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Natasha2012 [34]
2 years ago
13

 Pls someone help me

Mathematics
1 answer:
Firlakuza [10]2 years ago
8 0
Area of triangle = base x height over 2
triangle 1
10 x 12 = 120
120/2 = 60
triangle 2
10 x 12 = 120
120/2 = 60
triangle 3
18 x 12 = 216
216/2 = 108
triangle 4
18 x 12 = 216
216/2 = 108

add
108 + 108 + 60 + 60 = 336 square units
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Given: ⅓ (18x + 27) = 81 Prove: x = 12
sasho [114]

Answer:

Firstly, rewrite the equation:

⅓ (18 + 27) = 81

Substitute x for the given number of it's supposed equivalent.

In this case x = 12.

⅓ (18(12) + 27) = 81

Solve using PEMDAS and simplify what is in the parenthesis first. Then, multiply.

(18 x 12) + 27 = 243

Now, solving using PEMDAS, multiply the total of what you got that was originally in the parenthesis by ⅓ .

⅓ (243) = 81

When you multiple these number they are equivalent to 81.

81 = 81

Since the equation given, when substituted x for 12, is equivalent to 81, this proves that substituting x for 12 makes this equation true.

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Read 2 more answers
Solve the triangle.<br><br> a = 10, b = 23, C = 95°
gayaneshka [121]
C² = a² + b² - (2ab * cosC) 
<span>c² = 10² + 23² - (2 * 10 * 23 * cos95) </span>
<span>c² = 100 + 529 - (460 * -.08715) </span>
<span>c² = 629 - (-40.1) </span>
<span>c² = 669.1 </span>
<span>c = 25.87 </span>

<span>(Sin C) / C = (Sin A) / A </span>
<span>(Sin 95) / 25.87 = SinA / 10, Remember 0 < A < 85 </span>
<span>(10 * Sin95) / 25.87 = Sin A </span>
<span>A = arcsin ((10 * sin95) / 25.87) </span>
<span>A = 22.65º </span>

<span>B = 180 - A - C </span>
<span>B = 180 - 95 - 22.65 </span>
<span>B = 62.35º </span>

<span>I hope this helps. Have a good day.</span>
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3 years ago
How many real roots does a quadratic equation have if the discriminant is -3?
soldier1979 [14.2K]

\bf \qquad \qquad \qquad \textit{discriminant of a quadratic} \\\\\\ y=\stackrel{\stackrel{a}{\downarrow }}{a}x^2\stackrel{\stackrel{b}{\downarrow }}{+b}x\stackrel{\stackrel{c}{\downarrow }}{+c} ~~~~~~~~ \stackrel{discriminant}{b^2-4ac}= \begin{cases} 0&\textit{one solution}\\ positive&\textit{two solutions}\\ negative&\stackrel{\textit{no real roots}}{\textit{no solution}}\quad \checkmark \end{cases}

6 0
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