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s2008m [1.1K]
2 years ago
9

PLEASE HELP!!! I WiLL GIVE BRAINLIEST!!!!!

Mathematics
2 answers:
DIA [1.3K]2 years ago
6 0

Answer:

  • coefficient of x: 6
  • coefficient of y: 3
  • coefficient of z: 5

Step-by-step explanation:

The coefficient in an expression is the multiplier of a designated part of a term of the expression.

__

<h3>coefficients here</h3>

The given expression is

  5z -2 +6x +3y

It consists of four (4) terms, added together. One term is a constant, -2. The others each have a single variable that is multiplied by a constant. When we ask for the coefficient of the variable, we are asking for the value of that constant multiplier.

The term containing x is 6x. The multiplier of x in that term is 6, so we say ...

  the coefficient of x is 6.

The term containing y is 3y. The multiplier of y in that term is 3, so we say ...

  the coefficient of y is 3.

The term containing z is 5z. The multiplier of z in that term is 5, so we say ...

  the coefficient of z is 5.

__

<h3>other coefficients</h3>

In an expression such as πr²h, different parts of the expression will have different coefficients. If we ask for the coefficient of the expression, we are generally asking for the constant: π. (The Greek letter pi is the symbol used to represent the irrational constant ratio between the circumference and diameter of a circle. Is is about 3.1415926535....)

If we ask for the coefficient of h, we are asking for the product (πr²) that multiplies h in this expression. Similarly, the coefficient of r² is (πh). Note that we have specified a "designated part" of the expression for which we want the coefficient. This is important when we are solving equations.

__

The commutative property of multiplication means that parts of a product can be written in any order. That is, 5·z is the same product as z·5. When the constant and variable parts of the expression can be unambiguously identified, the order does not matter. In some cases, a constant coefficient may come in several parts: 3z·5·2 has a z-coefficient of 30, for example.

See below for a case where there may be some ambiguity.

__

<h3>exponents</h3>

In the expression r², the constant 2 that comes <em>after</em> the variable, written as a superscript, is called an "exponent," not a coefficient. Often on Brainly, and sometimes in curriculum materials, you will see this written as r2, because superscripts do not get properly rendered. If you're writing this in plain text, the form r^2 is preferred.

nadya68 [22]2 years ago
5 0

Answer:

x=5    y=3      z=5

Step-by-step explanation:

The coefficient is the number in front of the variable

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Step-by-step explanation:

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Please help! Can anyone help me out with this question?
TEA [102]

Answer:

\displaystyle h'(s) = 64s + 20

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
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  • Left to Right<u> </u>

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<u>Algebra I</u>

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<u>Calculus</u>

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]:                                                                                  \displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)

Step-by-step explanation:

<u>Step 1: Define</u>

\displaystyle h(s) = (-8s - 9)(-4s + 2)

<u>Step 2: Differentiate</u>

  1. [Derivative] Product Rule:                                                                               \displaystyle h'(s) = \frac{d}{ds}[(-8s - 9)](-4s + 2) + (-8s - 9)\frac{d}{ds}[(-4s + 2)]
  2. [Derivative] Basic Power Rule:                                                                       \displaystyle h'(s) = (1 \cdot -8s^{1 - 1} - 0)(-4s + 2) + (-8s - 9)(1 \cdot -4s^{1 - 1} - 0)
  3. [Derivative] Simplify:                                                                                         \displaystyle h'(s) = (-8)(-4s + 2) + (-8s - 9)(-4)
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  5. [Derivative] Combine like terms:                                                                     \displaystyle h'(s) = 64s + 20

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Derivatives

Book: College Calculus 10e

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