

1 answer:

MaRussiya [10]4 years ago

5 0

The associative property makes it so whichever which way the numbers are the answer will be the same but as shown in the picture this isn't true for this statement because the answers become completely different depending on where the numbers are in the equation.
6 divided by 3 is NOT equal to 3 divided by 6 which disproves that property.

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The equation of the hyperbola with center at (0, 0) opening horizontally, with a = 2, b = 5 is:

tresset_1 [31]

Answer:

$\frac{x^2}{4}-\frac{y^2}{25} =1$

Step-by-step explanation:

we are given

center =(0,0)

we can compare with

center=(h,k)=(0,0)

so, h=0 and k=0

a=2 and b=5

now, we can use equation of hyperbola formula

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2} =1$

now, we can plug values

$\frac{(x-0)^2}{2^2}-\frac{(y-0)^2}{5^2} =1$

we can simplify it

and we get equation of hyperbola as

$\frac{x^2}{4}-\frac{y^2}{25} =1$

5 0

3 years ago

12, 21, 30 and 39 are the first four terms in a sequence

vivado [14]

Answer:

12, 21, 30, 39, 48, 57, 66, 75, 84.

Ninth term is 84.

Step-by-step explanation:

The rule for this sequence is "add 9".

So, using this rule, we can do......

12, 21, 30, 39, 48, 57, 66, 75, 84.

Remember to add 9 each time!

Best of luck :3

4 0

3 years ago

Please help find domain and range

Oxana [17]

<h3>Answer: Choice A</h3>

Domain: x > 4
Range: y > 0

========================================================

Explanation:

We want to avoid having a negative number under the square root. Solving $x-4 \ge 0$ leads to $x \ge 4$

So it appears the domain could involve x = 4 itself; however, if we tried that x value, then we'd get a division by zero error.

So in reality, the domain is x > 4.

-------------

The range of y = sqrt(x) is the set of positive real numbers. So y > 0 is the range for this equation. Shifting left and right does not affect the range, so the range of y = sqrt(x-4) is also y > 0.

We are dividing a positive number (3) over some positive number in the denominator. Overall, the expression $\frac{3}{\sqrt{x-4}}$ is positive because positive/positive = positive.

Therefore, the range of the given equation is y > 0

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The graph is shown below. We have a vertical asymptote at x = 4 and a horizontal asymptote at y = 0. The green curve is fenced in the upper right corner (northeast corner).