Write the fraction equivalent (in simplest form) for each decimal.

Select the values of x that make the inequality x < -2 true.

ANTONII [103]

Answer:

A and B

Step-by-step explanation:

-3 and -2 are the only values less than or = to -2

the rest are greater than -2

8 0

3 years ago

Read 2 more answers

Fabian is purchasing a home for $89,500 and is financing 75% of it. The documentary stamp tax on the deed in his state is $0.70

ss7ja [257]

Answer: C) $462.96
Hope this will helpful.
Thank you.

6 0

3 years ago

DUE TODAY!

Leokris [45]

<em>Independent variables are variables of a quantity that are not affected by any conditions. </em>

<em>Dependent variables are variables of a quantity that change if conditions relative to that variable changes.</em>

For example, we generally we take x as independent variable by x variable and dependent variable by y variable.

To find the rate of change we get two values of independent variable (x's) and two values of dependent variables (y's) to get two coordinates in form of

(x,1,y1) and (x2,y2).

<h3>And we can find the rate of change by applying slope formula</h3>

$m= \frac{y_2-y_1}{x_2-x_1}$ .

3 0

3 years ago

Part A If their food and beverages cost $25.30 and there is an 8% meals tax, how much is the bill? (Find the 8% tax then add it

dmitriy555 [2]

Answer:

Step-by-step explanation:

Tax

Meal plus tax = 25.3 + 8/100 * 25.3

Meal plus tax = 25.3 + 2.024 Round later. Find the total.

Meal plus tax = 27.324

Meal plus tax = 27.32 rounded to the nearest cent.

Tip

Total cost = 27.324 + 27.324 * 18 /100

Total cost = 4.91832 + 27.324

Total cost = 32.242

Rounded to nearest cent = 32.24 You should always round at the last step.

3 0

3 years ago

A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min

sesenic [268]

$A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})$
$\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$
$A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4$

We're given that $A(0)=30$ . Multiply both sides by the integrating factor $e^{t/100}$ , then

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C$
$A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}$

Given that $A(0)=30$ , we have

$30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02$

so the amount of salt in the tank at time $t$ is

$A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}$

3 0

3 years ago