Ask question

Ask question

All categories

3 years ago

6

Show work and factor ?

1 answer:

kvasek [131]3 years ago

4 0

$5\cdot\dfrac{x^2-2x}{6}:\dfrac{3x-6}{x}=5\cdot\dfrac{x(x-2)}{6}\cdot\dfrac{x}{3x-6}\\\\=5\cdot\dfrac{x(x-2)}{6}\cdot\dfrac{x}{3(x-2)}=\dfrac{5\cdot x\cdot x}{6\cdot3}=\dfrac{5x^2}{18}$

You might be interested in

How much pure water must be mixed with 10 liters of a 25% acid solution to reduce it to a 10% acid solution.

Nadya [2.5K]

The answer is B

15 L of pure water must be mixed with 10 L of a 25% acid solution to reduce it to a 10% acid solution

3 0

4 years ago

Read 2 more answers

PLEASE SOMEONE HELP ANWSER FOR 23 PTS

SVEN [57.7K]

I’m sorry but I don’t understand what you need help with ?

4 0

3 years ago

Circle P is described by the equation (x−1)2+(y+6)2=9 and circle Q is described by the equation (x+4)2+(y+14)2=4. Select from th

xeze [42]

Answer:

(a) Circle Q is 9.4 units to the center of circle P

(b) Circle Q has a smaller radius

Step-by-step explanation:

Given

$P:(x - 1)^2 + (y + 6)^2 = 9$

$Q:(x + 4)^2 + (y + 14)^2 = 4$

Solving (a): The distance between both

The equation of a circle is:

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

Where

$Center: (h,k)$

$Radius:r$

P and Q can be rewritten as:

$P:(x - 1)^2 + (y + 6)^2 = 3^2$

$Q:(x + 4)^2 + (y + 14)^2 = 2^2$

So, for P:

$Center = (1,-6)$

$r = 3$

For Q:

$Center = (-4,-14)$

$r = 2$

The distance between them is:

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2$

Where:

$Center = (1,-6)$ --- $(x_1,y_1)$

$Center = (-4,-14)$ --- $(x_2,y_2)$

So:

$d = \sqrt{(1 - -4)^2 + (-6 - -14)^2$

$d = \sqrt{(5)^2 + (8)^2$

$d = \sqrt{25 + 64$

$d = \sqrt{89$

$d = 9.4$

Solving (b): The radius;

In (a), we have:

$r = 3$ --- circle P

$r = 2$ --- circle Q

By comparison

$2 < 3$

<em>Hence, circle Q has a smaller radius</em>

4 0

3 years ago

Simplify the following expression.<br><br><br> 3^11/5 / 3^-9/5

Bess [88]

Answer:

81

Step-by-step explanation:

$\frac{3^{11/5}}{3^{-9/5}}$ can be written as $3^{11/5}$ × $3^{9/5}$

= $3^{20/5}$

= $3^4$

= 81

8 0

2 years ago

GUYS PLEASE HELP ME PLEASE HELP HELP PLEASE HELP ASAP GUYS HELP!!!!!

slega [8]

Answer:

$SA=5\ yd^2$

Step-by-step explanation:

we know that

The surface area of the square pyramid is equal to the area of the square base plus the area of its four triangular faces

so

$SA=b^2+4[\frac{1}{2}(b)(h)]$

we have

$b=1\ yd\\h=2\ yd$

substitute

$SA=1^2+4[\frac{1}{2}(1)(2)]$

$SA=5\ yd^2$

5 0

3 years ago