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3 years ago

Express each product in the simplest form. A. wx/6 B. wx/3 C. 2/3/wx D. wx

Mathematics

1 answer:

____ [38]3 years ago

3 0

Answer:

A) $\frac{wx}{6}$

Step-by-step explanation:

$\frac{3wx}{6x}\cdot \frac{3wx}{9w}$

Multiply:

$\frac{3wx\cdot \:3wx}{6x\cdot \:9w}$

Cancel off the common factor "w":

$\frac{3x\cdot \:3wx}{6x\cdot \:9}$

Cancel off the common factor "x":

$\frac{3\cdot \:3wx}{6\cdot \:9}$

$\frac{9wx}{54}$

Cancel off the common factor "9":

$\frac{wx}{6}$

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Help please!!<br>Find the area of the figure!<br>

coldgirl [10]

Answer:

Step-by-step explanation:

(in X is 400) (in division is 0.000625) (in + its agin 400) ( and in minus is -40)

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3 years ago

In the figure below, Z is the center of the circle. Suppose that OR=4, ST= 4, UZ= 6, and VZ=4x-6. Find the following.

Ann [662]

Answer:

x = 3 units

SV = 2 units

Step-by-step explanation:

In $\odot Z$ QR and ST are chords such that:

QR = ST = 4

That is both the chords are equal in measure.

Equal chords are at equal distance from the center of the circle. Therefore,

VZ = UZ

4x - 6 = 6

4x = 6 + 6

4x = 12

x = 12/4

x = 3 units

Since, perpendicular dropped from the center of the circle bisects the chord.

$\therefore SV = \frac{1}{2} ST$

$\therefore SV = \frac{1}{2} \times 4$

$\therefore SV = 2\: units$

3 0

3 years ago

Solve 73 make sure to also define the limits in the parts a and b

Aleks04 [339]

73.

$f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}$

$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3$ $\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3$

Since we can't divide by zero, we need to find when:

$x^4-2x^2+144=0$

But before, we can factor the numerator and the denominator:

$\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}$

Now, we can conclude that the vertical asymptotes are located at:

$\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}$

so, for x = -3:

$\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty$ $\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty$

For x = 4:

$\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$ $\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$

4 0

1 year ago

A restaurant manager asks all of the customers this month to take an online survey (and get a free appetizer their next visit).

padilas [110]

Answer:

Which is the output of the formula =AND(12>6;6>3;3>9)?

TRUE

FALSE

Step-by-step explanation:

4 0

3 years ago

What is the square root of 69

Rasek [7]

The square root of 69 is 8.30.because 8.30*8.30 is 69

6 0

3 years ago

Read 2 more answers