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4 years ago

What is the perimeter of a10 unit square

Mathematics

2 answers:

Karo-lina-s [1.5K]4 years ago

7 0

It would be 40, because the square has 4 sides. 10×4=40

elena-s [515]4 years ago

4 0

40 square units 10+10+10+10

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Pedro is building a playground in the shape of a right triangle. He wants to know the area of the playground to help him decide

Bumek [7]

Answer:

Dimensions of rectangle:

Length =6 yards

Width=3 1/3 yards

Area of right triangle = 10 yd^2

Step-by-step explanation:

The right triangle has base as 6 yards and height as 3 1/3 of yards.

When we match this triangle with another right triangle to make the rectangle is another right triangle with sides 6 yards and 3 1/3 yards.

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3 years ago

Verify the identity. tan x plus pi divided by two = -cot x

kogti [31]

Recall that sin(π/2) = 1, and cos(π/2) = 0,

$\bf \textit{Sum and Difference Identities}
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sin(\alpha + \beta)=sin(\alpha)cos(\beta) + cos(\alpha)sin(\beta)
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sin(\alpha - \beta)=sin(\alpha)cos(\beta)- cos(\alpha)sin(\beta)
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cos(\alpha + \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta)
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cos(\alpha - \beta)= cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta)\\\\
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$

$\bf tan\left(x+\frac{\pi }{2} \right)=-cot(x)\\\\
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tan\left(x+\frac{\pi }{2} \right)\implies \cfrac{sin\left(x+\frac{\pi }{2} \right)}{cos\left(x+\frac{\pi }{2} \right)}
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\cfrac{sin(x)cos\left(\frac{\pi }{2} \right)+cos(x)sin\left(\frac{\pi }{2} \right)}{cos(x)cos\left(\frac{\pi }{2} \right)-sin(x)sin\left(\frac{\pi }{2} \right)}\implies \cfrac{sin(x)\cdot 0~~+~~cos(x)\cdot 1}{cos(x)\cdot 0~~-~~sin(x)\cdot 1}
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3 years ago

Identify the standard form of the equation by completing the square.

OLEGan [10]

Answer:

$\dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Step-by-step explanation:

Given equation:

$4x^2-9y^2-8x+36y-68=0$

This is an equation for a horizontal hyperbola.

To complete the square for a hyperbola

Arrange the equation so all the terms with variables are on the left side and the constant is on the right side.

$\implies 4x^2-8x-9y^2+36y=68$

Factor out the coefficient of the x² term and the y² term.

$\implies 4(x^2-2x)-9(y^2-4y)=68$

Add the square of half the coefficient of x and y inside the parentheses of the left side, and add the distributed values to the right side:

$\implies 4\left(x^2-2x+\left(\dfrac{-2}{2}\right)^2\right)-9\left(y^2-4y+\left(\dfrac{-4}{2}\right)^2\right)=68+4\left(\dfrac{-2}{2}\right)^2-9\left(\dfrac{-4}{2}\right)^2$

$\implies 4\left(x^2-2x+1\right)-9\left(y^2-4y+4\right)=36$

Factor the two perfect trinomials on the left side:

$\implies 4(x-1)^2-9(y-2)^2=36$

Divide both sides by the number of the right side so the right side equals 1:

$\implies \dfrac{4(x-1)^2}{36}-\dfrac{9(y-2)^2}{36}=\dfrac{36}{36}$

Simplify:

$\implies \dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Therefore, this is the standard equation for a horizontal hyperbola with:

center = (1, 2)
vertices = (-2, 2) and (4, 2)
co-vertices = (1, 0) and (1, 4)
$\textsf{Asymptotes}: \quad y = -\dfrac{2}{3}x+\dfrac{8}{3} \textsf{ and }y=\dfrac{2}{3}x+\dfrac{4}{3}$
$\textsf{Foci}: \quad (1-\sqrt{13}, 2) \textsf{ and }(1+\sqrt{13}, 2)$