All categories

3 years ago

Find the lcd of the pair of expressions

1 answer:

4 0

$\frac{2}{7} {x}^{10} {y}^{4} \: and \: \frac{2}{14} {x}^{8} {y}^{7} \\ lcd = 14 {x}^{10} {y}^{7}$
Answer is
$14 {x}^{10} {y}^{7}$

When it's powered numbers, remember to choose the one with the largest power. For example,
${x}^{10}$
And
${x}^{8}$
Choose 10, same goes to y, where you choose 7 not 4 since 7 is larger.

Hope this helps. - M

You might be interested in

Multiply integers -14(5)

Serga [27]

Negative times positive. The answer is -70

6 0

3 years ago

I need help on this question

dem82 [27]

The answer you are looking for is Catagorial Data.

6 0

3 years ago

The perimeter of the triangle is 4x+3y. Find the measure of the missing side.

Katena32 [7]

What picture theres no pic

5 0

3 years ago

A forestry company hopes to generate credits from the carbon sequestered in its plantations.The equation describing the forest w

Dmitry [639]

Answer:

Explanation:

Given:

The equation describing the forest wood biomass per hectare as a function of plantation age t is:

y(t) = 5 + 0.005t^2 + 0.024t^3 − 0.0045t^4

The equation that describes the annual growth in wood biomass is:

y ′ (t) = 0.01t + 0.072t^2 - 0.018t^3

To find:

a) The year the annual growth achieved its highest possible value

b) when does y ′ (t) achieve its highest value?

To determine the year the highest possible value was achieved, we will set the derivative y'(t) to zero. The values of t will be substituted into the second derivative to get the highest value

$\begin{gathered} 0\text{ = 0.01t + 0.072t}^2\text{ - 0.018t}^3 \\ using\text{ a graphing tool,} \\ t\text{ = -0.1343} \\ \text{t = 0} \\ t\text{ = 4.1343} \\ \\ Since\text{ we can't have time as a negative value, t = 0 or t = 4.1343} \end{gathered}$ $\begin{gathered} To\text{ ascertain which is the maximum, we take a second derivative} \\ y^{\prime}\text{'\lparen t\rparen = 0.01 + 0.144t - 0.054t}^2 \\ \\ substitute\text{ t = 0 and t = 4.13 respectively} \\ when\text{ t = 0 } \\ y^{\prime}\text{'\lparen t\rparen= 0.01 + 0.144\lparen0\rparen - 0.054\lparen0\rparen}^2 \\ y^{\prime}\text{'\lparen t\rparen= 0.01 + 0 - 0} \\ y^{\prime}\text{'\lparen t\rparen= 0.01 > 0} \\ \\ y^{\prime}\text{'\lparen t\rparen= 0.01 + 0.144\lparen4.13\rparen - 0.054\lparen4.13\rparen}^2 \\ y^{\prime}\text{'\lparen t\rparen= -0.316 < 0} \\ \\ if\text{ }y^{\prime}\text{'\lparen t\rparen > 0 , then it is minimum, } \\ if\text{ }y^{\prime}\text{'\lparen t\rparen < 0, then it is maximum} \end{gathered}$

SInce t = 4.13, gives y ′' (t) = -0.316 (< 0). This makes it the maximum value of t

The year the annual growth achieved its highest possible value to the nearest whole number will be

year 4

b) y ′ (t) will achieve its highest value, when we substitute the value of t that gives into the initial function.

Initial function: y(t) = 5 + 0.005t^2 + 0.024t^3 − 0.0045t^4

$undefined$

6 0

1 year ago

What inequality does the number line graph<br> represent?

Andrew [12]

Answer:

-4

-4 and so on

4 0

3 years ago