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2 years ago

Write the equation in standard form for the circle -20y = -x^2 - y^2 - 36

Mathematics

2 answers:

sasho [114]2 years ago

7 0

Answer:

(y-10)/64 - x^2/64=1

OLga [1]2 years ago

3 0

Answer:

x^2 + (y-10)^2 = 64 I think

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I will give brainliest to first (2x^2 − 7x^3 + 8x) + (−8x^3 − 3x^2 + 4)

cluponka [151]

Answer:-15x^3-x^2+8x+4

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Step-by-step explanation:

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2 years ago

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A line passes through (1, –5) and (–3, 7). a. Write an equation for the line in point-slope form. b. Rewrite the equation in slo

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3 years ago

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there were twenty one adults in line at a movie theater.That is three times the number of children in line. how many children we

V125BC [204]

You would do 21 divided by 3 and get 7.

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3 years ago

What is 7x^2-8x-9=-3x rounded to the nearest hundredth

jonny [76]

Answer: 1.55, and -0.83

Step-by-step explanation:

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2 years ago

Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

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3 years ago