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baherus [9]
3 years ago
14

Polynomials (3v^2+4v+6)-(v+5​

Mathematics
1 answer:
Allisa [31]3 years ago
4 0

Answer:

3v^2+3v+1

Step-by-step explanation:

add polynomials, we add the same terms together

the only v^2 term is 3v^2, so we learn it as 3v^2

4v and v are both the same terms, so we subtract them. 4v - v = 3v

6 and 5 are both the same terms, so we subtract them. 6-5 = 1

all together, it is 3v^2+3v+1

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Step-by-step explanation:

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4 years ago
A $33$-gon $P_1$ is drawn in the Cartesian plane. The sum of the $x$-coordinates of the $33$ vertices equals $99$. The midpoints
lisabon 2012 [21]

1.<span>
The midpoint </span>MPQ of PQ is given by  (a + c / 2, b + d / 2)<span>

2.
Let the x coordinates of the vertices of P_1 be : 

x1, x2, x3,…x33

the x coordinates of P_2 be :

</span>z1, x2, x3,…z33<span>

and the x coordinates of P_3 be:


w1, w2, w3,…w33</span>

<span>
3.
We are given with: 


</span>

X1 + x2 + x3… + x33 = 99

We also want to find the value of w1 + w2 + w3… + w33.<span>

4.

Now, based from the midpoint formula:</span>

 

Z1 = (x1 + x2) / 2

Z2 = (x2 + x3) / 2

Z3 = (x3 + x4) / 2

Z33 = (x33 + x1) / 2<span>

and 

</span>

<span>W1 = (z1 + z1) / 2


W2 = (z2 + z3) / 2</span>

<span>W3 = (z3 + z4) / 2

W13 = (z33 + z1) / 2

.
.

5.</span>

<span>W1 + w1 + w3… + w33 = (z1 + z1) / 2 +  (z2 + z3) / 2 + (z33 + z1) / 2 = 2 (z1 + z2 + z3… + z33) / 2</span>

<span>Z1 + z1 + z3… + z33 = (x1 + x2) / 2 + (x2 + x3) / 2 + (x33 + x1) / 2

</span>2 (x1 + x2 + x3… + x33) / 2 = (x1 + x2 + x3… + x33 = 99<span>


<span>Answer: 99</span></span>

8 0
3 years ago
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