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SpyIntel [72]
2 years ago
14

A piling for a high-rise building is pushed by two bulldozers at exactly the same time. One bulldozer exerts a force of 1250 pou

nds in a westerly direction. The other bulldozer pushes the piling with a force of 2650 pounds in a northerly direction. What is the magnitude of the resultant force upon the piling, to the nearest ten pounds?
Mathematics
2 answers:
Masja [62]2 years ago
8 0
The forces result in a right triangle. To obtain the resultant force, one can simply use the Pythagorean theorem. 1250 lbf and 2650 lbf both act as the legs of the triangle. Obtaining the hypotenuse via the theorem would yield the resultant force. This is done below:

c^2 = a^2 + b^2
c^2 = (1250)^2 + (2650)^2
c = 2930.017 lbf

Therefore, the magnitude of the resultant force is approximately equal to 2930 lbf.
Sav [38]2 years ago
7 0

Answer:

D edge

Step-by-step explanation:

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Step-by-step explanation:

3 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}\\
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 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}


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Ahat [919]

Answer:

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Step-by-step explanation:

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x =( -(-8)±√((-8)^2-4*1*15))/2(1)

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x=(8 + 2) / 2 = 10/5 = 5

solve with '-'

x=(8 - 2) / 2 = 6/2 = 3

x = (5,3)

5 0
2 years ago
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