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RideAnS [48]
3 years ago
12

Lucy is thinking of a number. the number is greater than two hundred twenty-five. her number is less than 2hundred, 2 tens,7 one

s. what is lucy number? howq do i solve this
Mathematics
1 answer:
mestny [16]3 years ago
6 0
2hundred, 2tens, 7ones is 227.
so the answer is 226 coz its smaller than 227 and larger than 225
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Divide the polynomials.<br> Your answer should be a polynomial.
Ulleksa [173]
I believe the answer is x^2 + 2

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What is the equivalent fraction for 1/2​
allochka39001 [22]
It could be:

2/4
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6 0
3 years ago
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A 24-foot ladder is placed against a vertical wall of a building, with the bottom of the ladder standing on level ground 6 feet
andrezito [222]

Answer:

23.24 feet

Step-by-step explanation:

Use the pythagorean theorem: a² + b² = c², where a and b are legs of the right triangle and c is the hypotenuse.

In this situation, the ladder is the hypotenuse of the triangle, and the distance from the base of the building is the long leg.

Plug in the ladder length as c and plug in the distance from the base of the building as a:

a² + b² = c²

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5 0
3 years ago
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I’m need help on these questions can any one answer them
LUCKY_DIMON [66]

Answer:

5. WAX and XAZ

6. WAZ and ZAX

7. WAX and YAU

8.  ZAY and YAU

Step-by-step explanation:

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6 0
3 years ago
How many strings can be formed by ordering the letters MISSISSIPPI which
Mazyrski [523]

Answer:

<em>1680 is the answer.</em>

Step-by-step explanation:

Here, we have 11 letters in the word <em>MISSISSIPPI.</em>

Repetition of letters:

M - 1 time

I - 4 times

S - 4 times

P - 2 times

As per question statement, we need a substring MISS in the resultant strings.

So, we need to treat MISS as one unit so that <em>MISS always comes together in all the strings</em>.

The resultant strings will look like:

<em>xxxxMISSxxx</em>

<em>xxMISSxxxxx</em>

<em>and so on.</em>

After we treat <em>MISS </em>as one unit, total letters = 8

Repetition of letters:

MISS - 1 time

I - 3 times

S - 2 times

P - 2 times

The formula for combination of letters with total of <em>n </em>letters:

\dfrac{n!}{p!q!r!}

where <em>p, q</em> and <em>r</em> are the number of times other letters are getting repeated.

<em>p = 3</em>

<em>q = 2</em>

<em>r = 2</em>

<em></em>

So, required number of strings that contain MISS as substring:

\dfrac{8!}{3!2!2!}\\\Rightarrow \dfrac{40320}{6\times 2 \times 2}\\\Rightarrow 1680

So, <em>1680 is the answer.</em>

3 0
3 years ago
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