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11 months ago

Zola is cutting a rope that is 3 yards long into pieces that are 4 inches long.

Mathematics

1 answer:

Inessa [10]11 months ago

4 0

Answer:

432. Zola will have 432 4-inch pieces of rope.

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find the centre and radius of the following Cycles 9 x square + 9 y square +27 x + 12 y + 19 equals 0

Citrus2011 [14]

Answer:

Radius: $r =\frac{\sqrt {21}}{6}$

$Center = (-\frac{3}{2}, -\frac{2}{3})$

Step-by-step explanation:

Given

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Solving (a): The radius of the circle

First, we express the equation as:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$r = radius$

$(h,k) =center$

So, we have:

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Divide through by 9

$x^2 + y^2 + 3x + \frac{12}{9}y + \frac{19}{9} = 0$

Rewrite as:

$x^2 + 3x + y^2+ \frac{12}{9}y =- \frac{19}{9}$

Group the expression into 2

$[x^2 + 3x] + [y^2+ \frac{12}{9}y] =- \frac{19}{9}$

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

Next, we complete the square on each group.

For $[x^2 + 3x]$

1: Divide the $coefficient\ of\ x\ by\ 2$

2: Take the $square\ of\ the\ division$

3: Add this $square\ to\ both\ sides\ of\ the\ equation.$

So, we have:

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

$[x^2 + 3x + (\frac{3}{2})^2] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Factorize

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Apply the same to y

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y +(\frac{4}{6})^2 ] =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ \frac{9}{4} +\frac{16}{36}$

Add the fractions

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{-19 * 4 + 9 * 9 + 16 * 1}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{21}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{7}{12}$

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

By comparison:

$r^2 =\frac{7}{12}$

Take square roots of both sides

$r =\sqrt{\frac{7}{12}}$

Split

$r =\frac{\sqrt 7}{\sqrt 12}$

Rationalize

$r =\frac{\sqrt 7*\sqrt 12}{\sqrt 12*\sqrt 12}$

$r =\frac{\sqrt {84}}{12}$

$r =\frac{\sqrt {4*21}}{12}$

$r =\frac{2\sqrt {21}}{12}$

$r =\frac{\sqrt {21}}{6}$

Solving (b): The center

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$r = radius$

$(h,k) =center$

From:

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

$-h = \frac{3}{2}$ and $-k = \frac{2}{3}$

Solve for h and k

$h = -\frac{3}{2}$ and $k = -\frac{2}{3}$

Hence, the center is:

$Center = (-\frac{3}{2}, -\frac{2}{3})$

6 0

2 years ago

Pls help Given: △ABC, CM⊥ AB, BC = 5, AB = 7 CA = 4 sqrt(2) Find: CM

babymother [125]

Answer:

The general plan is to find BM and from that CM. You need 2 equations to do that.

Step One

Set up the two equations.

(7 - BM)^2 + CM^2 = (4*sqrt(2) ) ^ 2 = 32

BM^2 + CM^2 = 5^2 = 25

Step Two

Subtract the two equations.

(7 - BM)^2 + CM^2 = 32

BM^2 + CM^2 = 25

(7 - BM)^2 - BM^2 = 7 (3)

Step three

Expand the left side of the new equation labeled (3)

49 - 14BM + BM^2 - BM^2 = 7

Step 4

Simplify And Solve

49 - 14BM = 7 Subtract 49 from both sides.

-49 - 14BM = 7 - 49

- 14BM = - 42 Divide by - 14

BM = -42 / - 14

BM = 3

Step Five

Find CM

CM^2 + BM^2 = 5^2

CM^2 + 3^2 = 5^2 Subtract 3^2 from both sides.

CM^2 = 25 - 9

CM^2 = 16 Take the square root of both sides.

sqrt(CM^2) = sqrt(16)

CM = 4 < Answer

Step-by-step explanation:

6 0

2 years ago

I have to find the area of a trapezoid using the formula A= 1/2 (b1 + b2) (h)

DedPeter [7]

Answer:

190

Step-by-step explanation:

1/2(6+13)(5)

6+ 13 = 19*5 = 95

95/ 1/2= 190

7 0

3 years ago

Expand 2y (2y + 8 )

Andrews [41]

Hello There!

2y x 2y = 4y²
2y x 8 = 16y.
Put it together and you get the answer of:
4y² + 16y.

Hope This Helps You!
Good Luck :)

- Hannah ❤

7 0

3 years ago

The graph of y=f(x) is above. If f(−4)=k, what is the value of f(k)?

cupoosta [38]

Answer:

The answer is B, which is -1.5.

3 0

3 years ago