A fence is 7 feet 6inches high 1 foot =12 INCHES WHAT IS THE HIEGHT OF THE FENCE IN CENTIMETRES

Which equation is written in slope-intercept form? Oy= 3x-2 O x=y+8 0 2x+ 5y = 12 O 3y = 8x-5

maw [93]

Answer:

Step-by-step explanation:

slope intercept form is

y-mx+b where m is the slope, b-is the y-intercept

y=3x-2

8 0

3 years ago

If AB = 15, BC = 12, and CA = 8, list the angles of abc in order from smallest to largest.

SIZIF [17.4K]

<span>angles of abc in order from smallest to largest.
<ABC, <CAB, <ACB

Hope it helps

p.s.
smallest angle opposites shortest side
largest angle opposites longest side</span>

6 0

3 years ago

Read 2 more answers

Gimme the answer because I have no idea how to do this it makes no sense to me

kati45 [8]

Answer:

C, D, B

Step-by-step explanation:

the mode is if a number is repeated more than one time

EX. 15, 23, 15, 23, 15

C because 15 is repeated 3 times and 19 is too.

D because 42 is repeated 2 times and so is 18.

B because 87 is repeated 2 times and 32 is too.

5 0

3 years ago

If K is the midpoint of HJ, HK = x+6, and HJ =4x-6, then KJ = ?

barxatty [35]

4x-6= x+6
4x= x + 12
5x= 12
X= 12/5
X= 2.5

6 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,

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

Since the parabola is symmetric around the y-axis, the 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

1 year ago