All categories

1 year ago

5p + 9c = p for c Help out with this problem

Mathematics

2 answers:

maria [59]1 year ago

6 0

SOLVING

First you need to subtract by 5P both sides

9c=p-5p

and combine like terms.

9c=-4p

Next divide 9 to both sides

$c=\frac{-4p}{9}$

$c=-\frac{4p}{9}$

Learn more;work harder

#Carryonlearning

Rom4ik [11]1 year ago

5 0

Answer:

Step-by-step explanation:

5p + 9c = p

9c = p - 5p

9c = -4p

c = -4p/9

c= $\frac{-4p}{9}$

You might be interested in

One half of the sum of 3 and q

ivann1987 [24]

Here you go let me know if you have questions

7 0

3 years ago

Find the distance between the two points to the nearest tenth. (0,-5) and (-4,2) 5 5.3 5.2 5.6

zubka84 [21]

Sqr (x2 - x1)^2 + (y2 - y1)^2
Sqr (-4-0)^2 + (2+5)^2
Sqr (16) + 49 = sqr 65
I got around 8.
(I might did something wrong, idk)

7 0

2 years ago

Can Someone Answer My Final Answers :) I’d appreciate it so much :)

DaniilM [7]

1. Remember that the perimeter is the sum of the lengths of the sides of a figure.To solve this, we are going to use the distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
where
$(x_{1},y_{1})$ are the coordinates of the first point
$(x_{2},y_{2})$ are the coordinates of the second point
Length of WZ:
We know form our graph that the coordinates of our first point, W, are (1,0) and the coordinates of the second point, Z, are (4,2). Using the distance formula:
$d_{WZ}= \sqrt{(4-1)^2+(2-0)^2}$
$d_{WZ}= \sqrt{(3)^2+(2)^2}$
$d_{WZ}= \sqrt{9+4}$
$d_{WZ}= \sqrt{13}$

We know that all the sides of a rhombus have the same length, so
$d_{YZ}= \sqrt{13}$
$d_{XY}= \sqrt{13}$
$d_{XW}= \sqrt{13}$

Now, we just need to add the four sides to get the perimeter of our rhombus:
$perimeter= \sqrt{13} + \sqrt{13} + \sqrt{13} + \sqrt{13}$
$perimeter=4 \sqrt{13}$
We can conclude that the perimeter of our rhombus is $4 \sqrt{13}$ square units.

2. To solve this, we are going to use the arc length formula: $s=r \alpha$
where
$s$ is the length of the arc.
$r$ is the radius of the circle.
$\alpha$ is the central angle in radians

We know form our problem that the length of arc PQ is $\frac{8}{3} \pi$ inches, so $s=\frac{8}{3} \pi$ , and we can infer from our picture that $r=15$ . Lest replace the values in our formula to find the central angle POQ:
$s=r \alpha$
$\frac{8}{3} \pi=15 \alpha$
$\alpha = \frac{\frac{8}{3} \pi}{15}$
$\alpha = \frac{8}{45} \pi$

Since $\alpha =POQ$ , We can conclude that the measure of the central angle POQ is $\frac{8}{45} \pi$

3. A cross section is the shape you get when you make a cut thought a 3 dimensional figure. A rectangular cross section is a cross section in the shape of a rectangle. To get a rectangular cross section of a particular 3 dimensional figure, you need to cut in an specific way. For example, a rectangular pyramid cut by a plane parallel to its base, will always give us a rectangular cross section.
We can conclude that the draw of our cross section is: