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3 years ago

Please help with all and show work

Mathematics

1 answer:

AnnyKZ [126]3 years ago

5 0

9) > cause 1 fl oz is .125 of a cup and multiply that by 96 which is 12 cups which is less than 13 cups
10)> cause 1 lb is 16 oz, multiply 16 by 25 which is 400, which is greater than 384 cups
11) = cause 1 yd is 36 in, so you would multiply 36 in by 8, which is 288 inches, which is the same

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Coefficient of (2x+y^2)^5

BigorU [14]

(2x + y2)5 heres you answer your so welcome

7 0

3 years ago

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

Graph the system of linear inequalities, please explain the process in solving this problem so I know how to do ones similar to

astra-53 [7]

Answer:

See attached. (Both inequalities graph to the same region.)

Step-by-step explanation:

To graph an inequality, you first replace the inequality symbol by an equal sign. Then you graph the line (or curve) described by the equation. If the inequality symbol includes the "or equal to" case, then the line (or curve) is solid and is part of the solution set. If not, then the line (or curve) is dashed and is not part of the solution set.

Here, the line x+y=5, will bound the solution set, but is not part of it.

Note that the line -x -y = -5 (from the second inequality) is the same ilne.

_____

After you have graphed the boundary of the solution region, you need to determine which side of the boundary has the solutions. I find it easy to do this by locating an x- or y-term with a positive coefficient and seeing which side of the inequality symbol that lies on. Here, both the x-term and the y-term lie on the "is greater than" side of the symbol, so the half-plane is shaded above and to the right of the line, where x- and y-values are greater than they are on the line.

In the second inequality, you can multiply the whole thing by -1 to get

... x + y > 5 . . . . same as the first inequality

or you can add (x+y) to both sides to get

... 0 < -5 +x +y

Once again, the x- and y-values that satisfy this are greater than those that are on the boundary line.

_____

The graph seems to show only one equation. Both are there, but the dashed blue line sits on top of the dashed red line, so you can't see it. The purple shading indicates both red and blue shading in the same area.

===== =====

Of course you graph the boundary line the way you would any linear equation. In this form, it is not too difficult to see that both the x- and y-intercepts are 5. (When x=0, y=5; and when y=0, x=5 for the boundary line's equation x+y=5.) Draw the dotted line through these intercept points, add shading above, and you have your graph.