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Sunny_sXe [5.5K]
2 years ago
9

Lines k and n intersect on the y-axis

Mathematics
1 answer:
avanturin [10]2 years ago
7 0

a) The equation of line k is:

y = -\frac{202}{167}x + \frac{598}{167}

b) The equation of line j is:

y = \frac{167}{202}x + \frac{1546}{202}

The equation of a line, in <u>slope-intercept formula</u>, is given by:

y = mx + b

In which:

  • m is the slope, which is the rate of change.
  • b is the y-intercept, which is the value of y when x = 0.

Item a:

  • Line k intersects line m with an angle of 109º, thus:

\tan{109^{\circ}} = \frac{m_2 - m_1}{1 + m_1m_2}

In which m_1 and m_2 are the slopes of <u>k and m.</u>

  • Line k goes through points (-3,-1) and (5,2), thus, it's slope is:

m_1 = \frac{2 - (-1)}{5 - (-3)} = \frac{3}{8}

  • The tangent of 109 degrees is \tan{109^{\circ}} = -\frac{29}{10}
  • Thus, the slope of line m is found solving the following equation:

\tan{109^{\circ}} = \frac{m_2 - m_1}{1 + m_1m_2}

-\frac{29}{10} = \frac{m_2 - \frac{3}{8}}{1 + \frac{3}{8}m_2}

m_2 - \frac{3}{8} = -\frac{29}{10} - \frac{87}{80}m_2

m_2 + \frac{87}{80}m_2 = -\frac{29}{10} + \frac{3}{8}

\frac{167m_2}{80} = \frac{-202}{80}

m_2 = -\frac{202}{167}

Thus:

y = -\frac{202}{167}x + b

It goes through point (-2,6), that is, when x = -2, y = 6, and this is used to find b.

y = -\frac{202}{167}x + b

6 = -\frac{202}{167}(-2) + b

b = 6 - \frac{404}{167}

b = \frac{6(167)-404}{167}

b = \frac{598}{167}

Thus. the equation of line k, in slope-intercept formula, is:

y = -\frac{202}{167}x + \frac{598}{167}

Item b:

  • Lines j and k intersect at an angle of 90º, thus they are perpendicular, which means that the multiplication of their slopes is -1.

Thus, the slope of line j is:

-\frac{202}{167}m = -1

m = \frac{167}{202}

Then

y = \frac{167}{202}x + b

Also goes through point (-2,6), thus:

6 = \frac{167}{202}(-2) + b

b = \frac{(2)167 + 202(6)}{202}

b = \frac{1546}{202}

The equation of line j is:

y = \frac{167}{202}x + \frac{1546}{202}

A similar problem is given at brainly.com/question/16302622

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Answer:

-3071/16

Step-by-step explanation:

Hope this helps, if you wrote it wrong or need help understanding please let me know!

4 0
2 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
What is 38% of 250 pls explain
cestrela7 [59]
95
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6 0
2 years ago
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3 0
2 years ago
How many outcome sequences are possible when a die is rolled 4 times where we say for instance that the outcome is 3 4 3 1 if th
STatiana [176]

Answer:

the possible outcome sequences when a die is rolled 4 times is 1296

Step-by-step explanation:

 Given the data in the question;

a die is rolled 4 times

and outcomes are { 3, 4, 3, 1 }

we know that; possible number of outcomes on a die is n = 6{ 1,2,3,4,5,6 }

Now when we roll a die lets say, r times

then the total number of possible outcomes will be;

N = n^r

given that; r = 4

Hence if we roll a die 4 times;

Total number of possible outcome N = 6⁴

N = 1296

Therefore, the possible outcome sequences when a die is rolled 4 times is 1296

3 0
3 years ago
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