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2 years ago

I REALLY NEED HELP ASAP WITH THESE 2 QUESTIONS!!!!!!!!!

Mathematics

1 answer:

DIA [1.3K]2 years ago

5 0

Answer:

4. Meena- 3

Jerry- 6

5a. 5

5b. 6

Step-by-step explanation:

We can use the distance formula to solve all of these.

$\sqrt{(x_{2} } -x_{1} )^2+(y_{2} -y_{1} )^2$

Plug the numbers in and make sure you understand how i got these answers :) Let me know if you see anything wrong about it, and i will try and fix it.

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An insurance policy sells for $800. Based on past data, an average of 1 in 50

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2 years ago

It takes approximately 4.65 quarts of milk to make a pound of cheese. Express this amount as a mixed number in simplest form.

ikadub [295]

The correct answer would be 4 and 13/20

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2 years ago

The perimeter of a rectangle is 150 cm. the length is 4 cm greater than the width find the width and length

Fudgin [204]

Let's call the width of our rectangle $w$ and the length $l$ . We can say $l = w + 4$ , since the length is equal to 4 cm greater than the width.

Also remember that the perimeter of a rectangle is the sum of two times the width and two times the length, or $P = 2l + 2w$ . To solve this problem, we can substitute in the information we know, as shown below:

$150 = 2(w + 4) + 2w$

$150 = 4w + 8$

$142 = 4w$

$w = \frac{71}{2}$

Now, we can substitute in the width we found into the formula for length, which is $l = w + 4$ :

$l = \frac{71}{2} + 4$

$l = \frac{79}{2}$

The width of our rectangle is $\boxed{\frac{71}{2} \,\, \textrm{cm}}$ cm and the length of our rectangle is $\boxed{\frac{79}{2} \,\, \textrm{cm}}$

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2 years ago

If a fraction can be written as a repeating decimal,only one digit can repeat over and over, without end. True or false?

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3 years ago

Determine the total number of proper subsets of the set of letters from the English alphabet a,b, c, ...,j. The number of proper

Inga [223]

$\{a,b,c,\ldots,j\}$ has ten elements, so any proper subset has at most nine of these elements.

The number of ways of taking any $0\le n\le 9$ letters from this set is given by the binomial coefficient,

$\dbinom{10}n=\dfrac{10!}{n!(10-n)!}$

and in particular, the total number of ways to picking proper subsets is

$\displaystyle\sum_{n=0}^9\binom{10}n=\dbinom{10}0+\dbinom{10}1+\cdots+\dbinom{10}9$

Without computing each term directly, let's instead use a direct result from the binomial theorem, which says

$\displaystyle\sum_{n=0}^N\binom Nna^nb^{N-n}=(a+b)^N$

If we replace $a=b=1$ , then we're left with

$\displaystyle\sum_{n=0}^N\binom Nn=(1+1)^N=2^N$

We can use this to evaluate our sum directly:

$\displaystyle\sum_{n=0}^9\binom{10}n=\sum_{n=0}^{10}\binom{10}n-\binom{10}{10}=2^{10}-1=1023$

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2 years ago