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grin007 [14]
2 years ago
9

Uhm, I'm a bit confused- ._. What's the X and Y for W, X, and V?

Mathematics
2 answers:
ELEN [110]2 years ago
7 0
2.5 and 6.5 for w 5.5 and 2.5 for v
MAVERICK [17]2 years ago
4 0

Answer:

x-x3.5,y1.5 w-x2.5,y6.5 v-x5.5,y6.5

Step-by-step explanation:

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How do I solve this equation: (3+2i)+(2+bi)=5-4i
oksian1 [2.3K]
(3+2i)+(2+bi)=5-4i

6 0
2 years ago
Given that the expression 2x^3 + mx^2 + nx + c leaves the same remainder when divided by x -2 or by x+1 I prove that m+n =-6
Alla [95]

Given:

The expression is:

2x^3+mx^2+nx+c

It leaves the same remainder when divided by x -2 or by x+1.

To prove:

m+n=-6

Solution:

Remainder theorem: If a polynomial P(x) is divided by (x-c), thent he remainder is P(c).

Let the given polynomial is:

P(x)=2x^3+mx^2+nx+c

It leaves the same remainder when divided by x -2 or by x+1. By using remainder theorem, we can say that

P(2)=P(-1)              ...(i)

Substituting x=-1 in the given polynomial.

P(-1)=2(-1)^3+m(-1)^2+n(-1)+c

P(-1)=-2+m-n+c

Substituting x=2 in the given polynomial.

P(2)=2(2)^3+m(2)^2+n(2)+c

P(2)=2(8)+m(4)+2n+c

P(2)=16+4m+2n+c

Now, substitute the values of P(2) and P(-1) in (i), we get

16+4m+2n+c=-2+m-n+c

16+4m+2n+c+2-m+n-c=0

18+3m+3n=0

3m+3n=-18

Divide both sides by 3.

\dfrac{3m+3n}{3}=\dfrac{-18}{3}

m+n=-6

Hence proved.

7 0
2 years ago
Arturo worked a 40 hour week at $12 per hour. He then received a raise of one dollar per hour and worked 830 hour week. How much
Mila [183]

Answer:

nobody can work 830 in one week; there are only 168 hours in a week

Step-by-step explanation:

3 0
3 years ago
Find the value of x​
Lubov Fominskaja [6]

Answer:

thanks<em> </em><em> </em><em>4</em><em>5</em>

Step-by-step explanation:

$$56-5($4+448

edeu

udyeu

rijdhxhchxjjxjsjduu73627

7 0
3 years ago
A certain positive integer has exactly 20 positive divisors. What is the smallest number of primes that could divide the integer
Alika [10]

As per my explanation to (b) above, the largest number of primes that could factor such a number is 4.

Note that  2,3,5 and 7   are the smallest primes, then use the reasoning from

(b) above. we are looking for four exponents, that, when 1 is added to each and all are multiplied together, would equal 20.

But no such integers  k, l, m, and n  exist such that (k + 1)(l + 1)(m + 1) (n + 1)  = 20 where k, l,m, and n ≥ 1 so this number, whatever it is, can't have 4 prime factors

Let's drop 7 out of the mix and suppose it has just 3 prime factors 2, 3, and 5 again we are looking for three exponents,  that, when 1 is added to each and all are multiplied together, would equal 20.  Put another way, we are looking for k, l and m ≥ 1   such that (k + 1)(l + 1)(m + 1) = 20

Note that the  only possibility here  is when we have 2 *2 *5  = 20 and the smallest possible product would be 2^(4 )* 3^(1) * 5^(1) =  2^4 * 3 * 5 = 240

Now......the only remaining possibility is  that this number is composed of the two smallest primes, 2 and 3,  and we are looking for  some k  and l ≥ 1 such that (k + 1)(l + 1) = 20 clearly, the only possibilities  are when k = 4 and l = 5, or vice-versa

So this number would factor as either 2^3 * 3^4   = 648  or 2^4 * 3^3 = 432 and both are > 240.

Learn more about positive integers at

brainly.com/question/1367050

#SPJ4

8 0
1 year ago
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