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3 years ago

12x+200=644 answer: x=37 whats the variable coefficient and the constant

Mathematics

1 answer:

givi [52]3 years ago

6 0

Hello!

The variable coefficient is a constant that is multiplied by the variable. As you can see, 12 is multiplied by x. Therefore, 12 is our variable coefficient.

Our constant is a number that is not multiplied by a variable. Our two constants are both 200 and 644.

Therefore, our answer is below.

Coefficient: 12

Constants: 200, 644

I hope this helps!

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-2n-13= - 3n - 5 pls solve!

Elis [28]

Answer:

n = 8

Step-by-step explanation:

-2n - 13 = -3n - 5

-2n + 3n = -5 + 13

n = 8

Hope this helps. Have a nice day!

8 0

3 years ago

Read 2 more answers

What are the solutions of the equation (2x + 3)2 + 8(2x + 3) + 11 = 0

Tom [10]

Answer:

$x=\frac{-7-\sqrt{5}}{2}$ or $x=\frac{-7+ \sqrt{5}}{2}$

Step-by-step explanation:

The given equation is

$(2x+3)^2+8(2x+3)+11=0$

Let us treat this as a quadratic equation in $(2x+3)$ .

where $a=1,b=8,c=11$

The solution is given by the quadratic formula;

$(2x+3)=\frac{-b\pm \sqrt{b^2-4ac} }{2a}$

We substitute these values into the formula to obtain;

$(2x+3)=\frac{-8\pm \sqrt{8^2-4(1)(11)} }{2(1)}$

$(2x+3)=\frac{-8\pm \sqrt{64-44} }{2}$

$(2x+3)=\frac{-8\pm \sqrt{20} }{2}$

$(2x+3)=\frac{-8\pm2\sqrt{5} }{2}$

$(2x+3)=-4\pm \sqrt{5}$

$(2x+3)=-4-\sqrt{5}$ or $(2x+3)=-4+ \sqrt{5}$

$2x=-3-4-\sqrt{5}$ or $2x=-3-4+ \sqrt{5}$

$2x=-7-\sqrt{5}$ or $2x=-7+ \sqrt{5}$

$x=\frac{-7-\sqrt{5}}{2}$ or $x=\frac{-7+ \sqrt{5}}{2}$

5 0

3 years ago

If u, v, and w are nonzero vectors in r 2 , is w a linear combination of u and v?

Tju [1.3M]

Not necessarily. $\mathbf u$ and $\mathbf v$ may be linearly dependent, so that their span forms a subspace of $\mathbb R^2$ that does not contain every vector in $\mathbb R^2$ .

For example, we could have $\mathbf u=(0,1)$ and $\mathbf v=(0,-1)$ . Any vector $\mathbf w$ of the form $(r,0)$ , where $r\neq0$ , is impossible to obtain as a linear combination of these $\mathbf u$ and $\mathbf v$ , since

$c_1\mathbf u+c_2\mathbf v=(0,c_1)+(0,-c_2)=(0,c_1-c_2)\neq(r,0)$

unless $r=0$ and $c_1=c_2$ .

8 0

3 years ago

Which of the following is 16 60 in simplest form

Dafna1 [17]

There are no options given, but I will give you the exact answer of 16/60 simplified.

$\frac{16}{60} \\ \\ GCF = 4 \\ \\ 16 \div 4 = 4 \\ \\ 60 \div 4 = 15 \\ \\ \frac{4}{15} \\ \\ Answer: \fbox {4/15} \ or \ \fbox {0.2667}$

6 0

3 years ago

Read 2 more answers

What is the derivative of 1/square root 4x.

Bumek [7]

Answer:

$\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{-1}{4x^\bigg{\frac{3}{2}}}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Exponential Properties

Exponential Property [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Property [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

Step 1: Define

Identify.

$\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg]$

Step 2: Differentiate

Simplify: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \bigg( \frac{1}{2\sqrt{x}} \bigg)'$
Rewrite [Derivative Property - Multiplied Constant]: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( \frac{1}{\sqrt{x}} \bigg)'$
Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( \frac{1}{x^\Big{\frac{1}{2}}} \bigg)'$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( x^\bigg{\frac{-1}{2}} \bigg)'$
Derivative Rule [Basic Power Rule]: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( \frac{-1}{2} x^\bigg{\frac{-3}{2}} \bigg)$
Simplify: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{-1}{4} x^\bigg{\frac{-3}{2}}$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{-1}{4x^\bigg{\frac{3}{2}}}$

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

Unit: Differentiation

5 0

2 years ago