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3 years ago

Is (-2,4) a solution of the graphed inequality?

Mathematics

2 answers:

k0ka [10]3 years ago

6 0

we need the equation or graph to figure that out

Mariana [72]3 years ago

5 0

You need a graph to get this question ..

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PLEASE HELP!! ABD and DBC are complementary angles. 70, what's the measurement of x?

Studentka2010 [4]

It is 45 because complementary means 90

6 0

3 years ago

Read 2 more answers

Find the counterclockwise circulation and outward flux of the field Fequals7 xy Bold i plus 2 y squared Bold j around and over t

77julia77 [94]

Answer:

The counterclockwise circulation is $\frac{7}{12}$ and the outward flux is $\frac{11}{15}$

Step-by-step explanation:

We are given the field $F(x,y) = (7xy,2y^2)$ . A picture of the region and the path we are considering is attached. Recalll the following theorems.

Given a field of the form F(x,y)=(f(x,y),g(x,y) with f,g having continous partial derivates, C is a closed path counterclockwise oriented, R is the region enclosed by C and n is the normal vector pointing outwards of the path C. Then

$\oint_C F\cdot dr =\iint_R \frac{\partial f}{dy}- \frac{\partial g}{dx} dA$ (this one calculates the counterclockwise circulation)

$\oint_C F\cdot n ds =\iint_R (\frac{\partial f}{dx}+ \frac{\partial g}{dy} dA$ (This one calculates the outward flux)

Then, recall that in our case f(x,y) = 7xy, g(x,y)=2y^2[/tex]. Then

$\frac{\partial f}{dx} = 7y,\frac{\partial f}{dy} = 7x$

$\frac{\partial g}{dx}=0, \frac{\partial g}{dy} = 4y$ .

Note that we just need to describe our region R. The region R lies between the parabola y=x^2 and the line y=x. Thus, one way to describe the region is as follows $0\leq x \leq 1, x^2\leq y \leq x$ . Then, using the previous results, we get that

$\oint_C F\cdot dr =\int_{0}^{1}\int_{x^2}^{x}7x-0dydx = 7\int_{0}^1x(x-x^2)dx = 7 \left.(\frac{x^3}{3}-\frac{x^4}{4})\right|_{0}^1 = 7(\frac{1}{3}-\frac{1}{4}) = \frac{7}{12}$ (circulation)

$\oint_C F\cdot n ds=\int_{0}^{1}\int_{x^2}^{x}7y+4ydydx = \frac{11}{2}\int_{0}^1\left.y^2\right_{x^2}^{x}dx = \frac{11}{2}\int_{0}^{1}x^2-x^4 dx = \frac{11}{2}\left(\frac{x^3}{3}-\frac{x^5}{5})\right|_{0}^{1}=\frac{11}{2}(\frac{1}{3}-\frac{1}{5})=\frac{11}{15}$ (flux)

5 0

3 years ago

What is the fair share?

andriy [413]

1 whole should be the answer

8 0

3 years ago

Read 2 more answers

Find the value of f(3) for the function. f(a)=2(a+4)-3 f(3)=___ (simplify your answer)

Nonamiya [84]

Answer:

<h3>Given => f(a)=2(a+4)-3</h3><h3>To Attain=> The value of f(3) for the function</h3>

Step-by-step explanation:

• By Putting the value of a = 3

•f(a)=2(a+4)-3

$f(3) = 2(3 + 4) - 3 \\ f(3) = 2 \times 7 - 3 \\ f(3) = 14 - 3 \\ f(3) = 11$

• Hence, 11 is the solution to the function

3 0

3 years ago

Read 2 more answers

In city A, the temperature rises 9° F from 8am then the temperature drops 8°F from 9am to 10am in city B the temperature drops 1

oksian1 [2.3K]

Answer:

For City A, the expression for temperature change is 9 + (-8)

The amount the temperature changes for City A = 1°F

For City B, the expression for temperature change is -1 + (-3) = -4°F

The amount the temperature changes for City B = -4°F

Step-by-step explanation:

The given information are;

In City A, the temperature rise from 8 am to 9 am = 9°F

In City A, the temperature drop from 9 am to 10 am = 8°F

In City B, the temperature drop from 8 am to 9 am = 1°F

In City B, the temperature drop from 9 am to 10 am = 3°F

The expression for that represents the temperature change from 8 am to 10 am is therefore;

For City A

Temperature change from 8 am to 10 am are

8 am to 9 am = +9°F

9 am to 10 am = -8°F

Sum, change from 8 am to 10 am = 9 + (-8) = 1°F

For City B

Temperature change from 8 am to 10 am are

8 am to 9 am = -1°F

9 am to 10 am = -3°F

Sum, change from 8 am to 10 am = -1 + (-3) = -4°F

4 0

4 years ago