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AveGali [126]
3 years ago
8

Arrange the following polynomial into descending order and identify the 3rd term.

Mathematics
2 answers:
Vera_Pavlovna [14]3 years ago
6 0
Thank you for posting your question here at brainly. Feel free to ask more questions.  

The best and most correct answer among the choices provided by the question is <span>18x</span>.<span>       
    </span><span>
Hope my answer would be a great help for you.</span>
mart [117]3 years ago
6 0

Answer with Step-by-step explanation:

We are given a polynomial:

18x + 2x^3 - 3x^2 + 10

We have to arrange it in descending order and then identify its third term

Polynomial in descending order is:

2x^3-3x^2+18x+10

As we can clearly see the third term is:

18x

Hence,

Polynomial in descending order is:

2x^3-3x^2+18x+10

and the third term is:

18x

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Find 3/7 of 21? Can you find this answer because my homework I do not know how to do so can you please find this question.
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Answer:

9

Step-by-step explanation:

3/7 multiplied by 21 = 3/7 x 21 = 9 = 3/7 x 21/1 = 63/7 = 9

Or if you can't do the math - Think of a  pizza with 21 slices . Now cut this pizza into 7 equal parts. You will find that each part contains exactly 3 slices. Now, eat 3 of these parts. With three slices in each part, you ate 9 slices - which is three sevenths of 21.

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During the summer, Tyler delivers packages and Miranda cares for animals on a farm. To show their earnings, Tyler makes a graph
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Miranda earns $2.20 more per your than Tyler. That is the right answer

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The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n
GarryVolchara [31]

Answer:

(-1,1),(4,-2)

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point (x,y)

Distance between points (x_1,y_1),(x_2,y_2) is given by \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]

Put (x,y)=(-1,1)

34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34

which is true. So, (-1,1) can be a vertex

Put (x,y)=(4,-2)

34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34

which is true. So, (4,-2) can be a vertex

Put (x,y)=(1,1)

34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22

which is not true. So, (1,1) cannot be a vertex

Put (x,y)=(2,-2)

34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22

which is not true. So, (2,-2) cannot be a vertex

Put (x,y)=(4,-1)

34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30

which is not true. So, (4,-1) cannot be a vertex

Put (x,y)=(-1,4)

34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70

which is not true. So, (-1,4) cannot be a vertex

So, possible points for the vertex are (-1,1),(4,-2)

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Answer:

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