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1 year ago

for what value of k, the line joining 3x-ky+7=0 is perpendicular to the line joining (4 ,3) and ( 5, -3).

Mathematics

1 answer:

Schach [20]1 year ago

4 0

Answer:

k = 18

=========

<h2>Given</h2>

3x - ky + 7 = 0

Passing through the points (4, 3) and (5, - 3)

The value of k, if the lines are perpendicular

<h2>Solution</h2>

We know the perpendicular lines have opposite reciprocal slopes, that is the product of their slopes is - 1.

Find the slope of line 1 by converting the equation into slope-intercept from standard form:

<em>standard form is ⇒ ax + by + c = 0, </em>
<em>slope - intercept form is ⇒ y = mx + b, where m is the slope</em>

3x - ky + 7 = 0
ky = 3x + 7
y = (3/k)x + 7/k

Its slope is 3/k.

Find the slope of line 2, using the slope formula:

m = (y₂ - y₁)/(x₂ - x₁) = (-3 - 3)/(5 - 4) = - 6/1 = - 6

We have both the slopes now. Find their product:

(3/k)*(- 6) = - 1
- 18/k = - 1
k = 18

So when k is 18, the lines are perpendicular.

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3 years ago

Assume ABC Company deposits $90,000 with First National Bank in an account earning interest at 6% per annum, compounded semi-ann

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3 years ago

Factor completely: 2x4 − 32.

vazorg [7]

Answer:

Option b) is correct.

The completed factor of given expression is $2(x-2)(x+2)(x^2+4)$

Step-by-step explanation:

Given expression is $2x^4-32$

To find the completed factor for the given expression:

$2x^4-32$ :

Taking the common number "2" outside to the above expression we get

$2x^4-32=2(x^4-16)$

Now rewritting the above expression as below

$=2(x^4-2^4)$ (since 16 can be written as the number 2 to the power of 4)

$=2((x^2)^2-(2^2)^2)$

The above expression is of the form $a^2-b^2=(a+b)(a-b)$

Here $a=x^2$ and $b=2^2$

Therefore it becomes

$=2(x^2+2^2)(x^2-2^2)$

$=2(x^2+4)(x^2-2^2)$

The above expression is of the form $a^2-b^2=(a+b)(a-b)$

Here $a=x$ and $b=2$

Therefore it becomes

$=2(x^2+4)(x+2)(x-2)$

$=2(x+2)(x-2)(x^2+4)$

Therefore $=2(x+2)(x-2)(x^2+4)$

$2x^4-32=2(x-2)(x+2)(x^2+4)$

Option b) is correct.

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3 years ago

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3 years ago