Convert 0.23cc into liters ?

Mathematics

2 answers:

suter [353]3 years ago

7 0

You just divide by 1000, you get 0.00023

Feliz [49]3 years ago

4 0

There are 1,000 cubic centimeters in each liter so you would divide 0.23 cc by 1,000 to get 0.00023.

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Text the answer 770-765-3026

Andre45 [30]

Answer:

A proportional relationship is there.

Step-by-step explanation:

Here in the table, the values of y are given corresponding to the values of x.

There are four sets of values of x and y.

We have to check whether the relationship is proportional or not.

Now, rate of change of y with respect to x will be from the first two pair of values is = $\frac{3.5 - (-1)}{-2 - (-8)} =0.75$

Again, rate of change of y with respect to x will be from the second two pair of values is = $\frac{12.5 - 3.5}{10 - (-2)} =0.75$

And, rate of change of y with respect to x will be from the third two pair of values is = $\frac{20-12.5}{20 - 10} =0.75$

So, the rate is always 0.75.

Therefore, a proportional relationship is there. (Answer)

6 0

3 years ago

Read 2 more answers

-16t squared + 16t + 480 t=time after jumping in seconds h=height in feet How many seconds did did it take her to hit the water

Agata [3.3K]

That is height
h=16t^2+16t+480
when h=0 is when it hits the water

0=16t^2+16t+480
undistribbute 16 for ease
0=16(t^2+t+30)
16*0=0 so assume
t^2+t+30=0
factor
(t+6)(t-5)=0
set to zero

t+6=0
minus 6
t=-6
false, since you cannot have negative time

t-5=0
add 5
t=5

at t=5 she hits the water

5 seconds

4 0

3 years ago

Find the equation in standard form of the line passing through the points 3,-4 and 5,1

aliina [53]

Answer:

5x - 2y - 23 = 0

Step-by-step explanation:

Line is passing through the points $(3,\:-4)= (x_1, \:y_1) \: \&\: (5,\: 1)= (x_2,\:y_2)$

Equation of line in two point form is given as:

$\frac{y -y_1 }{y_1 -y_2 } = \frac{x -x_1 }{x_1 -x_2 } \\ \\ \therefore \: \frac{y -( - 4) }{ - 4 -1} = \frac{x -3 }{3 -5 } \\ \\ \therefore \:\frac{y + 4 }{ - 5} = \frac{x -3 }{ - 2 } \\ \\ \therefore \: \frac{y + 4 }{ 5} = \frac{x -3 }{ 2 } \\ \\ \therefore \: 2(y + 4) = 5(x - 3) \\ \therefore \: 2y + 8 = 5x - 15 \\ \therefore \: 5x - 15 - 2y - 8 = 0 \\ \red{ \boxed{ \bold{\therefore \: 5x - 2y - 23 = 0}}} \\ is \: the \: required \: equation \: of \: line \: in \: \\ standard \: form.$