All categories

2 years ago

What is the solution to the following system of equation y=6x11 -2x3y=-7

Mathematics

1 answer:

Jlenok [28]2 years ago

3 0

Answer:

y=6x11−2x3y=−7

Step-by-step explanation:

You might be interested in

Solve for n.<br> n-26 = -13<br> n=1

Paraphin [41]

Answer:

I believe n equals 13.

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

EXAMPLE 5 If f(x, y, z) = x sin(yz), (a) find the gradient of f and (b) find the directional derivative of f at (1, 2, 0) in the

solong [7]

Answer:

<h2>a) f = sin(yz)i + xzcos(yz)j + xycos(yz)k</h2><h2>b) -2</h2>

Step-by-step explanation:

Given f(x, y, z) = x sin(yz), the formula for calculating the gradient of the function is expressed as ∇f(x, y, z) = fx(x, y, z)i+ fy(x, y, z)j+fz(x, y, z)k where;

fx, fy and fz are the differential of the functions with respect to x, y and z respectively.

a) ∇f(x, y, z) = sin(yz)i + xzcos(yz)j + xycos(yz)k

The gradient of f = sin(yz)i + xzcos(yz)j + xycos(yz)k

b) Directional derivative of f at (1,2,0) in the direction of v = i + 4j − k is expressed as ∇f(1, 2, 0)*v

∇f(1, 2, 0) = sin(2(0))i +1*0cos(2*0)j + 1*2cos(2*0)k

∇f(1, 2, 0) = sin0i +0cos(0)j + 2cos(0)k

∇f(1, 2, 0) = 0i +0j + 2k

Given v = i + 4j − k

∇f(1, 2, 0)*v (note that this is the dot product of the two vectors)

∇f(1, 2, 0)*v = (0i +0j + 2k)*(i + 4j − k )

Given i.i = j.j = k.k =1 and i.j=j.i=j.k=k.j=i.k = 0

∇f(1, 2, 0)*v = 0(i.i)+4*0(j.j)+2(-1)k.k

∇f(1, 2, 0)*v = 0(1)+0(1)-2(1)

∇f(1, 2, 0)*v =0+0-2

∇f(1, 2, 0)*v= -2

Hence, the directional derivative of f at (1, 2, 0) in the direction of v = i + 4j − k is -2

7 0

3 years ago

There is a a circle whose circumference is 34.54 cm.

Naily [24]

Answer:

$\huge\boxed{Diameter=11cm}$

Step-by-step explanation:

Important information:

Circumference of a circle= $\pi*diameter$

Method:

To work out the diameter, we would need to divide the circumference of 34.54 by pi, which in this case is 3.14, this gives us 11cm. This is because we are simply rearanging the formula of the circumference of a circle to make the diameter the subject.

Step By Step:

1) Rearange the formula to make the diameter the subject.

$diameter=\frac{circumference}{\pi}$