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USPshnik [31]
3 years ago
15

Solve 10(1 + 3b) = −20, justifying each step with an algebraic property.

Mathematics
1 answer:
Delicious77 [7]3 years ago
4 0
The correct answer is B = -1

10(1 + 3B) = -20

10 + 30B = -20

-20 - 10 = -30

30B = -30

B = -1


Hope this helps :)
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<h3>Given :</h3>
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  • Height of triangle = 10 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} }

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So:-

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

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\dashrightarrow \sf \: Area \: of \: triangle =  \dfrac{7 \times 10}{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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\dashrightarrow \sf \: Area \: of \: triangle =  \dfrac{7 \times 5 \times1 }{1}  \\

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\dashrightarrow \sf \: Area \: of \: triangle =7 \times 5

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

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\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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Answer:

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