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2 years ago

Find the equation of the line through point (8,−9) and perpendicular to 3+8=4.

1 answer:

4 0

The equation of the line through the point (8,−9) and perpendicular to 3x+8y=4 is given by 8x - 3y = 91.

We are aware that any straight line's equation can be expressed as

y = mx + c,

where m denotes the slope and c denotes a constant.

Also, two perpendicular lines' slopes are the negative reciprocals of one another.

Here, the equation of the given straight line is

3x+8y=4

i.e. 8y = 4 -3x

i.e. y = (4/8) - (3/8)x

Now the negative reciprocal of - 3/8 is 8/3.

Then we can write the equation of the perpendicular line is

y = (8/3)x + c ...(1)

Since (1) passes through the point (8, -9), so we can put x = 8 and y = -9 in (1) to get the value of c.

So, -9 = (8/3)*8 + c

i.e. -9 = 64/3 + c

i.e. c = -9 -64/3 = - (27 + 64)/3 = - 91/3

(1) can be written as

y = (8/3)x - (91/3)

i.e. 3y = 8x - 91

i.e. 8x - 3y = 91

Therefore the equation of the line through the point (8,−9) and perpendicular to 3x+8y=4 is given by 8x - 3y = 91.

Learn more about perpendicular lines here -

brainly.com/question/12209021

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Simplify the following expression.

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Answer:

Step-by-step explanation:

Remember PEMDAS!

So first parentheses:

1.6^2 = 2.56

8.7 + 9.3 = 18.

Because that is inside another parentheses you multipy them.

2.56 x 18 = 46.08

Now: 46.08 divided by 1.8 = 25.6

Now subtract:

25.6 - 3.7 - 1.9 which equals 20!

Hope that helped!

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3 0

3 years ago

Suppose the roots of the polynomial $x^2 - mx + n$ are positive prime integers (not necessarily distinct). Given that $m < 20

Vsevolod [243]

Answer:

18 values for n are possible.

Step-by-step explanation:

Given the quadratic polynomial:

$x^2 - mx + n$

such that:

Roots are positive prime integers and

$m < 20$

To find:

How many possible values of $n$ are there ?

Solution:

First of all, let us have a look at the sum and product of a quadratic equation.

If the quadratic equation is:

$Ax^{2} +Bx+C$

and the roots are: $\alpha$ and $\beta$

Then sum of roots, $\alpha+\beta = -\frac{B}{A}$

Product of roots, $\alpha \beta = \frac{C}{A}$

Comparing the given equation with standard equation, we get:

A = 1, B = -m and C = n

Sum of roots, $\alpha+\beta = -\frac{-m}{1} = m$

Product of roots, $\alpha \beta = \frac{n}{1} = n$

We are given that $m$

$\alpha$ and $\beta$ are positive prime integers such that their sum is less than 20.

Let us have a look at some of the positive prime integers:

2, 3, 5, 7, 11, 13, 17, 23, 29, .....

Now, we have to choose two such prime integers from above list such that their sum is less than 20 and the roots can be repetitive as well.

So, possible combinations and possible value of $n (= \alpha \times \beta)$ are:

$1.\ 2, 2\Rightarrow n = 2\times 2 = 4\\2.\ 2, 3 \Rightarrow n = 6\\3.\ 2, 5 \Rightarrow n = 10\\4.\ 2, 7\Rightarrow n = 14\\5.\ 2, 11 \Rightarrow n = 22\\6.\ 2, 13 \Rightarrow n = 26\\7.\ 2, 17 \Rightarrow n = 34\\8.\ 3, 3\Rightarrow n = 3\times 3 = 9\\9.\ 3, 5 \Rightarrow n = 15\\10.\ 3, 7 \Rightarrow n = 21\\$

$11.\ 3, 11\Rightarrow n = 33\\12.\ 3, 13 \Rightarrow n = 39\\13.\ 5, 5 \Rightarrow n = 25\\14.\ 5, 7 \Rightarrow n = 35\\15.\ 5, 11 \Rightarrow n = 55\\16.\ 5, 13 \Rightarrow n = 65\\17.\ 7, 7 \Rightarrow n = 49\\18.\ 7, 11 \Rightarrow n = 77$

So,as shown above 18 values for n are possible.

3 0

3 years ago