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2 years ago

6

The given line passes through the points (-4,-3) and (4,

1 answer:

Harlamova29_29 [7]2 years ago

6 0

Answer:

Step-by-step explanation:

first we gotta find the slope of the first line

(-4,-3),(4,1)

slope = (y2 - y1) / (x2 - x1) = (1- (-3) / (4 - (-4) = (1 + 3) / (4 + 4) = 4/8 = 1/2

so the slope is 1/2.....so we are looking for a line that is perpendicular....perpendicular lines have negative reciprocal slopes...all that means is flip the slope and change the sign....so the slope we need is :

1/2....flip it....2/1.....change the sign....-2.....we need a -2 slope

y - y1 = m(x - x1)

slope = -2

(-4,3)...x1 = -4 and y1 = 3

now sub

y - 3 = -2(x - (-4) =

y - 3 = -2(x + 4) <=====

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Find the exact value by using a half-angle identity. (6 points)

Elenna [48]

Answer:

c

Step-by-step explanation:

4 0

2 years ago

Help help i need help anyone ?

REY [17]

-9-6i/-3-2i
Factor -3 out of expression
-3(3+2i)/-3-2i
Then extract the negative sign out of the expression
-3(3+2i)/-(3+2i)
Reduce the fraction with -(3+2i)
-3/-1
-3*-1=3
Hope this helps

3 0

2 years ago

The difference between the two roots of the equation 3x^2+10x+c=0 is 4 2/3 . Find the solutions for the equation.

andrezito [222]

Answer:

Given the equation: $3x^2+10x+c =0$

A quadratic equation is in the form: $ax^2+bx+c = 0$ where a, b ,c are the coefficient and a≠0 then the solution is given by :

$x_{1,2} = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ......[1]

On comparing with given equation we get;

a =3 , b = 10

then, substitute these in equation [1] to solve for c;

$x_{1,2} = \frac{-10\pm \sqrt{10^2-4\cdot 3 \cdot c}}{2 \cdot 3}$

Simplify:

$x_{1,2} = \frac{-10\pm \sqrt{100- 12c}}{6}$

Also, it is given that the difference of two roots of the given equation is $4\frac{2}{3} = \frac{14}{3}$

i.e,

$x_1 -x_2 = \frac{14}{3}$

Here,

$x_1 = \frac{-10 + \sqrt{100- 12c}}{6}$ , ......[2]

$x_2= \frac{-10 - \sqrt{100- 12c}}{6}$ .....[3]

then;

$\frac{-10 + \sqrt{100- 12c}}{6} - (\frac{-10 + \sqrt{100- 12c}}{6}) = \frac{14}{3}$

simplify:

$\frac{2 \sqrt{100- 12c} }{6} = \frac{14}{3}$

or

$\sqrt{100- 12c} = 14$

Squaring both sides we get;

$100-12c = 196$

Subtract 100 from both sides, we get

$100-12c -100= 196-100$

Simplify:

-12c = -96

Divide both sides by -12 we get;

c = 8

Substitute the value of c in equation [2] and [3]; to solve $x_1 , x_2$

$x_1 = \frac{-10 + \sqrt{100- 12\cdot 8}}{6}$

or

$x_1 = \frac{-10 + \sqrt{100- 96}}{6}$ or

$x_1 = \frac{-10 + \sqrt{4}}{6}$

Simplify:

$x_1 = \frac{-4}{3}$

Now, to solve for $x_2$ ;

$x_2 = \frac{-10 - \sqrt{100- 12\cdot 8}}{6}$

or

$x_2 = \frac{-10 - \sqrt{100- 96}}{6}$ or

$x_2 = \frac{-10 - \sqrt{4}}{6}$

Simplify:

$x_2 = -2$

therefore, the solution for the given equation is: $-\frac{4}{3}$ and -2.

3 0

2 years ago

A tree grows in height by 21% per year. It is 2m tall after one year. After how many years will the tree be over 20m tall?

mariarad [96]

20/45 hope it helps have a good day

7 0

2 years ago

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Please help me with this

Stella [2.4K]

c is a correct answer you know

5 0

2 years ago

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