What’s the slope of these two lines if right u hey brainlist

Mathematics

1 answer:

MrRissso [65]2 years ago

7 0

Answer: they are both zero because they both are horizontal

Step-by-step explanation:

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The expression p/x^2 - 5x +6 simplifies to x+4/x-2. Which expression doees p represent?

gladu [14]

We know that
(ad)/(bd)=d/d time a/b=a/b since d's cancel
also
if a/b=c/d in simplest form, then a=c and b=d

we have
p/(x^2-5x+6)=(x+4)/(x-2)
therefor

p/(x^2-5x+6)=d/d times (x+4)/(x-2)
p/(x^2-5x+6)=d(x+4)/d(x-2)

therefor
p=d(x+4) and
x^2-5x+6=d(x-2)

we can solve last one
factor
(x-6)(x+1)=d(x-2)
divide both sides by (x-2)
[(x-6)(x+1)]/(x-2)=d
sub

p=d(x+4)
p=([(x-6)(x+1)]/(x-2))(x+4)
$p= \frac {(x-6)(x+1)(x+4)}{(x-2)}$

7 0

3 years ago

Find the coordinates of point X that lies along the directed line segment from Y(-8, 8) to T(-15, -13) and partitions the segmen

Nataly_w [17]

Answer:

C. (-13, -7)

Step-by-step explanation:

The location of a point O(x, y) that divides a line AB with location A $(x_1,y_1)$ and B $(x_2,y_2)$ in the ratio m:n is given by:

$x=\frac{m}{m+n} (x_2-x_1)+x_1\\\\y=\frac{m}{m+n} (y_2-y_1)+y_1$

Therefore the coordinates of point X That divides line segment from Y(-8, 8) to T(-15, -13) in the ratio 5:2 is:

$x=\frac{5}{5+2} (-15-(-8))+(-8)\\\\x=\frac{5}{7} (-15+8)-8=\frac{5}{7}(-7)-8=-5-8=-13 \\\\\\y=\frac{5}{5+2} (-13-8)+8\\\\y=\frac{5}{7} (-21)+8=5(-3)+8=-15+8=-7$

Therefore the coordinates of point X is at (-13, -7)

6 0

3 years ago

The polynomial of degree 5, P ( x ) has leading coefficient a=1, has roots of multiplicity 2 at x = 3 and x = 0 , and a root of

ser-zykov [4K]

Answer:

$p_{5} (t) = x^{5} - 5\cdot x^{4} + 3\cdot x^{3} +9\cdot x^{2}$ , for $r_{1} = 0$

Step-by-step explanation:

The general form of quintic-order polynomial is:

$p_{5}(t) = a\cdot x^{5} + b\cdot x^{4} + c\cdot x^{3} + d\cdot x^{2} + e \cdot x + f$

According to the statement of the problem, the polynomial has the following roots:

$p_{5} (t) = (x - r_{1})\cdot (x-3)^{2}\cdot x^{2} \cdot (x+1)$

Then, some algebraic handling is done to expand the polynomial:

$p_{5} (t) = (x - r_{1}) \cdot (x^{3}-6\cdot x^{2}+9\cdot x) \cdot (x+1)\\p_{5} (t) = (x - r_{1}) \cdot (x^{4}-5\cdot x^{3} + 3 \cdot x^{2} + 9 \cdot x)$