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3 years ago

Write an equation in slope-intercept form for a line that is (a) parallel (b) perpendicular to

Mathematics

1 answer:

ollegr [7]3 years ago

4 0

Answer:

A) parallel

It has the same slope.

y-intercept is 6.

So,

m = -5

b = + 6

y = -5x + 6

B) perpendicular

It has a negative reciprocal of the slope.

-5 = 1/5

y-intercept is 6

m = 1/5

b = + 6

y = 1/5x + 6

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Evaluate the expression Show your work b) -10 -5h for h = -6

vazorg [7]

Answer:

Step-by-step explanation:

-10-5x(-6)

-10+30

4 0

3 years ago

Find the surface area of the pyramid. 64in^2 56in^2 88in^2 118in^2

ziro4ka [17]

Answer:

To find the surface area of this pyramid, we must find the area of one triangle flap which we then multiply by 4 because there are 4 of those equal triangle flaps surrounding the base(square). Once we find the area of those triangles, we then find the base(square)'s area. Next, we add all those areas together to get the result of the surface area.

Formula for area of triangle;

A = BH x 1/2

Where 'B' represents the base, 'H' represents the height, and 1/2 is just dividing the product of those two lengths by 2.

Plug in what you know, given that the base of one triangle is 4, and the height is 6.

A = 4(6) x 1/2

A = 24 x 1/2

A = 24/2

A = 12, the area of one triangle is 12 inches.

Now, we find the area of all the triangles by multiplying the area of one triangle by 4.

12 = Area of triangle

4 triangles in total, so:

12(4)

= 48 inches is the area of all the triangles.

Now we find the area of the base(square) using the formula;

(vol of square formula)
A = $s^{2}$

Where 's' represents one side of the square which is being squared.

A = l · w

Where 'l' represents the length and 'w' represents the width.

Plug in what you know, given that one side of the square is 4 inches (as well as the length and width).

(I'll be using the formula $s^{2}$ )

A = $s^{2}$

A = 4^2 ← (The symbol ' ^ ' means raised to the power of.)

A = 16, the area of the base(square) is 16 inches.

Now we add both areas together:-

48 + 16

= 64 inches^2 is the surface area, your answer is A.

5 0

2 years ago

Equation of a line through (-1,-10) which is parallel to the line y=4x+3 has (need the slope and y intercept ) please help

valina [46]

Parallel lines have the same slope but different y intercepts

Slope: 4
y=4x+b
this is the new equation, plug in the point
-10=4(-1)+B
-10=-4+B
add 4
-6=B

Y intercept= -6
slope; 4

8 0

3 years ago

How to find the vertex calculus 2What is the vertex, focus and directrix of x^2 = 6y

son4ous [18]

Solution:

Given:

$x^2=6y$

Part A:

The vertex of an up-down facing parabola of the form;

$\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}$

Rewriting the equation given;

$\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\ \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\ \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\ \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\ \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}$

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

$\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}$

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

$\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}$

Therefore, the focus is;

$(0,\frac{3}{2})$

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

$\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}$

Therefore, the directrix is;

$y=-\frac{3}{2}$

3 0

11 months ago

Find the volume of the figure. A 1,053 M^3 B 2,106 M^3 C 513 M^3 D 58,5 M^3

vampirchik [111]

The answer is )A
OK THANK YOU I HOPE THIS HOPED

7 0

2 years ago