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devlian [24]
3 years ago
8

Item 20

Mathematics
1 answer:
STALIN [3.7K]3 years ago
3 0

Answer:

LLL

Step-by-step explanation:

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Draven's computer downloads files at a rate of 220 kilobytes per second. The computer has already downloaded the first 550 kilob
Effectus [21]

Answer:

no se

Step-by-step explanation:

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8 0
3 years ago
What value of x <br> in the solution set of 2(3x – 1) ≥ 4x – 6?
Brrunno [24]
First, you must distribute 2(3x-1).
To do that, you will multiply 2•3x, and 2•-1 because you are taking the number outside of the parentheses and multiplying (distributing) it to all the numbers inside

After distributing, the left side of your inequality will be 6x-2

Now you have
6x - 2 \geqslant 4x - 6

To find the value of x, you must subtract an x value from both sides of the equation, as well as a constant from each side.

so you have
6x - 2 \geqslant 4x - 6 \\ - 6x \geqslant + 6
And that will make the equation
- 4 \geqslant - 2x
Now, divide the variable side, ***BUT, because you are dividing by a negative number in an inequality, the inequality will switch sides.

Then, the value of x is greater than or equal to 2
5 0
3 years ago
Read 2 more answers
a reading teacher recorded the number of pages read in an hour by each of her students. The numbers are shown below: 54,39,29,43
Ilya [14]

Answer: The answer is 29, because that 29 would be the outlier of the numbers.

Step-by-step explanation:

8 0
3 years ago
Can you help me i don’t know the answers
Advocard [28]
1)

An irrational number is a number that a) can't be written as a fraction of two whole numbers AND b) is an infinite decimal without any sort of pattern.

For the first answer choice, clearly \frac{1}{3} does not pass the first criterion so we look at the second choice.

Let's come back to \sqrt{2} and \pi.

\frac{2}{9} doesn't meet our first criterion, and let's skip \sqrt{3} for now.

It is often easier to disprove an irrational number than to prove one. There are a few famous irrationals to know (although there is an infinite number of irrationals). The most common are \sqrt{2},  \pi, e,  \sqrt{3}. For now, it's just helpful to know these and recognize them.

So we can check off \sqrt{2},  \pi and \sqrt{3}.

2) 

For this next question, we know that \sqrt{64} = 8. Clearly this isn't irrational. Likewise, \frac{1}{2} isn't irrational. \frac{16}{4} =  \frac{4}{4} = 1, which is rational, leaving only \frac{ \sqrt{20}}{5} =  \frac{2 \sqrt{5} }{5}. By process of elimination, this is the correct answer. Indeed, \sqrt{5} is an irrational number.

3) This notation means that we have 0.3636363636... and so on, to an infinite number of digits. It is called a repeating decimal.

But it can be written as a fraction because its pattern repeats, unlike for an irrational number.

Let's say x=0.36363636.... Would you agree that 100x=36.36363636...? (We choose to multiply by 100 because there are two decimals that repeat. For 1, choose 10, for 3 choose 1,000, and so on.)

Now, let's subtract x from 100x and solve.

100x=36.36363636\\-x \ \ \ \ \ \ \ -0.36363636\\99x=36\\\\x= \dfrac{36}{99}= \dfrac{4}{11}

Voila!
4 0
3 years ago
Read 2 more answers
Write a unit test for addinventory(). call redsweater.addinventory() with parameter sweatershipment. print the shown error if th
Katena32 [7]
<span>#include <iostream> using namespace std; class InventoryTag { public: InventoryTag(); int getQuantityRemaining() const; void addInventory(int numItems); private: int quantityRemaining; }; InventoryTag::InventoryTag() { quantityRemaining = 0; } int InventoryTag::getQuantityRemaining() const { return quantityRemaining; } void InventoryTag::addInventory(int numItems) { if (numItems > 10) { quantityRemaining = quantityRemaining + numItems; } } int main() { InventoryTag redSweater; int sweaterShipment = 0; int sweaterInventoryBefore = 0; sweaterInventoryBefore = redSweater.getQuantityRemaining(); sweaterShipment = 25; cout << "Beginning tests." << endl; // FIXME add unit test for addInventory /* Your solution goes here */ cout << "Tests complete." << endl; return 0; }</span>
5 0
3 years ago
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