The Punic Wars had the GREATEST impact on the expansion ofes )
A)
the Roman Empire.
B)
the Roman Republic.
Egypt's "New Kingdom."
D)
Alexander the Great's Empire.
Submit

Answers

Answer 1
Answer:

Answer:

it b 100%

Step-by-step explanation:

Answer 2
Answer:

Answer:

A

Step-by-step explanation:


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Plz help I don’t get it
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Which functions could represent a reflection over the y-axis of the given function? Check all that apply. g(x) = –(4)x g(x) = 0.5(4)–x g(x) = 2(4)x g(x) = x g(x) = –x
Solve the following problem using only the numbers 13 4 8 5 10The difference between which two numbers is 6?

5/32 × 64y/100 please help :,)

Answers

Answer:  y/10

Step-by-step explanation:  step 1 = simplify y/100

                                              step 2 = (5/32 x 64) x y/100

                                              step 3 = simplify 5/32

                                              step 4 = (5/32 x 64) x y/100

                                              step 5 = The answer is y/10

Can anyone find out?

Answers

Answer:

i think its 85

Step-by-step explanation:

Answer:

D

Step-by-step explanation:

What is x2 + 2x + 9 = 0

Answers

Answer:

x has no real solution

Step-by-step explanation:

Our equation is qudratic equation so the method we will follow to solve it is using the dicriminant :

  • Let Δ be the dicriminant
  • a=1
  • b=2
  • c=9
  • Δ= 2²-4*1*9 =4-36=-32
  • we notice that Δ≤0⇒x has no real solution

Solve, check and show your work plz

Answers

Answer:

1. 3=x   2. no solution   3. infinate

Step-by-step explanation:

1. x=-5x+18

  +5x  +5x

6x=18

18/6=3

3=x

2.  3x-5=7+3x

     -3x       -3x

-5=7 No solution

3.   1/4(8x-12)=2x-3

        2x-3=2x-3

       -2x       -2x

     -3=-3

infinate

What is the solution to this system?

Answers

what system there’s no picture or equation

Answer:no picture/no question

Step-by-step explanation:

Suppose f(x,y)=xy, P=(−4,−4) and v=2i+3j. A. Find the gradient of f. ∇f= i+ j Note: Your answers should be expressions of x and y; e.g. "3x - 4y" B. Find the gradient of f at the point P. (∇f)(P)= i+ j Note: Your answers should be numbers C. Find the directional derivative of f at P in the direction of v. Duf= Note: Your answer should be a number D. Find the maximum rate of change of f at P. Note: Your answer should be a number E. Find the (unit) direction vector in which the maximum rate of change occurs at P. u= i+ j Note: Your answers should be numbers

Answers

Answers:

  • Gradient of f:    \nabla f =  y\hat{i} + x\hat{j}
  • Gradient of f at point p: \nabla f = -4\hat{i} -4\hat{j}
  • Directional derivative of f and P in direction of v: \nabla f(P)v = -20\n
  • The maximum rate of change of f at P:  | \nabla f(P)| =  4√(2)
  • The (unit) direction vector in which the maximum rate of change occurs at P is:  v =  -(1)/(√(2))\hat{i}-(1)/(√(2))\hat{j}

Step by step solutions:

Given that:

  • f(x,y) = xy
  • P = (-4,4)\n
  • v = 2i + 3j

A: Gradient of f

\nabla f = ((\partial f)/(\partial x), (\partial f)/(\partial y)) = (y,x) = y\hat{i} + x\hat{j}

B: Gradient of f at point P:

Just put the coordinates of p in above formula:

\nabla f = -4\hat{i} -4\hat{j}

C: The directional derivative of f and P in direction of v:

The directional derivative is found by dot product of \nabla f(P) \: \rm and \: \rm  v:

\nabla f(P)v = [-4,4][2,3]^T = -20\n

D: The maximum rate of change of f at P is calculated by evaluating the magnitude of gradient vector at P:

| \nabla f(P)| = √((-4)^2 + (-4)^2) = 4√(2)

E: The (unit) direction vector in which the maximum rate of change occurs at P is:

v = ((-4)/(4√(2)), (-4)/(4√(2))) = -(1)/(√(2))\hat{i}-(1)/(√(2))\hat{j}

That vector v is the needed unit vector in this case.

we divided by 4√(2) to make that vector as of unit length.

Learn more about vectors here:

brainly.com/question/12969462

Answer:

a) The gradient of a function is the vector of partial derivatives. Then

\nabla f=((\partial f)/(\partial x), (\partial f)/(\partial y))=(y,x)=y\hat{i} + x\hat{j}

b) It's enough evaluate P in the gradient.

\nabla f(P)=(-4,-4)=-4\hat{i} - 4 \hat{j}

c) The directional derivative of f at P in direction of V is the dot produtc of \nabla f(P) and v.

\nabla f(P) v=(-4,-4)\left[\begin{array}{ccc}2\n3\end{array}\right] =(-4)2+(-4)3=-20

d) The maximum rate of change of f at P is the magnitude of the gradient vector at P.

||\nabla f(P)||=√((-4)^2+(-4)^2)=√(32)=4√(2)

e) The maximum rate of change occurs in the direction of the gradient. Then

v=(1)/(4√(2))(-4,-4)=((-1)/(√(2)),(-1)/(√(2)))= (-1)/(√(2))\hat{i}-(1)/(√(2))\hat{j}

is the direction vector in which the maximum rate of change occurs at P.