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Vlad1618 [11]
1 year ago
5

This make no sense:

Mathematics
2 answers:
saveliy_v [14]1 year ago
5 0

Answer:

D is correct.

Step-by-step explanation:

Hoped this helped!

Hoped this helped!Hoped this helped!Hoped this helped!Hoped this helped!Hoped this helped!Hoped this helped!Hoped this helped!

Anika [276]1 year ago
4 0

Answer:

3.2

Step-by-step explanation:

Distance forumla

y = \sqrt{(x1 - x2)^2 + (y1-y2)^2}

y = \sqrt{(2 - (-1))^2 +(0-1)^2} \\ y = \sqrt{10} \\y =  3.2

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If they don't intersect then there is no solution and the lines are parrell.
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Find the value of x. *<br> Please help I’m really confused
mezya [45]

Due to the Triangle Angle Sum Theorem, we know that the sum of the interior angles of a triangle is 180 degrees.

90 + 2x - 2 + x + 5 = 180 degrees.

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4 0
2 years ago
Given P(x) = 2x3 - 11x2 + 18x - 15 is divided by x – 3. Use synthetic division to find the quotient and the remainder.
katrin [286]
You can find the remainder right away by simply plugging in x=3. The polynomial remainder theorem guarantees that the value of p(3) is the remainder upon dividing p(x) by x-3, but I digress...

Synthetic division yields

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.    |           6   -15     9
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which translates to

\dfrac{2x^3-11x^2+18x-15}{x-3}=2x^2-5x+3-\dfrac6{x-3}

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2 years ago
A particle P moves in a straight line such that, t s after leaving a point O, its velocity v ms^-1 is given by v=36t - 3t^2 for
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3 years ago
What is the coefficient of x2y3 in the expansion of (2x + y)5?
Zigmanuir [339]

Option C:

The coefficient of x^{2} y^{3} is 40.

Solution:

Given expression:

(2 x+y)^{5}

Using binomial theorem:

(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}

Here a=2 x, b=y

Substitute in the binomial formula, we get

(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}

                                                            $+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}

Let us solve the term one by one.

$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}

$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y

$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}

$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}

$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}

$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}

Substitute these into the above expansion.

(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}

The coefficient of x^{2} y^{3} is 40.

Option C is the correct answer.

5 0
2 years ago
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