24 is 33 1/3% of what number

(10v^4 + 30v^3 + 2v^2) / 10v

Zina [86]

Given expression is

$\frac{10v^4+30v^3+2v^2}{10v}$

Now we need to simplify this expression. We can begin by factoring numerator and denominator if possible.

$= \frac{2v^2(5v^2+15v+1)}{10v}$

divide numerator and denominator both by 2v

$= \frac{v(5v^2+15v+1)}{5}$

$= \frac{5v^3+15v^2+1v}{5}$

Hence final answer is $\frac{5v^3+15v^2+1v}{5}$

Or you can also keep factored form

$\frac{v(5v^2+15v+1)}{5}$

3 0

4 years ago

Line r is parallel to line c.

777dan777 [17]

Answer:

I think ∆5 is the correct Ans

3 0

3 years ago

Read 2 more answers

Y=−3(2.5)x

Hoochie [10]

Answer:

1. The equation represent an exponential decay

2. The rate of the exponential decay is -3×2.5ˣ·㏑(2.5)

Step-by-step explanation:

When a function a(t) = a₀(1 + r)ˣ has exponential growth, the logarithm of x grows with time such that;

log a(t) = log(a₀) + x·log(1 + r)

Hence in the equation -3 ≡ a₀, (1 + r) ≡ 2.5 and y ≡ a(t). Plugging in the values in the above equation for the condition of an exponential growth, we have;

log y = log(-3) + x·log(2.5)

Therefore, since log(-3) is complex, the equation does not represent an exponential growth hence the equation represents an exponential decay.

The rate of the exponential decay is given by the following equation;

$\frac{dy}{dx} =\frac{d(-3(2.5)^x)}{dx} = -\frac{d(3\cdot e^{x\cdot ln(2.5)})}{dx} = -3 \times 2.5^x\times ln(2.5)$

Hence the rate of exponential decay is -3×2.5ˣ × ㏑(2.5)

5 0

4 years ago

Who knows how to do this

Ivan

Area by Heron's formula:
semiperimeter = (15 + 20 + 25)/2 = 60/2 = 30 cm
$
Area = \sqrt{30(30-15)(30-20)(30-25)}= \sqrt{30 \cdot5\cdot10\cdot15}= \\\\ {} \ \ \ \ \ \ =\sqrt{22500} =150 \ cm^2$

The formula of the area of a triangle: Area = 1/2 * base * height
1/2 * 25 * height = 150
12.5 * height = 150
height = 150/12.5
height = 12 cm

3 0

3 years ago

The J.R. Ryland Computer Company is considering a plant expansion to enable the company to begin production of a new computer pr

Solnce55 [7]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the data:

Medium-Scale Large-Scale

Expansion Profit Expansion Profit

x f(x) y f(y)

Low 50 0.2 0 0.2

Demand Medium 150 0.5 100 0.5

High 200 0.3 300 0.3

a. Compute the expected value for the profit associated with the two expansion alternatives.

Which decision is preferred for the objective of maximizing the expected profit?

Expected value for medium scale expansion profit :

Expected value (E) = Σ(X) * f(x)

Σ[(50 * 0.2) + (150 * 0.5) + (200 * 0.3)]

= 145

Expected value for Large scale expansion profit :

Expected value (E) = Σ(X) * f(x)

Σ[(0 * 0.2) + (100 * 0.5) + (300 * 0.3)]

= 140

Medium scale expansion profit is preferred as it has the highest expected value.

b. Compute the variance for the profit associated with the two expansion alternatives.

Which decision is preferred for the objective of minimizing the risk or uncertainty?

Variance (V) = Σ(X - E)² * f(x):

Variance for medium scale expansion profit :

V = [((50-145)^2 * 0.2) + ((150-145)^2 * 0.5) + ((200-145)^2 * 0.3) = 2725

Variance for Large scale expansion profit :

V = [((0-140)^2 * 0.2) + ((100-140)^2 * 0.5) + ((300-140)^2 * 0.3) = 12400

Smaller variance is required to minimize risk, Hence, choose the medium scale expansion profit.

3 0

3 years ago