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3 years ago

Which of the following is NOT a solution to 2x+3y=12? A. (5,-3) B. (9,-2) C. (0,4) D. (6,0)

Mathematics

1 answer:

11Alexandr11 [23.1K]3 years ago

3 0

Answer:

Step-by-step explanation

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Calculus 3 chapter 16

o-na [289]

Evaluate $\vec F$ at $\vec r$ :

$\vec F(x,y,z) = x\,\vec\imath + y\,\vec\jmath + xy\,\vec k \\\\ \implies \vec F(\vec r(t)) = \vec F(\cos(t), \sin(t), t) = \cos(t)\,\vec\imath + \sin(t)\,\vec\jmath + \sin(t)\cos(t)\,\vec k$

Compute the line element $d\vec r$ :

$d\vec r = \dfrac{d\vec r}{dt} dt = \left(-\sin(t)\,\vec\imath+\cos(t)\,\vec\jmath+\vec k\bigg) \, dt$

Simplifying the integrand, we have

$\vec F\cdot d\vec r = \bigg(-\cos(t)\sin(t) + \sin(t)\cos(t) + \sin(t)\cos(t)\bigg) \, dt \\ ~~~~~~~~= \sin(t)\cos(t) \, dt \\\\ ~~~~~~~~= \dfrac12 \sin(2t) \, dt$

Then the line integral evaluates to

$\displaystyle \int_C \vec F\cdot d\vec r = \int_0^\pi \frac12\sin(2t)\,dt \\\\ ~~~~~~~~ = -\frac14\cos(2t) \bigg|_{t=0}^{t=\pi} \\\\ ~~~~~~~~ = -\frac14(\cos(2\pi)-\cos(0)) = \boxed{0}$

3 0

1 year ago

Dave has 12$ to spend on a 7 dollar book and two birthday cards (c) for his friends how much can he spend on each card if he buy

anzhelika [568]

Answer:

2.50

Step-by-step explanation:

So if you have a budget of 12 dollars, and the book is 7 dollars, subtract 12-7 and you get 5 dollars. And half of 5 dollars is 2.50. And if there is 2 cards, then that means that 1 card is 2.50

6 0

3 years ago

In a particular classroom, a student who did not study scored 60 points on a test. If that student scored 90 points when studyin

Oksanka [162]

30 points over 3 hours equals: 10 points/hr

5 0

2 years ago

How many possible committees of 4 people can be chosen from 15 men and 20 women so that at least two women should be on each com

Mnenie [13.5K]

If there has to be 2 men and 2 women, we know that we must take a group of 2 men out of the group of 15 men and a group of 2 women out of the group of 20 women. Therefore, we have:

(15 choose 2) x (20 choose 2)

(15 choose 2) = 105

(20 choose 2) = 190

190*105 = 19950

Therefore, there are 19950 ways to have a group of 4 with 2 men and 2women.

If there has to be 1 man and 3 women, we know that we must take a group of 1 man out of the group of 15 men and a group of 3 women out of the group of 20 women. Therefore, we have:

(15 choose 1) x (20 choose 3)

(15 choose 1) = 15

(20 choose 3) = 1140

15*1140 = 17100

Therefore, there are 17100 ways to have a group of 4 with 3 women and 1 man.

We now find the total outcomes of having a group with 4 women.

We know this is the same as saying (20 choose 4) = 4845

Therefore, there are 4845 ways to have a group of 4 with 4 women.

We now add the outcomes of 2 women, 3 women, and 4 women and get the total ways that a committee can have at least 2 women.

19950 + 17100 + 4845 = 41895 ways that there will be at least 2 women in the committee

5 0

2 years ago

5 x 8 + 6 / 6 - 12 x 2

Whitepunk [10]

Yes, you did it right. That is the correct answer! Nice job!

4 0

2 years ago