What is the slope of this line A -1/2 B 1/2 C 2

The height of a triangle is half the length of its base. The area of the triangle is 12.25cm. Find the height

pentagon [3]

Answer: The height of the triangle is: " 3.5 cm " .
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Note: 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 " .
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6 0

3 years ago

Find the zeros (roots, x-intercepts) for each po 3. f(x) = 5x^3- 20x

Aliun [14]

Answer:

v211

Step-by-step explanation:

6 0

3 years ago

Anyone know the answer to this algebra problem?

ikadub [295]

Answer: $\bold{a=1\qquad b=\dfrac{1}{16}\qquad c=\dfrac{1}{64}\qquad d=1\qquad e=\dfrac{4}{9}\qquad f=\dfrac{16}{81}}$

Step-by-step explanation:

$\begin{array}{c|l}\underline{\quad x\quad}&\underline{\quad 4^{-x}\qquad \qquad}\\-1&4^{-(-1)}=4^1=4\\\\0&4^{-(0)}=4^0=1\\\\2&4^{-(2)}=\dfrac{1}{4^2}=\dfrac{1}{16}\\\\4&4^{-(4)}=\dfrac{1}{4^4}=\dfrac{1}{64}\end{array}$

$\begin{array}{c|l}\underline{\quad \bigg x \quad}&\underline{\quad \bigg(\dfrac{2}{3}\bigg)^x\qquad \qquad}\\\\-1&\bigg(\dfrac{2}{3}\bigg)^{-1}=\dfrac{3}{2}\\\\0&\bigg(\dfrac{2}{3}\bigg)^{0}=1\\\\2&\bigg(\dfrac{2}{3}\bigg)^{2}=\dfrac{2^2}{3^2}=\dfrac{4}{9}\\\\4&\bigg(\dfrac{2}{3}\bigg)^{4}=\dfrac{2^4}{3^4}=\dfrac{16}{81}\end{array}$