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Alla [95]
4 years ago
14

The width of a rectangle is 3 less than twice the length, x. If the area is 43 square feet, which equation can be used to find t

he length in feet?
A. 2x(x-3) = 43
B. x(3-2x) = 43
C. 2x + 2(2x-3) = 43
D. X(2x-3) = 43
Mathematics
1 answer:
12345 [234]4 years ago
5 0
D) x*(2x-3)=43 would be the correct equation.
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Find the GCF of 40 and 60 listing factors for each number in order from least to greatest.
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40:1,2,4,5,8,10,20,40
60:1,2,3,4,5,6,10,12,15,20,30,60
 The GCF between 40 and 60 is 20
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3 years ago
PLEASE HELP BUT REALLY Find the value of x. Write your answer in simplest radical form, if necessary.
SCORPION-xisa [38]

Answer: x = √141

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x² = 141

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3 years ago
Sarah is currently making $31,000 a year at her job and has been offered a raise to $33,170 per year. What would be the percent
erastova [34]

Answer:

  7%

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The percentage increase is found as ...

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3 years ago
6)
earnstyle [38]

Answer:

It will take 10 months to build the same number of houses.

7 0
3 years ago
A baseball is thrown into the air with an upward velocity of 30 ft/s. its initial height was 6 ft, and its maximum height is 20.
marissa [1.9K]

Check the picture below.

where is the -16t² coming from?  that's Earth's gravity pull in feet.

\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{30}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{6}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\\\ h(t)=-16t^2+30t+6 \\\\[-0.35em] ~\dotfill

\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+30}t\stackrel{\stackrel{c}{\downarrow }}{+6} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)

\bf \left(-\cfrac{30}{2(-16)}~~,~~6-\cfrac{30^2}{4(-16)} \right)\implies \left( \cfrac{30}{32}~,~6+\cfrac{225}{16} \right)\implies \left(\cfrac{15}{16}~,~\cfrac{321}{16} \right) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (\stackrel{\stackrel{\textit{how many}}{\textit{seconds it took}}}{0.9375}~~,~~\stackrel{\stackrel{\textit{how many feet}}{\textit{up it went}}}{20.0625})~\hfill

7 0
4 years ago
Read 2 more answers
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