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sergiy2304 [10]
3 years ago
11

Which expression is a difference of cubes?

Mathematics
1 answer:
Tanzania [10]3 years ago
8 0

Answer:

Option 4 is correct.

Step-by-step explanation:

Given the following options

we have to choose the difference of cubes.

Option 1:

9w^{33}-y^{12}

9(w^{11})^{3}-(y^4)^3

which is not the difference of cubes

Option 2:

18p^{15}-q^{21}

18(p^5)^3-(q^7)^3

which is not the difference of cubes

Option 3:

36a^{22}-b^{16}

(6a^{11})^2-(b^8)^2

which is the difference of square

Option 4:

64c^{15}-a^{27}

(4c^5)^3-(a^9)^3

which is the difference of cubes.

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75 miles per 60 minutes.
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Use the description and table to graph the function, and determine the domain and range of f(x). Represent the domain and range
zmey [24]

The domain of the function is x > 0 and the range of the function is -∞ < f(x) < ∞

<h3>How to determine the domain?</h3>

The table represents a logarithmic function.

The domain represents the set of x values.

From the table, we can see that all the x values are positive values i.e. x > 0

Hence, the domain of the function is x > 0

<h3>How to determine the range?</h3>

From the table, the table outputs all zero, positive and negative numbers

This means that the range is the set of all real numbers.

Hence, the range of the function is -∞ < f(x) < ∞

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3 years ago
A number divided by 80 with a quotient of 7 and a remainder of 4
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3 years ago
Need help simplifying this
diamong [38]

The simplified answer is \frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}.

<u>Step-by-step explanation:</u>

$\frac{3 y+2 x}{z+2 x}-\frac{2 y-3 x}{3 x+y}-\frac{2 z(y+3 x)}{6 x^{2}+3 x z+2 x y+y z}

To add or subtract denominators of the fraction must be same.

If it is not the same, we must take LCM of the denominators. and so we can add the fractions.

To make the denominator same multiply the 1st term (\frac{3x+y}{3x+y}) and 2nd term by (\frac{z+2x}{z+2x})

= \frac{(3 y+2 x)(3 x+y)}{(z+2 x)(3 x+y)}-\frac{(2 y-3 x)(z+2 x)}{(3 x+y)(z+2 x)}-\frac{2 z(y+3 x)}{6 x^{2}+3 x z+2 x y+y z}

LCM of the denominators is 6x²+ 3xz + 2xy +yz.

Multiply the factors in the numerator.

= \frac{\left(6 x^{2}+3 y^{2}+11 x y\right)}{(z+2 x)(3 x+y)}-\frac{\left(2 y z+4 x y-3 x z-6 x^{2}\right)}{(3 x+y)(z+2 x)}-\frac{2 z y+6 x z}{6 x^{2}+3 x z+2 x y+y z}

Now, the denominators are same, you can subtract it.

= \frac{\left(6 x^{2}+6 x^{2}+11 x y-4 x y-2 y z-2 y z+3 x z-6 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}

= \frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}

Thus the simplified solution is  \frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}

4 0
3 years ago
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