Please help

Mathematics

2 answers:

dedylja [7]2 years ago

6 0

Answer:

C ( 8,-6)

Because you move the points according to the rule.

Lorico [155]2 years ago

6 0

It is c hope it helps

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Match each genre with its characteristic.

Klio2033 [76]

Answer:

tragedy= depicts the downfall of a high ranking hero

comedy= depicts trials of ordinary people

drama= ridicules human relationships

Step-by-step explanation:

5 0

1 year ago

At the bright-as-day light bulb factory, 5 out of each 136 bulbs produced are defective. If the daily production is 2720 bulbs,

lions [1.4K]

Answer:

100 lightbulbs

Step-by-step explanation:

Basically find the percentage of lightbulbs that are bad. 5/136. So about 3. 6 percent. I'm going to use a more exact form of this percent for my calculations though. Now use the decimal for of this (0.036....) and multiply it by 2720. Using my exact decimal, the answer just so happened to be exactly 100. So there will be 100 defective lightbulbs per day. (Teachers are a stickler for units, so don't forget them if it's for a teacher)

Hope this helps!

4 0

2 years ago

Please help find domain and range

Oxana [17]

<h3>Answer: Choice A</h3>

Domain: x > 4
Range: y > 0

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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.

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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).