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4 years ago

What is the value of the rational expression 2x+1/x^2 when x=5?

Mathematics

2 answers:

aleksandrvk [35]4 years ago

6 0

Answer:

{2x + 1}{ {x}^{2} } \\ = \dfrac{2(5) + 1}{ {5}^{2} } \\ = \{11}{25}

Answer is 11/25.

Step-by-step explanation:

Troyanec [42]4 years ago

5 0

Answer:

$\frac{11}{25}$

Step-by-step explanation:

Substitute x = 5 into the rational expression

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

= $\frac{10+1}{25}$

= $\frac{11}{25}$

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What is the total volume of glass used to make the award?

Sholpan [36]

152 inches cubes because when I done this in my class he gave us the answer

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3 years ago

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The parabola y = x² - 4 opens: <br> A.) up <br> B.) down<br> C.) right<br> D.) left

fomenos

Answer:

Step-by-step explanation:

Here the easy rules to remember the orientation of the parabolas are

a) If x is squared it opens up or down. And its coefficient of { $x^{2}[tex] is negative it opens down.b) If y is squared it opens side ways right or left. It its coefficient of [tex]y^{2}$

Hence in our equation of parabola

$y = x^ 2-4$

x is squared and its coefficient is positive , hence it opens up towards positive y axis.

3 0

3 years ago

The drama club sold bags of candy and cookies to raise money for the spring show. Bags of candy cost $5.50 and bags of cookies c

vaieri [72.5K]

Let's assume

the number of bags of candy =x

the number of bags of cookies =y

There were 5 more bags of cookies sold than candy

so, we get

$y=x+5$

Bags of candy cost $5.50

so, total cost of candy is

$=5.50x$

bags of cookies cost $3.50

so, total cost of cookies is

$=3.50y$

so,

total cost = (total cost of candy)+(total cost of cookies)

we get

total cost =5.50x+3.50y

so, we can plug

$5.50x+3.50y=53.50$

now, we can plug y

$5.50x+3.50(x+5)=53.50$

$5.5x\cdot \:10+3.5\left(x+5\right)\cdot \:10=53.5\cdot \:10$

$55x+35\left(x+5\right)=535$

$55x+35x+175=535$

$90x+175=535$

$90x+175-175=535-175$

$90x=360$

$x=4$

now, we can find y

$y=x+5$

$y=4+5$

$y=9$

so,

the number of bags of candy =4

the number of bags of cookies =9............Answer

5 0

3 years ago

Lesson: 1.08Given this function: f(x) = 4 cos(TTX) + 1Find the following and be sure to show work for period, maximum, and minim

Ber [7]

The given function is

$f(x)=4\cos \text{(}\pi x)+1$

The general form of the cosine function is

$y=a\cos (bx+c)+d$

a is the amplitude

2pi/b is the period

c is the phase shift

d is the vertical shift

By comparing the two functions

a = 4

b = pi

c = 0

d = 1

Then its period is

$\begin{gathered} \text{Period}=\frac{2\pi}{\pi} \\ \text{Period}=2 \end{gathered}$

The equation of the midline is

$y_{ml}=\frac{y_{\max }+y_{\min }}{2}$

Since the maximum is at the greatest value of cos, which is 1, then

$\begin{gathered} y_{\max }=4(1)+1 \\ y_{\max }=5 \end{gathered}$

Since the minimum is at the smallest value of cos, which is -1, then

$\begin{gathered} y_{\min }=4(-1)+1 \\ y_{\min }=-4+1 \\ y_{\min }=-3 \end{gathered}$

Then substitute them in the equation of the midline

$\begin{gathered} y_{ml}=\frac{5+(-3)}{2} \\ y_{ml}=\frac{2}{2} \\ y_{ml}=1 \end{gathered}$

The answers are:

Period = 2

Equation of the midline is y = 1

Maximum = 5

Minimum = -3

3 0

1 year ago

A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and

goldfiish [28.3K]

Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

$\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}$

Solve for S(t):

$\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}$

$e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}$

The left side is the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}$

Integrate both sides:

$e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt$

$e^{t/800}S(t)=64e^{t/800}+C$

$S(t)=64+Ce^{-t/800}$

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

$0=64+C\impleis C=-64$

and so (b) the amount of sugar in the tank at time t is

$S(t)=64\left(1-e^{-t/800}\right)$

As $t\to\infty$ , the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

7 0

4 years ago