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elixir [45]
3 years ago
7

Choose the correct sum of the polynomials (6x3 − 8x − 5) (3x3 6x 2). 9x3 − 2x − 3 3x3 2x 3 3x3 − 2x − 3 9x3 14x 3.

Mathematics
1 answer:
RUDIKE [14]3 years ago
5 0

Like terms have the same variable and power. The sum of the two polynomials (6x^3 -8x -5)+(3x^3+6x+2) is 3x^3 -2x -3.

<h3>What are Like terms?</h3>

Like terms are those terms that are having the same variables, also the variables are of the same order as well.

for example, 25x and 5x are like terms; 30xy and 7xy are like terms, 9x³ and 4x² are not like terms, etc.

We know that in order to find the sum of the two polynomials we need to add or subtract the like terms, therefore, the sum of the polynomials can be done as,

(6x^3 -8x -5)+(3x^3+6x+2)\\\\=  6x^3 -8x -5+3x^3+6x+2\\\\= 6x^3 +3x^3- 8x+6x-5+2\\\\ = 3x^3 -2x -3

Hence, the sum of the two polynomials (6x^3 -8x -5)+(3x^3+6x+2) is 3x^3 -2x -3.

Learn more about Like Terms:

brainly.com/question/2513478

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0.002 is 1/10 (fraction) of......
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<span>0.002 is the 1/10 of what number.
Find the number that applies to this.
=> 0.002 has a thousandths value since it is placed next to the decimal points
=> To be able to get the answer. Let’s convert 1/10 into a decimal.
=> 1 / 10 = 0.1
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3 years ago
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Write 230% in simplest form
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3 0
4 years ago
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In triangle $ABC$, let angle bisectors $BD$ and $CE$ intersect at $I$. The line through $I$ parallel to $BC$ intersects $AB$ and
Umnica [9.8K]

Answer:

41

Step-by-step explanation:

If you work through a series of obscure calculations involving area and the radius of the incircle, they boil down to a simple fact:

... For MN║BC, perimeter ΔAMN = perimeter ΔABC - BC = AB+AC

.. = 17+24 = 41

_____

Wow! Thank you for an interesting question with a not-so-obvious answer.

_____

<em>A little more detail</em>

The point I that you have defined is the incenter—the center of an inscribed circle in the triangle. Its radius is the distance from I to any side, such as BC, for example.

If we use "Δ" to represent the area of the triangle and "s" to represent the semi-perimeter, (AB+BC+AC)/2, then the incircle has radius Δ/s. The area Δ can be computed from Heron's formula by ...

... Δ = √(s(s-a)(s-b)(s-c)) . . . . where a, b, c are the side lengths

For this triangle, the area is Δ = √38480 ≈ 196.1632 units². That turns out to be irrelevant.

The altitude to BC will be 2Δ/(BC), so the altitude of ΔAMN = (2Δ/(BC) -Δ/s). Dividing this by the altitude to BC gives the ratio of the perimeter of ΔAMN to the perimeter of ΔABC, which is 2s.

Putting these ratios and perimeters together, we get ...

... perimeter ΔAMN = (2Δ/(BC) -Δ/s)/(2Δ/(BC)) × 2s

... = (2/(BC) -1/s) × BC × s = 2s -BC

... perimeter ΔAMN = AB +AC

8 0
3 years ago
Answer ASAP please, I need help.
Darya [45]

Step-by-step explanation:

martha-12

0+6+12

i think thats the answer

8 0
3 years ago
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