I need 7-10 pictured

If a/2 = b/3, then b/a = ? A. 3/2 B. 2/3

Paul [167]

It would be A because b+a

4 0

3 years ago

If u help me your dreams will come true ya know

wariber [46]

Answer:

Area of larger circle = $\pi *r^2=3.14(11^2)=379.94$

Area of smaller circle = $\pi *r^2=3.14(3^2)=28.26$

Probability that it lands in the shaded area (smaller circle):

28.26 ÷ 379.94 = 0.07438... = 0.07 (nearest hundredth)

8 0

2 years ago

There are 12 inches per foot. How many feet is 96 inches? Enter only the number.

Ierofanga [76]

8, the answer needs to be more than a certain amount of characters.

7 0

2 years ago

Read 2 more answers

Which equation can be used to find the length of Line segment A C? Triangle A B C is shown. Angle A C B is 90 degrees and angle

taurus [48]

Answer:

AC = 10sin(40°)

Explanation:

The diagram representing the question is shown in the attached image

Since the given triangle is a right-angled triangle, we can apply the special trig functions

These functions are as follows:

sin(θ) = opposite / hypotenuse

cos(θ) = adjacent / hypotenuse

tan(θ) = opposite / adjacent

Now, in the given diagram:

θ = 40°

AC is the side opposite to θ

AB = 10 in is the hypotenuse

Based on these givens, we will use the sin(θ) function

Therefore:

$sin(40) = \frac{AC}{10}\\ \\AC = 10sin(40)$

5 0

3 years ago

Read 2 more answers

A 500-gallon tank initially contains 220 gallons of pure distilled water. Brine containing 5 pounds of salt per gallon flows int

Wittaler [7]

Answer: The amount of salt in the tank after 8 minutes is 36.52 pounds.

Step-by-step explanation:

Salt in the tank is modelled by the Principle of Mass Conservation, which states:

(Salt mass rate per unit time to the tank) - (Salt mass per unit time from the tank) = (Salt accumulation rate of the tank)

Flow is measured as the product of salt concentration and flow. A well stirred mixture means that salt concentrations within tank and in the output mass flow are the same. Inflow salt concentration remains constant. Hence:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = \frac{d(V_{tank}(t) \cdot c(t))}{dt}$

By expanding the previous equation:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt} + \frac{dV_{tank}(t)}{dt} \cdot c(t)$

The tank capacity and capacity rate of change given in gallons and gallons per minute are, respectivelly:

$V_{tank} = 220\\\frac{dV_{tank}(t)}{dt} = 0$

Since there is no accumulation within the tank, expression is simplified to this:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt}$

By rearranging the expression, it is noticed the presence of a First-Order Non-Homogeneous Linear Ordinary Differential Equation:

$V_{tank} \cdot \frac{dc(t)}{dt} + f_{out} \cdot c(t) = c_0 \cdot f_{in}$ , where $c(0) = 0 \frac{pounds}{gallon}$ .

$\frac{dc(t)}{dt} + \frac{f_{out}}{V_{tank}} \cdot c(t) = \frac{c_0}{V_{tank}} \cdot f_{in}$

The solution of this equation is:

$c(t) = \frac{c_{0}}{f_{out}} \cdot ({1-e^{-\frac{f_{out}}{V_{tank}}\cdot t }})$

The salt concentration after 8 minutes is:

$c(8) = 0.166 \frac{pounds}{gallon}$

The instantaneous amount of salt in the tank is:

$m_{salt} = (0.166 \frac{pounds}{gallon}) \cdot (220 gallons)\\m_{salt} = 36.52 pounds$

3 0

3 years ago