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Anestetic [448]
2 years ago
10

In which quadrant is the point ( - 2, 3) located?

Mathematics
1 answer:
stepan [7]2 years ago
8 0

ANSWER:

Quadrant ll

________________________________

________________________________HOPE IT HELPS

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Complete the explanation of how you would shade a model to show 2 x 2/3
Alekssandra [29.7K]

Answer:

B.

Step-by-step explanation:

3 0
3 years ago
What is the difference in the surface area of two cubes, one with edge 15 and one with an edge of 12
Lilit [14]

Remark

A cube has six sides, all of them equal. The formula for 1 side is s^2. The formula for all six = 6s^2.


Step One

Find the surface area of the larger cube.

Area = 6 *s^2

s = 15

Area = 6 *15^2

Area = 6 * 225

Area = 1350


Step Two

Find the area of the smaller cube

Area = 6s^2

Area = 6 * 12^2

Area = 6 * 144

Area = 864


Step Three

Find the difference

Area1 - Area2 = difference

1350 - 864 = 486

The difference is area = 486 units^2 <<<<< Answer


There is a slightly shorter way. Take out the common factor of 6

Difference = 6 * (15^2 - 12^2)

Difference = 6 * (225 - 144)

Difference = 6 * (81)

Difference = 486

6 0
3 years ago
How to solve for x where (2^x)*x =n?
Molodets [167]
Tough call.  
You may have to use Newton's Method or something like that.
Graph y=2^x and then graph n/x on the same set of axes.  You may have to assign some arbitrary value to n to make this work.  From the graph you can read off the approximate coordinates of the point of intersection.
4 0
3 years ago
The numbers 0,1/2, -53 and 0.433 all belong to which category ?
kirill [66]
0, 1/2, -53, and 0.433 all belong in rational category
3 0
3 years ago
Find the derivative of the function at P 0 in the direction of A. ​f(x,y,z) = 3 e^x cos(yz)​, P0 (0, 0, 0), A = - i + 2 j + 3k
Alik [6]

The derivative of f(x,y,z) at a point p_0=(x_0,y_0,z_0) in the direction of a vector \vec a=a_x\,\vec\imath+a_y\,\vec\jmath+a_z\,\vec k is

\nabla f(x_0,y_0,z_0)\cdot\dfrac{\vec a}{\|\vec a\|}

We have

f(x,y,z)=3e^x\cos(yz)\implies\nabla f(x,y,z)=3e^x\cos(yz)\,\vec\imath-3ze^x\sin(yz)\,\vec\jmath-3ye^x\sin(yz)\,\vec k

and

\vec a=-\vec\imath+2\,\vec\jmath+3\,\vec k\implies\|\vec a\|=\sqrt{(-1)^2+2^2+3^2}=\sqrt{14}

Then the derivative at p_0 in the direction of \vec a is

3\,\vec\imath\cdot\dfrac{-\vec\imath+2\,\vec\jmath+3\,\vec k}{\sqrt{14}}=-\dfrac3{\sqrt{14}}

3 0
4 years ago
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