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Slav-nsk [51]
3 years ago
14

I NEED HELP PLEASE HELP. SHOW ME HOW YOU DID IT

Mathematics
1 answer:
victus00 [196]3 years ago
5 0
Sin = - 4/8

Quadrant IV = only cosine is positive

a = height (4)
b = base ( 8^2-4^2=b^2
b = 6.93 @
\sqrt{48}

c = hypothenuse(8)

cos =
\sqrt{48}
/8

tan = - 4/
\sqrt{48}

sec = 1/cos
1/cos = 1/ (
\sqrt{48}
/8)
sec = 8/
\sqrt{48}

csc = 1/sin
1/sin = 1/(-4/8)
csc = - 2

cot = 1/tan
1/tan = 1/(-4/
\sqrt{48}
)
cot = -
\sqrt{48}
/4

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Answer:

  1. r = 3V/(2πh²)
  2. h = 3V/b²
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Step-by-step explanation:

1. Multiply both sides of the equation by the reciprocal of the coefficient of r.

V\cdot\dfrac{3}{2\pi h^2}=r\\\\r=\dfrac{3V}{2\pi h^2}

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2. Multiply both sides of the equation by the reciprocal of the coefficient of h.

V\cdot\dfrac{3}{b^2}=h\\\\h=\dfrac{3V}{b^2}

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3. Solve the circumference formula for r, then substitute the given information.

C=2\pi r\\\\r=\dfrac{C}{2\pi}\qquad\text{divide by the coefficient of r}\\\\r=\dfrac{50\,\text{cm}}{2\pi}=\dfrac{25}{\pi}\,\text{cm}\approx 7.9577\,\text{cm}

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4. Solve the perimeter formula for width, the substitute the given information and do the arithmetic.

P=2(L+W)\\\\\dfrac{P}{2}=L+W\qquad\text{divide by 2}\\\\\dfrac{P}{2}-L=W\qquad\text{subtract L}\\\\\dfrac{40\,\text{cm}}{2}-5\,\text{cm}=W=15\,\text{cm}

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In general, solving for a particular variable involves "undoing" what has been done to the variable, usually in the reverse order. In part 4, the variable W has L added and the sum is multiplied by 2. We "undo" those operations, last operation first, by dividing by 2 and subtracting L.

The properties of equality say you can do what you like to an equation as long as you do the same thing to both sides of the equation. So, when we say "divide by 2", we mean "divide both sides of the equation by 2." Likewise, "subtract L" means "subtract L from both sides of the equation."

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