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3 years ago

Convert 2/800 as a decimal using an equivalent fraction

Mathematics

1 answer:

grin007 [14]3 years ago

8 0

Answer:

The answer I got was 0.003

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Find the equation of the line passing through the point (–1, 0) and parallel to the line y = –1∕2x + 3∕7

PilotLPTM [1.2K]

Answer:

y = -1/2x - 1/2

Step-by-step explanation:

Slope-Intercept Form: y = mx + b

m - slope

b - y-intercept

Step 1: Define

y = -1/2x + 3/7

Point (-1, 0)

Step 2: Find parallel line

<em>Parallel lines have the same slope as the original but different y-intercepts.</em>

<em>m</em> = -1/2

y = -1/2x + b

0 = -1/2(-1) + b

0 = 1/2 + b

b = -1/2

Step 3: Write parallel line

y = -1/2x - 1/2

7 0

4 years ago

A Rhombus with diagonal 1 equal to 5 ft and diagonal 2 equal to 8 ft

Minchanka [31]

Answer:

Area = 20 ft²

Step-by-step explanation:

Area of a thrombus

½ × d1 × d2

½ × 5 × 8

4 0

3 years ago

Read 2 more answers

Gretchen paid 2,5, and 3 dollors if she paid 4 dollors for each at a sale

VladimirAG [237]

40 I am pretty sure that is the right answer

5 0

4 years ago

Put the following equation in standard form and determine the quadratic, linear, and constant coefficients. -3x2 - 8 = 5x - 7

marusya05 [52]

Answer:

-3x² -5x -1 = 0
-3 (quadratic)
-5 (linear)
-1 (constant)

Step-by-step explanation:

The equation will be in standard form when terms are listed in order of decreasing degree, and the right side of the equation is 0.

<h3>Standard form</h3>

We can subtract the right-side expression from both sides to get standard form.

-3x² -8 -(5x -7) = 5x -7 -(5x -7)

-3x² -8 -5x +7 = 0 . . . . . . . simplify a bit

-3x² -5x -1 = 0 . . . . . . . . . collect terms

The standard form equation can be written ...

-3x² -5x -1 = 0

<h3>Coefficients</h3>

The quadratic coefficient is the coefficient of the term with degree 2. The quadratic coefficient is -3.

The linear coefficient is the coefficient of the term with degree 1. The linear coefficient is -5.

The constant coefficient is the coefficient of the term with no variables. The constant is -1.

<em>Additional comment</em>

We can make the leading coefficient positive by multiplying the equation by -1. This gives ...

3x² +5x +1 = 0

with quadratic, linear, and constant coefficients 3, 5, 1.

This is a legitimate answer to this question. In the case of linear equations, the "standard form" has the constant on the right side of the equal sign, and the leading coefficient is required to be positive. A negative leading coefficient can sometimes lead to errors (when the sign is overlooked), so having a positive leading coefficient is often preferred.

8 0

2 years ago

10 points, please help me and explain how to do this with answers!

8_murik_8 [283]

$\bf f(x)=log\left( \cfrac{x}{8} \right)\\\\
-----------------------------\\\\
\textit{x-intercept, setting f(x)=0}
\\\\
0=log\left( \cfrac{x}{8} \right)\implies 0=log(x)-log(8)\implies log(8)=log(x)
\\\\
8=x\\\\
-----------------------------$

$\bf \textit{y-intercept, is setting x=0}\\
\textit{wait just a second!, a logarithm never gives 0}
\\\\
log_{{ a}}{{ b}}=y \iff {{ a}}^y={{ b}}\qquad\qquad 
% exponential notation 2nd form
{{ a}}^y={{ b}}\iff log_{{ a}}{{ b}}=y 
\\\\
\textit{now, what exponent for "a" can give you a zero? none}\\
\textit{so, there's no y-intercept, because "x" is never 0 in }\frac{x}{8}\\
\textit{that will make the fraction to 0, and a}\\
\textit{logarithm will never give that, 0 or a negative}\\\\
$

$\bf -----------------------------\\\\
domain
\\\\
\textit{since whatever value "x" is, cannot make the fraction}\\
\textit{negative or become 0, , then the domain is }x\ \textgreater \ 0\\\\
-----------------------------\\\\
range
\\\\
\textit{those values for "x", will spit out, pretty much}\\
\textit{any "y", including negative exponents, thus}\\
\textit{range is }(-\infty,+\infty)$
p, li { white-space: pre-wrap; }

----------------------------------------------------------------------------------------------

now on 2)

$\bf f(x)=\cfrac{3}{x^4}$ if the denominator has a higher degree than the numerator, the horizontal asymptote is y = 0, or the x-axis,

in this case, the numerator has a degree of 0, the denominator has 4, thus y = 0

vertical asymptotes occur when the denominator is 0, that is, when the fraction becomes undefined, and for this one, that occurs at $x^4=0\implies x=0$ or the y-axis

----------------------------------------------------------------------------------------------

now on 3)

$\bf f(x)=\cfrac{1}{x}$

now, let's see some transformations templates

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\

\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\mathbb{R}^{{{ B}}x+{{ C}}}+{{ D}}
\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\
\bullet \textit{ vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\
\qquad if\ {{ D}}\textit{ is positive, upwards}
\end{array}$

now, let's take a peek at g(x)

$\bf \begin{array}{lcllll}
g(x)=&-&\cfrac{1}{x}&+3\\
&\uparrow &&\uparrow \\
&\textit{upside down}&&
\begin{array}{llll}
\textit{vertical shift up}\\
\textit{by 3 units}
\end{array}
\end{array}$

3 0

3 years ago