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Greeley [361]
3 years ago
11

Evaluate 2a + 6b for a = 4 and b = 5.12a+60 =2a + 6b=​

Mathematics
2 answers:
Kobotan [32]3 years ago
7 0

Step-by-step explanation:

12×4= 48

48+60= 108

2×4= 8

6×5= 30

8+30= 38

12a+60= 108

2a+6b= 38

galina1969 [7]3 years ago
5 0

Answer:

38 and 108

Step-by-step explanation:

2a + 6b

2 (4) + 6 (5) Plug in values

8+30 = 38

12a + 60

12 (4) + 60 = 108

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A semicircle is drawn onto one of the shorter sides of a rectangle. The shorter side of the rectangle measures 4 centimeters. Th
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<h3><u>Answer</u><u>:</u></h3>

  • 8.78 cm
<h3><u>Solution</u><u>:</u></h3>

<em>We are given that :</em>

  • A semicircle is drawn onto the shorter side of rectangle.
  • Shorter side of rectangle measures 4cm. i.e; the diameter of the circle is 4cm
  • Area of the figure is 41.4 cm²

<em><u>Using </u><u>Formulas</u><u>:</u></em>

Area of rectangle:

\quad\hookrightarrow\quad{\pmb{ \mathfrak {length\times breadth }}}

Area of semicircle:

\quad\hookrightarrow\quad{\pmb{ \mathfrak{ \dfrac{\pi r^2 }{2}}}}

  • <u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>length </u><u>of </u><u>longer </u><u>side </u><u>of </u><u>rectangle</u><u>!</u>

Here, we can know that the figure is composed of one rectangle and one semicircle. Therefore by combining the areas of rectangle and circle and taking the length of rectangle as a variable , we will finds its value ;

\quad\dashrightarrow\quad \sf {Area_{\tiny { total}} = Area_{\tiny {rectangle}}+ Area_{\tiny {semicircle}}}

\quad\dashrightarrow\quad \sf {A = ( l \times b ) + \dfrac{\pi r^2 }{2} }

\quad\dashrightarrow\quad \sf {41.4 =  ( l \times 4 ) +\dfrac{3.14\times 2^2}{2} }

\quad\dashrightarrow\quad \sf { 41.4= ( l \times 4 )+ \dfrac{ 3.14\times 4}{2}}

\quad\dashrightarrow\quad \sf { 41.4=( l \times 4 )+ \dfrac{12.56}{2}}

\quad\dashrightarrow\quad \sf { 41.4= (l \times 4 )+ 6.28}

\quad\dashrightarrow\quad \sf {l \times 4 = 41.4-6.28 }

\quad\dashrightarrow\quad \sf { l\times 4 = 35.12}

\quad\dashrightarrow\quad \sf { l =\dfrac{35.12}{4}}

\quad\dashrightarrow\quad \underline{\underline{\pmb{\sf {l = 8.78\:cm}}} }

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