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3 years ago

Find the LCM (least common multiple) of (5x-9) and (3x+ 8)

1 answer:

5 0

In order to get the LCM of two binomial expressions (5x - 9) and (3x + 8), you need to factor first each binomial if factorable.

In this given binomial expression (5x - 9) and (3x + 8), you do not need to factor both because they are not factorable.

Next thing to do is to multiply the two binomial expressions by using FOIL method.
(5x - 9)(3x + 8) = 15x^2 + 40x - 27x - 72 = 15x^2 + 13x - 72

Therefore, the LCM of (5x - 9) and (3x + 8) is 15x^2 + 13x- 72.

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What is the area of a circle that has a diameter if 6 inches?

Taya2010 [7]

Answer:

28.27

Step-by-step explanation:

A=(1/4) πd^2

6 0

3 years ago

Read 2 more answers

Simplify and write the trigonometric expression in terms of sine and cosine:

asambeis [7]

Answer:

we have the expression as;

1/sin u cos u

Step-by-step explanation:

tan u = sin u/cos u

cot u = cos u/sin u

Thus;

sin u/cos u + cos u/sin u

The lcm is sin u cos u

Thus, we have that;

(sin^2 u + cos^2 u)/sin u cos u

But ; sin^2 u + cos^2 u = 1

so we have ;

1/sin u cos u

4 0

3 years ago

A box designer has been charged with the task of determining the surface area of various open boxes (no lid) that can be constru

Viktor [21]

Answer:

1) $S = 2\cdot w\cdot l - 8\cdot x^{2}$ , 2) The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$ , 3) $S = 176\,in^{2}$ , 4) $x \approx 4.528\,in$ , 5) $S = 164.830\,in^{2}$

Step-by-step explanation:

1) The function of the box is:

$S = 2\cdot (w - 2\cdot x)\cdot x + 2\cdot (l-2\cdot x)\cdot x +(w-2\cdot x)\cdot (l-2\cdot x)$

$S = 2\cdot w\cdot x - 4\cdot x^{2} + 2\cdot l\cdot x - 4\cdot x^{2} + w\cdot l -2\cdot (l + w)\cdot x + l\cdot w$

$S = 2\cdot (w+l)\cdot x - 8\cdpt x^{2} + 2\cdot w \cdot l - 2\cdot (l+w)\cdot x$

$S = 2\cdot w\cdot l - 8\cdot x^{2}$

2) The maximum cutout is:

$2\cdot w \cdot l - 8\cdot x^{2} = 0$

$w\cdot l - 4\cdot x^{2} = 0$

$4\cdot x^{2} = w\cdot l$

$x = \frac{\sqrt{w\cdot l}}{2}$

The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$

3) The surface area when a 1'' x 1'' square is cut out is:

$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1\,in)^{2}$

$S = 176\,in^{2}$

4) The size is found by solving the following second-order polynomial:

$20\,in^{2} = 2 \cdot (8\,in)\cdot (11.5\,in)-8\cdot x^{2}$

$20\,in^{2} = 184\,in^{2} - 8\cdot x^{2}$

$8\cdot x^{2} - 164\,in^{2} = 0$

$x \approx 4.528\,in$

5) The equation of the box volume is:

$V = (w-2\cdot x)\cdot (l-2\cdot x) \cdot x$

$V = [w\cdot l -2\cdot (w+l)\cdot x + 4\cdot x^{2}]\cdot x$

$V = w\cdot l \cdot x - 2\cdot (w+l)\cdot x^{2} + 4\cdot x^{3}$

$V = (8\,in)\cdot (11.5\,in)\cdot x - 2\cdot (19.5\,in)\cdot x^{2} + 4\cdot x^{3}$

$V = (92\,in^{2})\cdot x - (39\,in)\cdot x^{2} + 4\cdot x^{3}$

The first derivative of the function is:

$V' = 92\,in^{2} - (78\,in)\cdot x + 12\cdot x^{2}$

The critical points are determined by equalizing the derivative to zero:

$12\cdot x^{2}-(78\,in)\cdot x + 92\,in^{2} = 0$

$x_{1} \approx 4.952\,in$

$x_{2}\approx 1.548\,in$

The second derivative is found afterwards:

$V'' = 24\cdot x - 78\,in$

After evaluating each critical point, it follows that $x_{1}$ is an absolute minimum and $x_{2}$ is an absolute maximum. Hence, the value of the cutoff so that volume is maximized is: