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4 years ago

What is the LCD of 1/2 and 3/5? A.2 B.7 C.10 D.20

Mathematics

2 answers:

sashaice [31]4 years ago

8 0

The least common denominator is 10.

10 / 2 = 5
10 / 5 = 2

There isn't a smaller number that 2 and 5 can get into.

Tcecarenko [31]4 years ago

4 0

The answer is C hope you get a A++++++ on this test and have a great day :D

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The population is decreasing at a rate of 5% per year. If the population is 30,000 today, what will the population be in 12 year

Lera25 [3.4K]

Answer:12,000

Step-by-step explanation:

.05*30,000=1,500

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4 years ago

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Help please solve<br> <img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B6x%5E5%2B11x%5E4-11x-6%7D%7B%282x%5E2-3x%2B1

Shkiper50 [21]

Answer:

$\displaystyle -\frac{1}{2} \leq x < 1$

Step-by-step explanation:

<u>Inequalities</u>

They relate one or more variables with comparison operators other than the equality.

We must find the set of values for x that make the expression stand

$\displaystyle \frac{6x^5+11x^4-11x-6}{(2x^2-3x+1)^2} \leq 0$

The roots of numerator can be found by trial and error. The only real roots are x=1 and x=-1/2.

The roots of the denominator are easy to find since it's a second-degree polynomial: x=1, x=1/2. Hence, the given expression can be factored as

$\displaystyle \frac{(x-1)(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)^2(x-\frac{1}{2})^2} \leq 0$

Simplifying by x-1 and taking x=1 out of the possible solutions:

$\displaystyle \frac{(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)(x-\frac{1}{2})^2} \leq 0$

We need to find the values of x that make the expression less or equal to 0, i.e. negative or zero. The expressions

$(6x^3+14x^2+10x+12)$

is always positive and doesn't affect the result. It can be neglected. The expression

$(x-\frac{1}{2})^2$

can be 0 or positive. We exclude the value x=1/2 from the solution and neglect the expression as being always positive. This leads to analyze the remaining expression

$\displaystyle \frac{(x+\frac{1}{2})}{(x-1)} \leq 0$

For the expression to be negative, both signs must be opposite, that is

$(x+\frac{1}{2})\geq 0, (x-1)$

$(x+\frac{1}{2})\leq 0, (x-1)>0$

Note we have excluded x=1 from the solution.

The first inequality gives us the solution

$\displaystyle -\frac{1}{2} \leq x < 1$

The second inequality gives no solution because it's impossible to comply with both conditions.

Thus, the solution for the given inequality is

$\boxed{\displaystyle -\frac{1}{2} \leq x < 1 }$

7 0

3 years ago

A line segment endpoints at (3, 2) and (2, -3). Which reflection will produce an image with endpoints at (3, -2) and (2, 3)

givi [52]

A reflection across the x-axis. Hope this helps!

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3 years ago

Which phrase is represented by the expression 6/r+5?

ANTONII [103]

Answer:

D.five more than the quotient of 6 and a number,

Step-by-step explanation:

To evaluate the the phrase written as :

6/r+5

Here, there is no bracket to make any distinction, hence using the BODMAS mode of evaluation, we deal with the division first before the sum. Hence we interpr te the equation as :

6 / r = quotient ; then, 5 is added to the quotient obtained from the divison of 6 by r.

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3 years ago