What is x in the given diagram?

A frost has caused a temporary spike in the price of orange juice. the price is now $10 a gallon. before the frost, the price wa

Tema [17]

It is $6 more now than before...maybe you were looking for percent increase?

%change=100(final-initial)/initial

%change=100(10-4)/4=150% increase

3 0

2 years ago

Read 2 more answers

Help me with B please

salantis [7]

Area of each piece of paper is $8\frac{1}{2}\times11=\frac{17}{2}\times11=\frac{187}{2}$

Greatest number of possible piece of paper is $[1000\div\frac{187}{2}]=[1000\times\frac{2}{187}]=[\frac{2000}{187}]=10$

4 0

3 years ago

$34,300 loan at 3.5% for 4 months.

cestrela7 [59]

keeping in mind that 4 months is not even a year, since there are 12 months in a year, 4 months is then 4/12 years.

let's assume is simple interest.

$\bf ~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$34300\\ r=rate\to 3.5\%\to \frac{3.5}{100}\dotfill &0.035\\ t=years\to \frac{4}{12}\dotfill &\frac{1}{3} \end{cases} \\\\\\ A=34300\left[ 1+(0.035)\left( \frac{1}{3} \right) \right]\implies A= 34300(1.011\overline{6})\implies A=34700.1\overline{6}$

5 0

2 years ago

Tim installs 50 square feet of his floor in 45 minutes. At this rate, how long does it take him to install 495 square feet? (Ple

Alenkinab [10]

Answer:

445.5min=595²ft

Step-by-step explanation:

y=0.9x

y=0.9*495

y=445.5

5 0

3 years ago

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A 15 kilogram object is suspended from the end of a vertically hanging spring stretches the spring 1/3 meters. At time t = 0, th

Yuri [45]

Answer:

$15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) =$ ± $170 cos(5t)$

$y(0)=0$ , $y'(0)=0$

Step-by-step explanation:

See the attached image

This problem involves Newton's 2nd Law which is: ∑F = ma, we have that the acting forces on the mass-spring system are: $F_{r} (t)$ that correspond to the force of resistance on the mass by the action of the spring and $F(t)$ that is an external force with unknown direction (that does not specify in the enounce).

For determinate $F_{r} (t)$ we can use Hooke's Law given by the formula $F_{r} (t) = k y(t)$ where $k$ correspond to the elastic constant of the spring and $y(t)$ correspond to the relative displacement of the mass-spring system with respect of his rest state.

We know from the problem that an 15 Kg mass stretches the spring 1/3 m so we apply Hooke's law and obtain that...

$k = \frac{F_{r}}{y} = \frac{mg}{y} = \frac{15 Kg (9.81 \frac{m}{s^{2} } )}{\frac{1}{3} m} = 441.45 \frac{N}{m}$

Now we apply Newton's 2nd Law and obtaint that...

$F_{r} (t)$ ± $F(t)$ = $ma(t)$

$F_{r} (t) = ky(t) = 441.45y(t)$

$F(t) = 170 cos(5t)$

$m = 15 kg$

$a(t) = \frac{d^{2}y(t)}{dt^{2} }$

Finally... $15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) =$ ± $170 cos(5t)$

We know from the problem that there's not initial displacement and initial velocity, so... $y(0)=0$ and $y'(0)=0$

Finally the Initial Value Problem that models the situation describe by the problem is

$\left \{ 15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) = \frac{+}{} 170 cos(5t) \atop {y(0)=0, y'(0)=0\right.$

6 0

2 years ago