Write an equation in point-slope form for the line through the given point with the given slope.

Mathematics

1 answer:

Gennadij [26K]2 years ago

6 0

Answer:

A. y + 5 = negative two-fifths (x + 3)

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What is the equation, in point-slope form, for a line that goes through

TiliK225 [7]

Answer:

B. $y+4=-\frac{5}{6}(x-8)$

Step-by-step explanation:

The point slope form of a line is $y-y_1=m(x-x_1)$ . Since the slope is -5/6 then m=-5/6. Since the line goes through point (8,-4) then substitute $x_1=8\\y_1=-4$ and simplify.

The equation becomes $y--4=-\frac{5}{6}(x-8)\\y+4=-\frac{5}{6}(x-8)$

7 0

3 years ago

Solve the following system of equations 2x+7y=-57 and -6x-8y=80

hram777 [196]

I would use the process of elimination. Divide 13 from both sides
2x+7y=-57 y=-7 then plug in y to an equation
-6x-8y=80 2x+7(-7)=-57 multiple
multiple the first equation by 3 2x-49=-57 add 49 to both sides
6x+21y=-171 2x=-8 divide two from both sides
-6x-8y=80 x=-4
combine like terms So your things would be x=-4 and y=-7
13y=-91 And itd make the solution (-4,-7).

7 0

3 years ago

If the second term of a geometric sequence of real numbers is -2 and the fifth term is 16 then what is the fourteenth term?

Agata [3.3K]

<h3>Given</h3>

A geometric sequence such that ...

$a_2=-2\quad\text{and}\quad a_5=16$

$a_{14}$

<h3>Solution</h3>

We can use the ratio of the given terms to find the common ratio of the sequence, then use that to find the desired term from one of the given terms. We don't actually need the common ratio (-2). All we need is its cube (-8).

$a_2=a_1r^{(2-1)}=a_1r^1\\a_5=a_1r^{(5-1)}=a_1r^4\\a_{14}=a_1r^{(14-1)}=a_1r^13=a_5r^9\\\\\dfrac{a_5}{a_2}=\dfrac{a_1r^4}{a_1r^1}=r^3=\dfrac{16}{-2}=-8\\\\r^9=\left(r^3\right)^3=(-8)^3=-512\\a_{14}=a_5(-512)\\\\a_{14}=-8192$