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EleoNora [17]
2 years ago
14

The rule that refers to grouping is blank property

Mathematics
2 answers:
baherus [9]2 years ago
5 0

Answer:

associative property

Step-by-step explanation:

Grouping like this

a + (b + c) = (a + b) + c

or

a(bc) = (ab)c

is the associative property

Degger [83]2 years ago
5 0

Answer= Associative Property

Step-by-step explanation:

Affiliate. The word "associative" comes from "associate" or "group"; the Associative Property is the rule that refers to grouping. For addition, the rule is "a + (b + c) = (a + b) + c"; in numbers, this means 2 + (3 + 4) = (2 + 3) + 4.

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"Parameterize the plane through the point (−1,−3,1) with the normal vector ⟨−4,4,3⟩
Makovka662 [10]
The equation of the required plane can be obtained thus:
-4(x + 1) + 4(y + 3) + 3(z - 1) = 0
-4x - 4 + 4y + 12 + 3z - 3 = 0
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Let x = 1, y = 2, then 4(1) - 4(2) - 3z = 5
z = (4 - 8 - 5)/3 = -9/3 = -3
Thus, point (1, 2, -3) is a point on the plane.

Let a = (a1, a2, a3) and b = (b1, b2, b3) be vectors parallel to the plane.
Then, -4a1 + 4a2 + 3a3 = 0 and -4b1 + 4b2 + 3b3 = 0
Let a1 = 2, a2 = -1, then a3 = (4(2) - 4(-1))/3 = (8 + 4)/3 = 12/3 = 4 and let b1 = -1 and b2 = 2, then b3 = (4(-1) - 4(2))/3 = (-4 - 8)/3 = -12/3 = -4
Thus a = (2, -1, 4) and b = (-1, 2, -4)

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3 0
4 years ago
What is the position of 9 in the number 932,805?
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(In the European notation it would in in the hundreds place).</span>
4 0
4 years ago
Read 2 more answers
Find dx/dt when y=2 and dy/dt=1, given that x^4=8y^5-240<br><br> dx/dt=
katrin2010 [14]

Answer:

The value of \frac{dx}{dt} is \frac{160}{x^3}.

Step-by-step explanation:

The given equation is

x^4=8y^5-240

We need to find the value of \frac{dx}{dt}.

Differentiate with respect to t.

4x^3\frac{dx}{dt}=8(5y^4)\frac{dy}{dt}-0              [\because \frac{d}{dx}x^n=nx^{n-1},\frac{d}{dx}C=0]

4x^3\frac{dx}{dt}=40y^4\frac{dy}{dt}

It is given that y=2 and dy/dt=1, substitute these values in the above equation.

4x^3\frac{dx}{dt}=40(2)^4(1)

4x^3\frac{dx}{dt}=40(16)(1)

4x^3\frac{dx}{dt}=640

Divide both sides by 4x³.

\frac{dx}{dt}=\frac{640}{4x^3}

\frac{dx}{dt}=\frac{160}{x^3}

Therefore the value of \frac{dx}{dt} is \frac{160}{x^3}.

4 0
3 years ago
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