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Misha Larkins [42]
2 years ago
7

Layna can paint a 350-square foot room in 4 hours. It takes Bebe 7 hours to paint the same room. Which equation can be used to f

ind the time in hours, t, it takes Layna and Bebe to paint the room together?
Rate
(Rooms per Hour)
Time
(Hours)
Fraction Completed
Layna
One-fourth

Bebe
One-seventh

One-fourth t + one-seventh t = 1
One-fourth t times one-seventh t = 1
4 t + 7 t = 1
4 t minus 7 t = 1
Mathematics
1 answer:
ohaa [14]2 years ago
8 0

The equation can be used to find the time in hours, \dfrac{t}{4}+\dfrac{t}{7}=1 and the time will be 2.5 hours. So the correct answer is option A.

<h3>What is an equation?</h3>

It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

Layna can paint a 350-square-foot room in 4 hours.It takes Bebe 7 hours to paint the same room.

Let t be the total time taken by both to paint.They are completing 1 job so the equation will be :

\dfrac{t}{4}+\dfrac{t}{7}=1

Further solving it we get;

7 t + 4 t = 28

11t  =  28

t = 2.5 hours.

Therefore the equation can be used to find the time in hours, \dfrac{t}{4}+\dfrac{t}{7}=1 and the time will be 2.5 hours. So the correct answer is option A.

To know more about equations follow

brainly.com/question/2972832

#SPJ1

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<h3>Given :</h3>
  • Base of triangle = 7 yd
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\\  \\

<h3>To find:</h3>
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\\  \\

We know:-

When base and height of triangle is given we use this formula:

\bigstar \boxed{ \rm Area \: of \: triangle =  \frac{base \times height}{2} }

\\  \\

So:-

\\  \\

\dashrightarrow \sf \: Area \: of \: triangle =  \dfrac{base \times height}{2}  \\

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\dashrightarrow \sf \: Area \: of \: triangle =  \dfrac{7 \times 5 \times2 }{2}  \\

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\dashrightarrow \sf \: Area \: of \: triangle =  \dfrac{7 \times 5 \times\cancel2 }{\cancel2}  \\

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\\  \\

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\\  \\

\dashrightarrow \bf \: Area \: of \: triangle =35 {yd}^{2}  \\

\\  \\

\therefore  \underline{\textsf{ \textbf {\: Area \: of \: triangle = \red{35}}} {  \red{\bf{yd} }^{ \red2} }}

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<h3>know more :-</h3>

\small\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf \small{Formulas\:of\:Areas:-}}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}\end{gathered}]

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