All categories

2 years ago

Select the equation of the line that passes through the point (3, 5) and is perpendicular to the line x = 4. (2 points)

1 answer:

7 0

Perpendicular lines have opposite reciprocal slopes.

x = 4 has a slope of undefined since it is a vertical line.

It's undefined because 1/0 = undefined. We can't divide by 0.

The opposite reciprical of 1/0 is -0/1, which equals to 0.

Any line with a slope of 0 is:

y = 0x + b
y = b

This line goes through the point of (3, 5).

Since the equation starts with y = something, we have to use the y-value of the point it goes through. The point is in the form of (x, y).

5 is in the y-value.

The equation of the perpendicular line is y = 5.

You might be interested in

A. 8pi B. pi C. 2pi D. 4pi

mestny [16]

Answer: Option C

$AC = 2\pi$

Step-by-step explanation:

The arc length is calculated as

$L = \theta * R$

Then

$AC = \theta * R$

We know that

$BC = 24\ ft$

If BC is the diameter of the circumference then the radius R is:

$R = \frac{BC}{2}$

$R = \frac{24}{2}$

$R = 12\ ft$

Now we convert the anglo from degrees to radians $\theta= 30\° * \frac{\pi}{180\°}\\\\\theta=\frac{1}{6}\pi$

Finally

$AC = \frac{1}{6}\pi * 12$

$AC = 2\pi$

7 0

2 years ago

Read 2 more answers

In a research article, you find that r is reported to be 4.8. How would you interpret this finding?

borishaifa [10]

Answer:

...............

Step-by-step explanation:

3 0

3 years ago

Please help with the screenshot

oksian1 [2.3K]

Answer:

x = 10.73

Step-by-step explanation:

$\sin(59) = \frac{9.2}{x}$

$x = \frac{9.2}{ \sin(59) } = 10.73$

7 0

2 years ago

Read 2 more answers

Detmermine the best method to solve the following equation, then solve the equation. (3x-5)^2=-125

liq [111]

For the given equation;

$(3x-5)^2=-125$

We shall begin by expanding the parenthesis on the left side, after which we would combine all terms on and move all of them to the left side, which shall yield a quadratic equation. Then we shall solve.

Let us begin by expanding the parenthesis;

$\begin{gathered} (3x-5)^2\Rightarrow(3x-5)(3x-5) \\ (3x-5)(3x-5)=9x^2-15x-15x+25 \\ (3x-5)^2=9x^2-30x+25 \end{gathered}$

Now that we have expanded the left side of the equation, we would have;

$\begin{gathered} 9x^2-30x+25=-125 \\ \text{Add 125 to both sides and we'll have;} \\ 9x^2-30x+25+125=-125+125 \\ 9x^2-30x+150=0 \end{gathered}$

We shall now solve the resulting quadratic equation using the quadratic formula as follows;

$\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{Where;} \\ a=9,b=-30,c=150 \\ x=\frac{-(-30)\pm\sqrt[]{(-30)^2-4(9)(150)}}{2(9)} \\ x=\frac{30\pm\sqrt[]{900-5400}}{18} \\ x=\frac{30\pm\sqrt[]{-4500}}{18} \\ x=\frac{30\pm\sqrt[]{-900\times5}}{18} \\ x=\frac{30\pm\sqrt[]{-900}\times\sqrt[]{5}}{18} \\ x=\frac{30\pm30i\sqrt[]{5}}{18} \\ \text{Therefore;} \\ x=\frac{30+30i\sqrt[]{5}}{18},x=\frac{30-30i\sqrt[]{5}}{18} \\ \text{Divide all through by 6, and we'll have;} \\ x=\frac{5+5i\sqrt[]{5}}{3},x=\frac{5-5i\sqrt[]{5}}{3} \end{gathered}$

ANSWER:

$x=\frac{5+5i\sqrt[]{5}}{3},x=\frac{5-5i\sqrt[]{5}}{3}$

3 0

3 months ago

Two adjacent supplementary angles are

mars1129 [50]

Adds up to 180 degrees. For two angles to be non-supplementary the must sum to anything other than 180 degrees, to be non-adjacent they can't share a common side (Ray)

3 0

2 years ago

Read 2 more answers