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2 years ago

A wise man once said, "400 reduced by 3 times my age is 142." What is his age

Mathematics

1 answer:

seropon [69]2 years ago

8 0

Answer: 40 / 3 = 142

Step-by-step explanation:

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The profit when selling 'x' pairs of shoes is P(x)=-4x^2+32x-39. How many pairs of shoes should be sold in order to maximize the

Vitek1552 [10]

Answer: $4$ pairs

Step-by-step explanation:

Given

Profit on selling 'x' pairs of shoes is $P(x)=-4x^2+32x-39$

Find the derivative of the function to find out the critical points

$\Rightarrow -8x+32=0\\\Rightarrow x=4$

Thus, for 4 pair of shoes profit is maximize.

5 0

2 years ago

9 more than 2 multiplied by f

goblinko [34]

In expression form this is:
9 + 2f
Hope this helps!

4 0

3 years ago

Read 2 more answers

Which answer choice shows the correct way to write the phrase?

timurjin [86]

Answer:

y + 1

Step-by-step explanation:

if y = the total balls

then y + 1 = total balls plus 1

8 0

2 years ago

Find the slope of each line and then determine if the lines are parallel, perpendicular or neither. If a value is not an integer

Aleonysh [2.5K]

As given by the question

There are given that the point of two-line

$\begin{gathered} (1,\text{ 7) and (5, 5)} \\ (-1,\text{ -3) and (1, 1)} \end{gathered}$

Now,

From the condition of a parallel and perpendicular line

If the slopes are equal then the lines are parallel

If the slopes are negative reciprocal then the lines are perpendicular

If the slopes are neither of the above are true then lines are neither

Then,

First, find the slope of both of line

So,

For first-line, from the formula of slope

$\begin{gathered} m_1=\frac{y_2-y_1}{x_2-x_1} \\ m_1=\frac{5_{}-7_{}}{5_{}-1_{}} \\ m_1=-\frac{2}{4} \\ m_1=-\frac{1}{2} \end{gathered}$

Now,

For second-line,

$\begin{gathered} m_1=\frac{y_2-y_1}{x_2-x_1} \\ m_1=\frac{1_{}-(-3)_{}}{1-(-1)_{}} \\ m_1=\frac{4}{2} \\ m_1=2 \end{gathered}$

The given result of the slope is negative reciprocal because

$\begin{gathered} -\frac{1}{2}=-(-\frac{2}{1}) \\ -\frac{1}{2}=2 \end{gathered}$

Hence, the slope of line1 is -1/2, and slope of line2 is 2 and the lines are perpendicular.

5 0

11 months ago

Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de

Ganezh [65]

Answer:

a. $\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

b. $\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

Step-by-step explanation:

The initial value problem is given as:

$y' +y = 7+\delta (t-3) \\ \\ y(0)=0$

Applying laplace transformation on the expression $y' +y = 7+\delta (t-3)$

to get $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

$y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$