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3 years ago

16+36= distributive property

Mathematics

2 answers:

lutik1710 [3]3 years ago

8 0

Answer:

4(4 + 9)

Step-by-step explanation:

Well... you can divide each by 4 and you get 4(4 + 9)

That's about all you can do for the distributive property.

steposvetlana [31]3 years ago

4 0

Answer:

a + b = b + a Commutative

Step-by-step explanation:

a = 16

b = 36

16 + 36 = 36 +16

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(9+8)+6/3-7×2 how to slove

klemol [59]

Answer:

Step-by-step explanation:

There is a specific order to do these kind of problems..

first, we do everything in parenthesis...

(9 + 8) + 6/3 - 7 * 2

17 + 6/3 - 7 * 2

now, starting from the LEFT side, do all multiplication and division in the order it comes

17 + 2 - 14

now, starting from the LEFT side, do all addition and subtraction in the order it comes

19 - 14 = 4

so ur answer is 4

Learn this...order of operations....this is the order u do these problems in..

PEMDAS

P = parenthesis

E = exponents

M = multiplication

D = division

A = addition

S = subtraction

keep in mind, multiplication/division are done from left to right....u do not have to do multiplication before u do division....it just depends on which comes first when starting from the left side.....that also applies to addition/subtraction

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At what point on the paraboloid y = x2 + z2 is the tangent plane parallel to the plane 3x + 2y + 7z = 2? (if an answer does not

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If $f(x, y, z) = c$ represent a family of surfaces for different values of the constant $c$ . The gradient of the function $f$ defined as $\nabla f$ is a vector normal to the surface $f(x, y, z) = c$ .

Given the paraboloid

$y = x^2 + z^2$ .

We can rewrite it as a scalar value function f as follows:

$f(x,y,z)=x^2-y+z^2=0$

The normal to the paraboloid at any point is given by:

$\nabla f= i\frac{\partial}{\partial x}(x^2-y+z^2) - j\frac{\partial}{\partial y}(x^2-y+z^2) + k\frac{\partial}{\partial z}(x^2-y+z^2) \\ \\ =2xi-j+2zk$

Also, the normal to the given plane $3x + 2y + 7z = 2$ is given by:

$3i+2j+7k$

Equating the two normal vectors, we have:

$2x=3\Rightarrow x= \frac{3}{2} \\ \\ -1=2 \\ \\ 2z=7\Rightarrow z= \frac{7}{2}$

Since, -1 = 2 is not possible, therefore there exist no such point on the paraboloid $y = x^2 + z^2$ such that the tangent plane is parallel to the plane 3x + 2y + 7z = 2.

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