What is the slop for run:(4,5) rise: (7,5)

Solve the equation p=2q + 2w for q.

elena-s [515]

Answer:

q = 1 /2 p − w

Step-by-step explanation:

4 0

4 years ago

Lcm of 2/5 and 5/18?

SSSSS [86.1K]

Answer:

10

Step-by-step explanation:

4 0

3 years ago

Which equations represent circles that have a diameter of 12 units and a center that lies on the y-axis

Tju [1.3M]

The equations representing the circle having a diameter of 12 units and a center lying on y-axis are x² + (y – 3)² = 36, and x² + (y + 8)² = 36, making options 1, and 5 the two choices.

In the question, we are looking for the equations of the circle having a diameter of 12 units and a center lying on y-axis.

The standard form of the equation of a circle is (x - h)² + (y - k)² + r², where (h, k) is the center of the circle, and r is its radius.

A circle having a diameter of 12 units, will have a radius of 12/2 = 6 units. Thus, r = 6.

A circle having the center on the y-axis will have a center as (0, k) as all points on the y-axis have x-coordinate 0.

Thus, the equation of the circle, with a diameter of 12 units and the center lying on the y-axis takes the form:

x² + (y - k)² = 6², or, x² + (y - k)² = 36.

From the given options, the equations x² + (y – 3)² = 36, and x² + (y + 8)² = 36, are of this form.

Thus, the equations representing the circle having a diameter of 12 units and a center lying on y-axis are x² + (y – 3)² = 36, and x² + (y + 8)² = 36, making options 1, and 5 the two choices.

Learn more about equations of circle at

brainly.com/question/16039879

#SPJ4

The provided question is incomplete. The complete question is:

"Which equations represent circles that have a diameter of 12 units and a center that lies on the y-axis? Select two options.

x² + (y – 3)² = 36
x² + (y – 5)² = 6
(x – 4)² + y² = 36
(x + 6)² + y² = 144
x² + (y + 8)² = 36."

8 0

2 years ago

PLEASE HELP ME.....

Virty [35]

Answer:

They would meet each other at $01 : 24$ PM.

Step-by-step explanation:

Erik took a trip to see his friend Mike who lives 308 miles away.

He left his place at 10 AM driving at 70 mph.

In 2 hours, his friend Mike left his place driving towards Erik at an average speed of 50 mph.

As Erik left his house at 10 AM driving at 70 mph and in 2 hours, his friend Mike left his place driving towards Erik at an average speed of 50 mph. It means

Erik left his place at 12:00 PM noon.

so, at 12:00 PM Erik had already traveled:

$2\times 70=140\:\:miles$

Miles left $=308-140=168\:\:miles$

Let 't' be the time when they meet

so

$70t\:+\:50t\:=168$

$120t=168$

$\mathrm{Divide\:both\:sides\:by\:}120$

$\frac{120t}{120}=\frac{168}{120}$

$t=\frac{7}{5}$

so

$t=\frac{7}{5}=1.4\:\:hours$

or $1$ hour and $24$ minutes after $12:00$ noon

i.e.

$12\:+\:1.4\:=1\:O'clock\:+\:24\:minutes$

Therefore, they would meet each other at $01 : 24$ PM.

7 0

4 years ago

Help!!!!!!!!!!!!!!!!!!!!

lapo4ka [179]

Well... the segment FG = 4x + 3, the segment GH = 7x - 12, now, the

segment FH = FG + GH, those two fellows added up, and we know that FG and GH are both equal halves because G is the midpoint.

thus, FG = GH
or

$\bf 4x+3=7x-12\implies 3+12=7x-4x\implies 15=3x
\\\\\\
\cfrac{15}{3}=x\implies \boxed{5=x}\\\\
-------------------------------\\\\
thus\qquad 
\begin{cases}
\overline{FG}=4x+3\\
4(5)+3\\
23\\
------\\
\overline{GH}=7x-12\\
7(5)-12\\
23
\end{cases}\impliedby equal\ halves
\\\\\\
\overline{FH}=23+23\implies \overline{FH}=46\implies 6y-2=46
\\\\\\
6y=46+2\implies y=\cfrac{48}{6}$

and fairly sure, you know how much that is.

7 0

3 years ago