2 years ago

Find the equation of the line that is perpendicular to the given line and passes through the given point.

Mathematics

1 answer:

liberstina [14]2 years ago

4 0

Y= -1/5x-3
Y= -1/5x0-3
Y=-3

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Please help I only have 15 minutes to complete this last one

Natalka [10]

Answer:

the slope of MN= -5-1/3-(-6)=-2/3

two lines are parallel>> slopes are equal

equation of line = (y-y'=m(x-x'))

therefore, y-1= -2/3(x-6)

y-1 = -2/3x + 4

y+2/3x=5

6 0

3 years ago

For which value of x is the figure a rhombus?

zysi [14]

Answer:

D. x = 7

Step-by-step explanation:

The labeled angles are the acute angles of a right triangle, so they are complementary. Their measures add up to 90 deg.

8x + 7 + 3x + 6 = 90

11x + 13 = 90

11x = 77

x = 7

3 0

3 years ago

Read 2 more answers

(X^4)-(45x^2)+324 what are the zeroes? And how many turning points?

dmitriy555 [2]

Zeroes:

We must solve

$x^4-45x^2+324=0$

To do so, we define the auxiliary variable $t=x^2$ . The equation becomes

$t^2-45t+324=0$

The quadratic formula yields the solutions

$t=9,\quad t=36$

Substituting back $t=x^2$ gives

$x^2=9 \iff x=\pm 3,\quad x^2=36 \iff x=\pm 6$

So, the zeroes are -6, -3, 3, 6.

Turning points:

Turning points are points where a function stops being increasing to become decreasing, or vice versa. Since functions are increasing when their first derivative is positive and decreasing when it's negative, turning points are points where the first derivative is zero.

We have

$f(x)=x^4-45x^2+324 \implies f'(x)=4x^3-90x$