5.75 an hour good luck lollllllll
Answer:The distance from a point to a line is
distance(ax+by+c=0,(x0,y0)) = abs(ax0+by0+c)/sqrt(a^2+b^2)
So first convert the formula for the line to standard form
5x+2y=4
5x+2y-4=0
a=5, b=2, c=-4, x0=15, y0=-21
Step-by-step explanation:
Answer: The height of the triangle is: " 3.5 cm " .
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<u>
Note</u>: The formula/equation for the area, "A" , of a triangle is:
A = (1/2) * b * h ; or write as: A = (b * h) / 2 ;
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in which: "A = area of the triangle" ;
"b = base length" ;
"h = "[perpendicular] height" ;
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Given: h = (b/2) ;
A = 12.25 cm²
{Note: Let us assume that the given area was "12.25 cm² " .}.
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We are to find the height, "h" ;
The formula for the Area, "A", is: A = (b * h) / 2 ;
Let us rearrange the formula ;
to isolate the "h" (height) on one side of the equation;
→ Multiply EACH side of the equation by "2" ; to eliminate the "fraction" ;
2*A = [ (b * h) / 2 ] * 2 ;
to get: " 2A = b * h " ;
↔ " b * h = 2A " ;
Divide EACH SIDE of the equation by "b" ; to isolate "h" on one side of the equation:
→ (b * h) / b = (2A) / b ;
to get:
→ h = 2A / b ;
Since "h = b/2" ; subtitute "b/2" for "h" ;
Plug in: "12.25 cm² " for "A" ;
→ b/2 = 2A/b ; → Note: " 2A/b = [2* (12.25 cm²) ] / b " ;
Note: " 2* (12.25 cm²) = 24.5 cm² ;
Rewrite as:
→ b/2 = (24.5 cm²) / b ;
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Cross-multiply: b*b = (24.5 cm²) *2 ;
to get: b² = 49 cm² ;
Take the "positive square root" of each side of the equation" ;
to isolate "b" on one side of the equation ; & to solve for "b" ;
→ +√(b²) = +√(49 cm²) ;
→ b = 7 cm ;
Now, we want to solve for "h" (the height) :
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→ h = b / 2 = 7 cm / 2 = 3.5 cm ;
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Answer: The height of the triangle is: " 3.5 cm <span>" .
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