567+4+10. What is the answer

What is equivalent to (1/4) to the -3 power

Lyrx [107]

Exact Form: -3/4 (fraction)

Decimal Form: -0.75

3 0

4 years ago

Read 2 more answers

Please help!<br> What percentage of students scored between 70 and 90?image^

goblinko [34]

Answer:

50%

Step-by-step explanation:

50% of the data, in theory, resides between the first and third quartiles

3 0

3 years ago

Solve using the quadratic formula. Show all work. Write each solution in simplest form. No decimals.

alexira [117]

Answer:

Option B

Step-by-step explanation:

Given quadratic equation is,

12a² + 9a + 7 = 0

By comparing this equation with standard quadratic equation,

hx² + kx + c = 0

h = 12, k = 9 and c = 7

By using quadratic formula,

a = $\frac{-k\pm\sqrt{k^2-4hc}}{2h}$

= $\frac{-9\pm\sqrt{9^2-4(12)(7)}}{2(12)}$

= $\frac{-9\pm\sqrt{81-336}}{2(12)}$

= $\frac{-9\pm\sqrt{-255}}{24}$

= $\frac{-9\pm i\sqrt{255}}{24}$

a = $\frac{-9+ i\sqrt{255}}{24},\frac{-9- i\sqrt{255}}{24}$

Therefore, Option B will be the correct option.

3 0

3 years ago

5. A tank initially contains 20 lb of salt dissolved in 200-gallon of water. Starting at time t = 0, a solution containing 0.5 l

sladkih [1.3K]

Answer:

(a) Amount of salt as a function of time

$A(t)=100-80e^{-0.02t}$

(b) The time at which the amount of salt in the tank reaches 50 lb is 23.5 minutes.

(c) The amount of salt when t approaches to +inf is 100 lb.

Step-by-step explanation:

The rate of change of the amount of salt can be written as

$\frac{dA}{dt} =rate\,in\,-\,rate\,out\\\\\frac{dA}{dt}=C_i*q_i-C_o*q_o=C_i*q_i-\frac{A(t)}{V}*q_o\\\\\frac{dA}{dt}=0.5*4-\frac{A(t)}{200}*4\\\\\frac{dA}{dt} =\frac{100-A(t)}{50}=-(\frac{A(t)-100}{50})$

Then we can rearrange and integrate

$\frac{dA}{dt}= -(\frac{A(t)-100}{50})\\\\\int \frac{dA}{A-100}=-\frac{1}{50} \int dt\\\\ln(A-100)=-\frac{t}{50}+C_0\\\\ A-100=Ce^{-0.02*t}\\\\\\A(0)=20 \rightarrow 20-100=Ce^0=C\\C=-80\\\\A=100-80e^{-0.02t}$

Then we have the model of A(t) like

$A(t)=100-80e^{-0.02t}$

(b) The time at which the amount of salt reaches 50 lb is

$A(t)=100-80e^{-0.02t}=50\\\\e^{-0.02t}=(50-100)/(-80)=0.625\\\\-0.02*t=ln(0.625)=-0.47\\\\t=(-0.47)/(-0.02)=23.5$

(c) When t approaches to +infinit, the term e^(-0.02t) approaches to zero, so the amount of salt in the solution approaches to 100 lb.

7 0

3 years ago

If it takes 1,800 foot- pounds of work to push a rock fifteen feet, what amount of force was used?

ella [17]

Answer:

120 lbs

Step-by-step explanation:

Work = force * distance

1800 ft-lb = force * 15 ft

force = 1800 ft-lb / 15 ft

=1800/15 ft-lb/ft

=120lbs

6 0

4 years ago