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2 years ago

PLZ HELP ME it's due at 2:45 p.m.

Mathematics

1 answer:

Georgia [21]2 years ago

4 0

Answer:

16 uniforms, assuming the mail fee is for the entire order.

Step-by-step explanation:

Money to spend: $475

Cost per uniform: $29

16 uniforms: 16 x 29 = $464

Adding mailing fee: $464 + $6.50 = $469.50 out of $475 total

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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

What is 30/60 simplified and what do you multiply to get that?

maksim [4K]

1/2
Divide each number by 30

7 0

2 years ago

Explain whether the following given sets are closed under addition:

Ivan

Answer:

Step-by-step explanation:

a) Yes.

$\forall\ a,b\ \in\ B,\ a+b\in B\ (since\ 0+0=0)$

b) Yes

$W=\{0,1,2,3,..,\}\ (whole\ numbers)\\\forall\ a,b\ \in\ 3W,\ a+b\in 3W\\since:\\a\ \in\ 3W \Longrightarrow\ \exists\ k_1\in W | a=3*k_1\\\\b\ \in\ 3W \Longrightarrow\ \exists\ k_2\in W | b=3*k_2\\\\a+b=3*k_1+3*k_2=3*(k_1+k_2) \in 3W\\$

c) Yes

$\forall n\ \in\ \mathbb{N}: \Longrightarrow\ n+1 \in\ \mathbb{N}\\$

d) No

$3\in V , 3+3=6\notin V\\$

7 0

2 years ago

Factoriza 4x²-y²-6y-9

sveta [45]

$4 x^{2} -y^{2} -6y-9 \\ 4 x^{2} = y^{2} +6y+9 \\ 4 x^{2} =(y+3)(y+3) \\ \sqrt{ 4x^{2} } = \sqrt{(y+3)(y+3)} =\\2x=y+3\\2x-3=y$

y=2x-3

4 0

2 years ago

Read 2 more answers

Factor this expression.<br> x2 + 9x + 8

Ket [755]

Answer:

(x+1)(x+8) When factoring squares whose squared coefficient is one the roots must add up to the coefficient of the slope and multiply out to the intercept value.

6 0

2 years ago