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Vlada [557]
4 years ago
12

What the dependant, independent, and equation is

Mathematics
1 answer:
Anestetic [448]4 years ago
7 0

Answer:

Dependent Value: Cost (C)

Independent Value: Bushels (B)

Step-by-step explanation:

You can make the letter D out of the side of the bar graph and the letter I on the bottom, this tells you the label on the side is the dependent and the label on the bottom is the independent. If you dont have a line graph then think about this sentence starter: ...... effects.......

Quote to make you happy:

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What is the probability of spinning a 2 on a spinner that has 8 equal sides numbered 1-8 and rolling a 3 on a die
BlackZzzverrR [31]

Answer:

1/48

Step-by-step explanation:

To find the probability of 2 independent events, we multiply their probabilities together.


Probability of getting a 2 on the spinner, 1 of 8 spots

1/8  =

Probability of getting a 3 on the die, 1 of 6

1/6 =


Multiply these probabilities together

1/8 * 1/6 = 1/48

7 0
4 years ago
-6= b/18 please help me solve it
irina [24]
-6 = \frac{b}{18}   Multiply both sides by 18
-108 = b   Switch the sides to make it easier to read
b = -108
8 0
4 years ago
PLEASE HELP! WILL UPVOTE!
hram777 [196]
Standard form is
ax+by=c
where a and b and c are integers
and a is normally positivie

y+1/5=3x
minus 3x both sides
-3x+y+1/5=0
minus 1/5 both sides
-3x+y=-1/5
times -5 both sides
15x-5y=1
not listed
8 0
4 years ago
12 ft
lara31 [8.8K]

Answer:

942.48

Step-by-step explanation:

hope this helps

3 0
2 years ago
Find the limit............​
liubo4ka [24]

Answer: -\frac{9}{x^2}

====================================================

Work Shown:

\displaystyle \lim_{\Delta x \to 0^{+}} \frac{\frac{9}{x+\Delta x} - \frac{9}{x}}{\Delta x}\\\\\\\displaystyle \lim_{\Delta x \to 0^{+}} \frac{\frac{9x}{x(x+\Delta x)} - \frac{9(x+\Delta x)}{x(x+\Delta x)}}{\Delta x}\\\\\\\displaystyle \lim_{\Delta x \to 0^{+}} \frac{\frac{9x-9(x+\Delta x)}{x(x+\Delta x)}}{\Delta x}\\\\\\\displaystyle \lim_{\Delta x \to 0^{+}} \frac{9x-9x-9\Delta x}{x\Delta x(x+\Delta x)}\\\\\\

\displaystyle \lim_{\Delta x \to 0^{+}} \frac{-9\Delta x}{x\Delta x(x+\Delta x)}\\\\\\\displaystyle \lim_{\Delta x \to 0^{+}} \frac{-9}{x(x+\Delta x)}\\\\\\\displaystyle \frac{-9}{x(x+0)}\\\\\\\displaystyle -\frac{9}{x^2}\\\\\\

The trick here is to first simplify the expression so that the \Delta x term in the very lower denominator cancels out. This is to avoid dividing by zero. Once this division by zero error is avoided, we can replace the delta x term with 0 and evaluate the limit as \Delta x approaches 0 from the right or positive direction.

5 0
3 years ago
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