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Solve this simultaneous equation by using substitution method 3x-2y=12 and x+3y=-7

Answers

Answer 1

Answer:

Answer:

Solving this simultaneous equation by using substitution method 3x-2y=12 and x+3y=-7 we get x=2 and y=-3

Solution set = {2,-3}

Step-by-step explanation:

We need to solve this simultaneous equation by using substitution method 3x-2y=12 and x+3y=-7

let:

$3x-2y=12--eq(1)\nx+3y=-7 --eq(2)$

Finding value of x from equation 2 and putting in eq(1)

$From \ eq(2)\nx+3y=-7\nx=-7-3y$

Putting value of x in eq(1)

$3x-2y=12\nPut \ x=-7-3y\n3(-7-3y)-2y=12\nSolving\n-21-9y-2y=12\n-11y=12+21\n-11y=33\ny=(33)/(-11)\ny=-3$

So, we get y=-3

Now put value of y in equation 2 to find value of x

$x+3y=-7\nx+3(-3)=-7\nx-9=-7\nx=-7+9\nx=2$

So, we get value of x =2

So, solving this simultaneous equation by using substitution method 3x-2y=12 and x+3y=-7 we get x=2 and y=-3

Solution set = {2,-3}

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Consider the following function. f ( x ) = 1 − x 2 / 3 Find f ( − 1 ) and f ( 1 ) . f ( − 1 ) = f ( 1 ) = Find all values c in ( − 1 , 1 ) such that f ' ( c ) = 0 . (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) c = Based off of this information, what conclusions can be made about Rolle's Theorem? This contradicts Rolle's Theorem, since f ( − 1 ) = f ( 1 ) , there should exist a number c in ( − 1 , 1 ) such that f ' ( c ) = 0 . This does not contradict Rolle's Theorem, since f ' ( 0 ) = 0 , and 0 is in the interval ( − 1 , 1 ) . This does not contradict Rolle's Theorem, since f ' ( 0 ) does not exist, and so f is not differentiable on ( − 1 , 1 ) . This contradicts Rolle's Theorem, since f is differentiable, f ( − 1 ) = f ( 1 ) , and f ' ( c ) = 0 exists, but c is not in ( − 1 , 1 ) . Nothing can be concluded.

Answers

Answer:

(E)Nothing can be concluded.

Step-by-step explanation:

Given the function $f(x)=1-x^{(2)/(3)}$

$f(-1)=1-(-1)^{(2)/(3)}=1.5-0.86603i\nf(1)=1-1^{(2)/(3)}=1-1=0$

$f'(x)=-(2)/(3)x^{-(1)/(3)}\nf'(x)=-\frac{2}{3\sqrt[3]{x} }$

If the derivative is set equal to zero, the function is undefined.

Nothing can be concluded since $f(1)\neq f(-1)$ and no such c in (-1,1) exists such that $f'(c)=0$

THEOREM

Rolle's theorem states that any real-valued differentiable function that attains equal values at two distinct points must have at least one stationary point somewhere between them—that is, a point where the first derivative is zero.

Final answer:

The function f(x) = 1 - x^2/3 has f(-1) = f(1) = 2/3. The derivative f'(x) = -2x/3 equals zero at x=0, which is in the interval (-1, 1). Therefore, this does not contradict Rolle's Theorem.

Explanation:

The function given is f ( x ) = 1 - x ^ 2 /3. To find the values f(-1) and f(1), we simply substitute these values into the function. Therefore, f(-1) = 1 - (-1) ^ 2 /3 = 1 - 1/3 = 2/3 and f(1) = 1 - 1^2/3 = 2/3. As you can see, f(-1) = f(1).

Now, to find the value 'c' such that f'(c) = 0, first we need to determine the derivative of the function, f'(x) = -2x/3. Setting this equal to zero gives the equation 0 = -2x/3, which has the solution x = 0. Therefore, f'(c) = 0 at c = 0, which is within the interval (-1, 1).

Finally, regarding Rolle's Theorem which states that if a function is continuous on the closed interval [a, b], differentiable on the open interval (a, b), and f(a) = f(b), then there exists at least one c in the interval (a, b) such that f'(c) = 0, our results are consistent with Rolle's Theorem, since f is differentiable, f(-1) = f(1), and a 'c' value exists in the interval (-1, 1) such that f'(c) = 0.

Learn more about Rolle's Theorem here:

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A manufacturing company has two retail outlets. It is known that 40% of all potentialcustomers buy products from outlet 1 alone, 35% buy from outlet 2 alone, and 10% buy fromboth 1 and 2. LetAdenote the event that a potential customer, randomly chosen, buys fromoutlet 1, and letBdenote the event that the customer buys from outlet 2. Suppose a potentialcustomer is chosen at random. For each of the following events, represent the eventsymbolically and then find its probability.Required:
a. The customer buys from outlet 1.
b. The customer does not buy from outlet 2.
c. The customer does not buy from outlet 1 or does not buy from outlet 2.
d. The customer does not buy from outlet 1 and does not buy from outlet 2.

Answers

Answer:

$P(A) = 0.50$

$P(B') = 0.55$

$P(A' U B') = 0.775$

$P(A'B') = 0.275$

Step-by-step explanation:

Given

$P(A\bar B) = 40\%$ --- From outlet 1 alone

$P(\bar A B) = 35\%$ --- From outlet 2 alone

$P(AB) = 10\%$ --- From both

Solving (a): Buys from outlet 1;

This is represented as: P(A) and the solution is:

$P(A) = P(A\bar B) + P(A B)$

$P(A) = 40\% + 10\%$

$P(A) = 50\%$

$P(A) = 0.50$

Solving (b): Does not buy from outlet 2;

This is represented as: P(B'):

First, calculate the probability that the customer buys from 2

$P(B) = P(\bar AB) + P(A B)$

$P(B) = 35\% + 10\%$

$P(B) = 45\%$

$P(B) = 0.45$

Using the complement rule, we have:

$P(B') = 1 - P(B)$

$P(B') = 1 - 0.45$

$P(B') = 0.55$

Solving (c): Does not buy from 1 or does not buy from 2

This is represented as: $P(A' U B')$

And the solution is:

$P(A' U B') = P(A') + P(B') - P(A'B')$

$P(A' U B') = P(A') + P(B') - P(A') * P(B')$

Using complement rule:

$P(A') = 1 - P(A)$

$P(A') = 1 - 0.50$

$P(A') = 0.50$

The equation becomes:

$P(A' U B') = 0.50 + 0.55 - 0.50 * 0.55$

$P(A' U B') = 0.775$

Solving (d): Does not buy from 1 or does not buy from 2

This is represented as:

$P(A'B')$

And it is calculated as:

$P(A'B') = P(A') * P(B)'$

$P(A'B') = 0.50 * 0.55$

$P(A'B') = 0.275$

The attendance at Treeton Middle School was 2,465 on Monday. The attendance was tracked for the next two days. The change in attendance from the previous day is shown in the table.Which equation shows the attendance on Tuesday?

Answers

Answer:

2107.575

Step-by-step explanation:

Final answer:

To find the attendance on Tuesday, we need to add the change in attendance from Monday to Tuesday to the attendance on Monday.

Explanation:

The attendance at Treeton Middle School was 2,465 on Monday. The table shows the change in attendance from the previous day. To find out the attendance on Tuesday, we need to add the change in attendance from Monday to Tuesday to the attendance on Monday.

Change in attendance from Monday to Tuesday: -53
Attendance on Monday: 2,465

Attendance on Tuesday = Attendance on Monday + Change in attendance from Monday to Tuesday
Attendance on Tuesday = 2,465 + (-53) = 2,412

Therefore, the equation that shows the attendance on Tuesday is: Attendance on Tuesday = 2,412.

Learn more about Finding attendance on Tuesday here:

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in a recent year a baseball team paid five times to their ace pitcher as they paid to their shortstop. if they paid $12 million for the two salaries, what did they pay each player ? btw this is something with consecutive