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Aleonysh [2.5K]
3 years ago
12

Three times a number minus another number is -3. The sum of the numbers is 11. Find the numbers

Mathematics
1 answer:
MatroZZZ [7]3 years ago
8 0

The first equation is

3x-y=-3

The second equation is

x+y=11

If you sum the two equations you'll cancel y:

4x=8 \iff x=2

And since the sum must be 11, we have y=9.

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I need help with this question
mr Goodwill [35]

Answer:

It would be 1.5m+4 is greater than or equal to 7. This is because 1.5m is the regular rate of growth, plus the 4cm already grown. She wants the hair to be no shorter than 7cm. So, the 15m+4 has to be greater than the 7cm.

Step-by-step explanation:

5 0
3 years ago
For f(x) = 3x + 1 and g(x) = x^2 - 7, find f(g(4)).
kodGreya [7K]

First, let's find g(4). After we find it, we can substitute the value into f(x).


To find g(4), we need to substitute x = 4 into the function g(x), as shown below:

g(x) = x^2 - 7

g(4) = 4^2 - 7

g(4) = 9


Now that we have found g(4) = 9, we can substitute this value in for x in f(x).

f(x) = 3x + 1

f(9) = 3(9) + 1

f(9) = 28


f(g(4)) is equal to 28.

4 0
3 years ago
two buses leave towns 604 mi apart at the same time and travel toward eachother. One bus travels 9 mi/h faster than the other on
muminat

Answer:

s = 71 mph, and s+9 = f = 80 mph

Step-by-step explanation:

The distances the two busses travel add up to 604 mi.

Letting f be the faster speed and s the slower.  Then f = s + 9 (mph).

Then (s + 9)(mph)(4 hr) + (s)(mph)(4 hr) = 604 mi.  Solve this for s:

4s + 36 + 4s = 604 mi, or 8s = 568.  Finally, s = 71 mph, and s+9 = f = 80 mph.


3 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
Find AB I need help ASAP please
kykrilka [37]

Answer:

AB= 94

Step-by-step explanation:

43+43= 86

180-86= 94

6 0
3 years ago
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