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zaharov [31]
4 years ago
8

The slope of a line is negative for use the cordons of the labeled point to find a point-slope equation of the line.

Mathematics
1 answer:
Alexeev081 [22]4 years ago
5 0
For slope m and point (x₁, y₁), the point-slope equation of the line can be written as
  y -y₁ = m(x -x₁)

You have m = -4 and (x₁, y₁) = (4, -9), so the equation of your line is
  D. y+9 = -4(x-4)
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What are the zeros of the quadratic function represented by this graph?​
mario62 [17]

Answer:

Step-by-step explanation:

answer : C  when this graph passes trough x-axis

6 0
3 years ago
10 more than a number c is 3
Eddi Din [679]

Answer:

c is -7

Step-by-step explanation:

-7 + 10 = 3

:D

8 0
3 years ago
Evaluate |bc|, given a = 5, b = -3, and c = -2.
Anni [7]

Answer:

6

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Algebra I</u>

  • |Absolute Value| - makes any number positive

Step-by-step explanation:

<u>Step 1: Define</u>

|bc|

a = 5

b = -3

c = -2

<u>Step 2: Evaluate</u>

  1. Substitute:                    |-3 · -2|
  2. Multiply:                        |6|
  3. Evaluate:                       6
7 0
3 years ago
For each of the following vector fields F , decide whether it is conservative or not by computing curl F . Type in a potential f
Phantasy [73]

The key idea is that, if a vector field is conservative, then it has curl 0. Equivalently, if the curl is not 0, then the field is not conservative. But if we find that the curl is 0, that on its own doesn't mean the field is conservative.

1.

\mathrm{curl}\vec F=\dfrac{\partial(5x+10y)}{\partial x}-\dfrac{\partial(-6x+5y)}{\partial y}=5-5=0

We want to find f such that \nabla f=\vec F. This means

\dfrac{\partial f}{\partial x}=-6x+5y\implies f(x,y)=-3x^2+5xy+g(y)

\dfrac{\partial f}{\partial y}=5x+10y=5x+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=10y\implies g(y)=5y^2+C

\implies\boxed{f(x,y)=-3x^2+5xy+5y^2+C}

so \vec F is conservative.

2.

\mathrm{curl}\vec F=\left(\dfrac{\partial(-2y)}{\partial z}-\dfrac{\partial(1)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x)}{\partial z}-\dfrac{\partial(1)}{\partial z}\right)\vec\jmath+\left(\dfrac{\partial(-2y)}{\partial x}-\dfrac{\partial(-3x)}{\partial y}\right)\vec k=\vec0

Then

\dfrac{\partial f}{\partial x}=-3x\implies f(x,y,z)=-\dfrac32x^2+g(y,z)

\dfrac{\partial f}{\partial y}=-2y=\dfrac{\partial g}{\partial y}\implies g(y,z)=-y^2+h(y)

\dfrac{\partial f}{\partial z}=1=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=z+C

\implies\boxed{f(x,y,z)=-\dfrac32x^2-y^2+z+C}

so \vec F is conservative.

3.

\mathrm{curl}\vec F=\dfrac{\partial(10y-3x\cos y)}{\partial x}-\dfrac{\partial(-\sin y)}{\partial y}=-3\cos y+\cos y=-2\cos y\neq0

so \vec F is not conservative.

4.

\mathrm{curl}\vec F=\left(\dfrac{\partial(5y^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial x}\right)\vec\jmath+\left(\dfrac{\partial(5y^2)}{\partial x}-\dfrac{\partial(-3x^2)}{\partial y}\right)\vec k=\vec0

Then

\dfrac{\partial f}{\partial x}=-3x^2\implies f(x,y,z)=-x^3+g(y,z)

\dfrac{\partial f}{\partial y}=5y^2=\dfrac{\partial g}{\partial y}\implies g(y,z)=\dfrac53y^3+h(z)

\dfrac{\partial f}{\partial z}=5z^2=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=\dfrac53z^3+C

\implies\boxed{f(x,y,z)=-x^3+\dfrac53y^3+\dfrac53z^3+C}

so \vec F is conservative.

4 0
4 years ago
Find the greatest possible error for the measurement 4.7 cm.
gtnhenbr [62]
How is it an error? Need more explatinion or an example
4 0
3 years ago
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