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3 years ago

Determine whether each equation defines y as function of x, 2x+y=8

Mathematics

1 answer:

solong [7]3 years ago

6 0

Answer:

Yes. This equation defines y as function of x.

Step-by-step explanation:

Step 1:

Solve the equation for y (this puts the equations in terms of y)

2x+y=8 becomes

y = 2x - 8

This is a linear equation, which is a funtion.

*Another way to solve this is to look at the graph. Graphing this equation gives us a line. Use the vertical line test to see if it's a function.

To pass the vertical line test (VLT), there can only be one y value to each x value along the graph. If it passes the VLT, the equation is a function.

If it fails the VRT, the equation is not a function.

You might be interested in

Classify each number as an integer or not -32/4 yes or no -27/9 yes or no 5/11 yes or no -73.27 yes or no -36 yes or no

muminat

Answers:

-32/4 = yes
-27/9 = yes
5/11 = no
-73.27 = no
-36 = yes

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

Explanation:

Use long division or a calculator to find that -32/4 turns into -8. This is an integer because it's a negative whole number.

The same goes for -27/9 because it turns into -3

The value -36 is also an integer as well

Something like 5/11 is not an integer because 5/11 = 0.4545 approximately which isn't a positive whole number. Integers are either positive or negative whole numbers, or 0.

For similar reasons, we can see that -73.27 isn't an integer either because it's not a negative whole number.

3 0

2 years ago

The annual inventory cost C for a manufacturer is given below, where Q is the order size when the inventory is replenished. Find

shtirl [24]

Answer:

Change in annual cost is 1.63 (decreasing).

Instantaneous rate of change is 1.65 (decreasing).

Step-by-step explanation:

Given that,

$C=\dfrac{1017000}{Q}+6.8\:Q$

The annual change in cost is given by,

$Change\:in\:C=C\left(348\right)-C\left(347\right)$

Calculate the value of cost at Q = 347 and Q =348 by substituting the value in cost function.

Calculating value of C at Q=347,

$C=\dfrac{1017000}{347}+6.8\left(347\right)$

$C=5290.44$

Calculating value of C at Q=348,

$C=\dfrac{1017000}{348}+6.8\left(348\right)}$

$C= 5288.81$

Substituting the value,

$Change\:in\:C=5288.81-5290.44$

$Change\:in\:C=-1.63$

Negative sign indicate that there is decrease in annual cost when Q is increased from 347 to 348

Therefore, change is annual cost is 1.63 (decreasing).

Instantaneous rate of change is given by the formula,

$f’\left(x\right)=\lim_{h\rightarrow 0}\dfrac{f\left(x+h\right)-f\left(x\right)}{h}$

Rewriting,

$C’\left(Q\right)=\lim_{h\rightarrow 0}\dfrac{C\left(Q+h\right)-C\left(Q\right)}{h}$

Now calculate \dfrac{C\left(Q+h\right) by substituting the value Q+h in cost function,

$C\left(Q+h\right)= \dfrac{1017000}{Q+h}+6.8\left(Q+h\right)$

Therefore,

$C’\left(Q\right)= \lim _{h\to 0}\:\dfrac{\dfrac{1017000}{Q+h}+6.8\left(Q+h\right)-\left(\dfrac{1017000}{Q}+6.8Q\right)}{h}$

By using distributive law,

$C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{1017000}{Q+h}+6.8Q+6.8h-\dfrac{1017000}{Q}-6.8Q}{h}$

Cancelling out common factors,

$C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{1017000}{Q+h}-\dfrac{1017000}{Q}+6.8h}{h}$

Now, LCD of $\left(Q+h\right)$ and $Q$ is $Q(Q+h)$ . So multiplying first term by $Q$ and second term by

Therefore,

$C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{1017000}{Q+h}\times\dfrac{Q}{Q}-\dfrac{1017000}{Q}\times\dfrac{Q+h}{Q+h}+6.8h}{h}$

Simplifying,

$C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{1017000Q}{\left(Q+h\right)\left(Q\right)}-\dfrac{1017000\left(Q+h\right)}{Q\left(Q+h\right)}+6.8h}{h}$

$C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{1017000Q-1017000\left(Q+h\right)}{\left(Q+h\right)\left(Q\right)}+6.8h}{h}$

$C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{1017000Q-1017000Q-1017000h}{\left(Q+h\right)\left(Q\right)}+6.8h}{h}$

$C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{-1017000h}{\left(Q+h\right)\left(Q\right)}+6.8h}{h}$

Factoring out h from numerator,

$C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{h\left(\dfrac{-1017000}{\left(Q+h\right)\left(Q\right)}+6.8\right)}{h}$

Cancelling out h,

$C'\left(Q\right) = \lim_{h\to 0}\: -\dfrac{1017000}{\left(Q+h\right)Q}+6.8$

Calculating the limit by plugging value h = 0,

$C'\left(Q\right) = -\dfrac{1017000}{\left(Q+0\right)Q}+6.8$

$C'\left(Q\right) = -\dfrac{1017000}{Q^{2}}+6.8$

Given that Q=347,

$C'\left(347\right) = -\dfrac{1017000}{347^{2}}+6.8$

$C'\left(347\right) = -1.65$

Negative sign indicate that there is decrease in instantaneous rate when Q is 347

Therefore, instantaneous rate of change at Q=347 is 1.65 (decreasing).

4 0

2 years ago

PLEASE HELP!!!! Find m

astra-53 [7]

Answer:

I couldn't find it its confusing

3 0

2 years ago

Solve for t(advanced rational equations)

Mariulka [41]

t² -7 + 12t⁻² = 0

Multiply each side by t² :

t⁴ -7t² + 12 = 0

This is a quadratic equation in the variable ' t² '.
For just a moment, to avoid confusion, let U = t².
Then the equation is

U² - 7U + 12 = 0
whence
(U - 4) (U - 3) = 0
and
U = 4 and U = 3 .

Now we can go back to ' t² ' in place of U :

t² = 4
t = +2 and t = -2

t² = 3
t = +√3 and t = -√3

5 0

2 years ago

Read 2 more answers

The spinner is spun twice. whats the probability of the arrow landing on three and then an odd number?

GenaCL600 [577]

Answer:

A lot

Step-by-step explanation:

Spin it
See what it lands on

8 0

2 years ago

Determine whether each equation defines y as function of x, 2x+y=8​

Determine whether each equation defines y as function of x, 2x+y=8