Need the answer to x and y !
Need the answer to x and y ! - 1

Answers

Answer 1
Answer:

x = 2 y = 7

Step-by-step explanation:

the explan at the pic


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What’s the complex number in polar form with argument theta between 0 and 2pie?
8+8i

Answers

9514 1404 393

Answer:

  45. (8√2)∠π/4

  46. (7√2)∠3π/4

Step-by-step explanation:

The polar form of a+bi is ...

  a +bi ⇔ (√(a²+b²))∠arctan(b/a)

__

45. 8 +8i = (√(8² +8²))∠arctan(8/8) = (8√2)∠π/4

__

46. -7 +7i = (√((-7)² +7²))∠arctan(7/-7) = (7√2)∠3π/4

_____

Additional comment

The arctan function usually returns a value in the range -π/2 to π/2. The quadrant of the desired result can be determined by the signs of the components of the vector. For negative real parts, π needs to be added to the value that is usually returned by the arctan function.

PLEASE I NEED HELP!!! FIRST ANSWER IS BRANLIEST!!! Apply the distributive property to create an equivalent expression. 6(5x-3)

Answers

Answer:

30x - 18

Step-by-step explanation:

6(5x - 3)

Apply the distributive property.

6(5x) + 6(-3)

30x + - 18

Answer:

30x - 18 is your final answer

Which lst of ordered pairs represents solutions to x+y= 2? (4.6), (0, 2), (4, 2) (4,6), (0, 2). (4, 2) (4,-6), (0, 2), (4, 2) ​

Answers

answer: x+y=2

0+2=²

( 0 , 2 )

Answer:

(0,2)

Step-by-step explanation:

0 + 2 = 2

Where x = 0, y = 2

Hello I am needing help on this word problem please. I think it's 30 but I'm questioning myself

Answers

Solution:

Since 5 pounds of meat will feed about 26 people.

Thus;

\begin{gathered} (5)/(26)lb\text{ would feed 1 person} \n  \n =0.1923 \end{gathered}

If she is expecting 156 people, she should prepare;

\begin{gathered} (0.1923*156)lb\text{ of meat} \n  \n \approx30lb \end{gathered}

ANSWER: 30lb

Find a solution to the following initial-value problem: dy dx = y(y − 2)e x , y (0) = 1.

Answers

This equation is separable, as

(\mathrm dy)/(\mathrm dx)=y(y-2)e^x\implies(\mathrm dy)/(y(y-2))=e^x\,\mathrm dx

Integrate both sides; on the left, expand the fraction as

\frac1{y(y-2)}=\frac12\left(\frac1{y-2}-\frac1y\right)

Then

\displaystyle\int(\mathrm dy)/(y(y-2))=\int e^x\,\mathrm dx\implies\frac12(\ln|y-2|-\ln|y|)=e^x+C

\implies\frac12\ln\left|\frac{y-2}y\right|=e^x+C

Since y(0)=1, we get

\frac12\ln\left|\frac{1-2}1\right|=e^0+C\implies C=-1

so that the particular solution is

\frac12\ln\left|\frac{y-2}y\right|=e^x-1\implies\boxed{y=\frac2{1-e^(2e^x-2)}}

The directional derivative of f(x, y) at (2, 1) in the direction going from (2, 1) toward the point (1, 3) is −2/ √ 5, and the directional derivative at (2, 1) in the direction going from (2, 1) toward the point (5, 5) is 1. Compute fx(2, 1) and fy(2, 1

Answers

Answer:

the partial derivatives are

fx =5/9

fy =(-13/18)

Step-by-step explanation:

defining the vector v (from (2,1) to (1,3))

v=(1,3)-(2,1) = (-1,2)

the unit vector will be

v'=(-1,2)/√5 = (-1/√5,2/√5)

the directional derivative is

fv(x,y) = fx*v'x + fy*v'y = fx*(-1/√5)+fy(2/√5) =-2/√5

then defining the vector u ( from (2, 1) toward the point (5, 5) )

u=(5,5)-(2,1) = (3,4)

the unit vector will be

u'=(3,4)/5 = (3/5,4/5)

the directional derivative is

fu(x,y) = fx*ux + fy*uy = fx*(3/5)+fy(4/5)=1

thus we have the set of linear equations

-fx/√5*+2*fy/√5 =(-2/√5) → -fx + 2*fy = -2

(3/5) fx+(4/5)*fy=1 → 3* fx+4*fy = 5

subtracting the first equation twice to the second

 3*fx+4*fy -(- 2fx)*-4*fy = 5 -2*(-2)

5*fx=9

fx=5/9

thus from the first equation

-fx + 2*fy = -2

fy= fx/2 -1 = 5/18 -1 = -13/18

thus we have

fx =5/9

fy =(-13/18)