Ask question

Ask question

All categories

3 years ago

12

In the literal equation below, what do C and F represent?

1 answer:

Zolol [24]3 years ago

3 0

Answer: Variables

F and C are placeholders for numbers. They are called variables because they are allowed to vary, or change. If you change C then it affects F, and vice versa. If the values for the placeholders is not allowed to change, yet it holds a number, then it is considered a constant. In this case, we don't have constants or else the formula isn't too useful.

You might be interested in

A scatter plot was made to show the value of certain cars based on the age of the car in years. The equation of the scatter plot

Mashutka [201]

Am not sure but I think it C

6 0

3 years ago

Read 2 more answers

PLEASE HELP!!!!!!!!!!!!!!!!!!!!!

Shalnov [3]

Answer:

D i think sorry if wrong

Step-by-step explanation:

8 0

3 years ago

What is the range of the function f(x) = –2|x + 1|

Oduvanchick [21]

Answer:

$f\left(x\right)\le \:0$

Step-by-step explanation:

$\mathrm{The\:range\:of\:an\:absolute\:function\:of\:the\:form}\:-c|ax+b|+k\:\mathrm{is}\:\:f\left(x\right)\le \:k$

$f(x) = -2|x + 1|+0$

$k=0$

$f\left(x\right)\le \:0$

8 0

3 years ago

Please help me it needs to be turned in tonight

Katarina [22]

Answer:

Step-by-step explanation:

1) x=14 y=40

2)a=10 b=37

3)u=62 v=59

4)s=9 t=14

5)u=66 v=38

hope this helps

6 0

3 years ago

The coordinates of the endpoints of AB and CD are A(2,

Ratling [72]

Answer:

Option 1: CD is a perpendicular bisector of AB

Step-by-step explanation:

Let us find out the slopes of various line segments and the Distances and then we will draw the conclusions accordingly.

Formula to find slope

$m= \frac{y_2-y_1}{x_2-x_1}$

Formula to Find Distance between two points

$D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

mAB ( represents , Slope of AB )

1. $mAC= \frac{3-2}{2-5}=\frac{1}{-3}=-\frac{1}{3}$

2. $mBC=\frac{2-1}{5-8}=\frac{1}{-3}=-\frac{1}{3}$

3. $mCD=\frac{5-2}{6-5}=\frac{3}{1}=3$

4. $AC=\sqrt{(3-2)^2+(2-5)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}$

5. $BC=\sqrt{(2-1)^2+(5-8)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}$

mAC = mBC , and C is common point , hence these three are collinear points making a straight line whole slope is $-\frac{1}{3}$

$mAB=-\frac{1}{3}$

$mCD=3$

$mAB \times mCD = -\frac{1}{3} \times 3 = -1$

Hence CD ⊥ AB

Also

From Point 4 and point 5 above , we see that

AC = CB

Hence CD bisect AB at C, also CD ⊥ AB

There fore

CD is a perpendicular bisector of AB

Therefor option 1 is true

4 0

3 years ago