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Vedmedyk [2.9K]
3 years ago
12

Please Answer the question with the formula and simplification

Mathematics
2 answers:
Mariulka [41]3 years ago
8 0

<em>answer = \frac{1}{2}   \\ solution \\  \frac{ - 4}{5}  \times  \frac{3}{7}   \times  \frac{15}{16}  \times ( \frac{ - 14}{9} ) \\  =   \frac{ - 4 \times 3 \times 15 \times ( - 14)}{5 \times 7 \times 16 \times 9}  \\  =  \frac{2520}{5040}  \\  = divide \: 5040 \: by \: 2520 \:  \\  =  \:  \frac{1}{2}  \\ hope \: it \: helps</em>

dybincka [34]3 years ago
3 0

Answer:

1/2

Step-by-step explanation:

-4    3     15    -14

--- *--- *   --- *------

5   7      16     9

Rewriting

-4    3     5*3   -7*2

--- *--- *   --- *------

5   7      4*4     3*3

Canceling like terms

-1    1     1*1   -1*2

--- *--- *   --- *------

1   1      1*4     1*1

This leaves

2

---

4

1/2

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Ms. Bell's mathematics class consists of 12 sophomores, 8 juniors, and 9 seniors. How many different ways can Ms. Bell create a
shtirl [24]

Answer:

792

Step-by-step explanation:

It's a combination question. The order is of no consequence. Also the fact that there are juniors and seniors is not important either.

So the answer is

12C5

12!

====

(12 - 5)! * 5!

12 * 11 * 10 * 9 * 8

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

5 * 4 * 3 * 2 * 1

792

7 0
3 years ago
Which of the following square root of -80
Marta_Voda [28]

You can factor -80 as

-80 = (-1)\cdot 16 \cdot 5

So, we have

\sqrt{-80} = \sqrt{(-1)\cdot 16 \cdot 5}

The square root of a product is the product of the square roots:

\sqrt{-80} = \sqrt{(-1)}\sqrt{16}\sqrt{5}

Since i^2=-1 and 4^2=16, we have

\sqrt{-80} = 4i\sqrt{5}

8 0
3 years ago
If p is a polynomial show that lim x→ap(x)=p(a
Lostsunrise [7]

Let p(x) be a polynomial, and suppose that a is any real number. Prove that

lim x→a p(x) = p(a) .

 

Solution. Notice that

 

2(−1)4 − 3(−1)3 − 4(−1)2 − (−1) − 1 = 1 .

 

So x − (−1) must divide 2x^4 − 3x^3 − 4x^2 − x − 2. Do polynomial long division to get 2x^4 − 3x^3 − 4x^2 – x – 2 / (x − (−1)) = 2x^3 − 5x^2 + x – 2.

 

Let ε > 0. Set δ = min{ ε/40 , 1}. Let x be a real number such that 0 < |x−(−1)| < δ. Then |x + 1| < ε/40 . Also, |x + 1| < 1, so −2 < x < 0. In particular |x| < 2. So

 

|2x^3 − 5x^2 + x − 2| ≤ |2x^3 | + | − 5x^2 | + |x| + | − 2|

= 2|x|^3 + 5|x|^2 + |x| + 2

< 2(2)^3 + 5(2)^2 + (2) + 2

= 40

 

Thus, |2x^4 − 3x^3 − 4x^2 − x − 2| = |x + 1| · |2x^3 − 5x^2 + x − 2| < ε/40 · 40 = ε.

3 0
3 years ago
3.30, -√10, 17/5, √11 least to greatest
Vilka [71]

Answer:

-√10,√11 , 3.30, 17/5

Step-by-step explanation:

-√10,√11 , 3.30, 17/5

17/5 is 3.4

the 10 is a negative so its the smallest

11 square root is a imiganery

7 0
4 years ago
Read 2 more answers
Help thank youuu -Eric chavez
Sphinxa [80]

Answer:

y = 4x + 5

Step-by-step explanation:

5 0
3 years ago
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