All categories

3 years ago

What is the sum or difference of 517 37/50 + 312 3/100

Mathematics

1 answer:

Romashka [77]3 years ago

8 0

The sum is 829.77, and the difference is 205.44

You might be interested in

a construction worker has 15 feet of pipe he needs to divide the pipe into 9 equal parts which location on a number line represe

Maslowich

SENT THE SOLUTION WITH DRAWINGS THE HELP.

7 0

3 years ago

Diana works at a clothing store. she sold 1/5 of the total number of green shirts on Monday and 3/12 of the total number of gree

Doss [256]

Answer:

Step-by-step explanation:

3/12 really is 1/4 (Divide top and bottom by 3) 3/3 = 1 and 12/3 = 4

1/5 + 1/4 = 4/20 + 5/20 = 9/20

There is something we don't know about this question. If you put a remark, I will come back and look at it.

6 0

3 years ago

How do you solve these equations? I don't want you to answer all of them, just tell me how to solve each type of equation on the

miss Akunina [59]

Answer:

8. Identify the common denominator; express each fraction using that denominator; combine the numerators of those rewritten fractions and express the result over the common denominator. Factor out any common factors from numerator and denominator in your result. (It's exactly the same set of instructions that apply for completely numerical fractions.)

9. As with numerical fractions, multiply the numerator by the inverse of the denominator; cancel common factors from numerator and denominator.

10. The method often recommended is to multiply the equation by a common denominator to eliminate the fractions. Then solve in the usual way. Check all answers. If one of the answers makes your multiplier (common denominator) be zero, it is extraneous. (10a cannot have extraneous solutions; 10b might)

Step-by-step explanation:

For a couple of these, it is helpful to remember that (a-b) = -(b-a).

$\dfrac{5}{x+2}+\dfrac{25-x}{x^2-3x-10}=\dfrac{5(x-5)}{(x+2)(x-5)}+\dfrac{25-x}{(x+2)(x-5)}\\\\=\dfrac{5x-25+25-x}{(x+2)(x-5)}=\dfrac{4x}{x^2-3x-10}$

___

$\displaystyle\frac{\left(\frac{x}{x-2}\right)}{\left(\frac{2x}{2-x}\right)}=\frac{x}{x-2}\cdot\frac{-(x-2)}{2x}=\frac{-x(x-2)}{2x(x-2)}=-\frac{1}{2}$

___

$\dfrac{3}{x-1}+\dfrac{6}{x^2-3x+2}=2\\\\\dfrac{3(x-2)}{(x-1)(x-2)}+\dfrac{6}{(x-1)(x-2)}=\dfrac{2(x-1)(x-2)}{(x-1)(x-2)}\\\\3x-6+6=2(x^2-3x+2) \qquad\text{multiply by the denominator}\\\\2x^2-9x+4=0 \qquad\text{subtract 3x}\\\\(2x-1)(x-4)=0 \qquad\text{factor; x=1/2, x=4}$

Neither solution makes any denominator be zero, so both are good solutions.

8 0

3 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

HELP, i have no idea what the answers are

qaws [65]

I Dont have any idea

6 0

3 years ago