A circular garden has a circumference of

Calculate the surface area of a piston with a diameter of 2.25inches

Talja [164]

i cant tell you that it is from a test

6 0

3 years ago

If BOTH equations are in slope-intercept form then the--------------? method would be best, but the------------? method would al

nadezda [96]

Answer:

Step-by-step explanation:

If BOTH equations are in slope-intercept form then the-graphing-? method would be best, but the-substitution-? method would also be effective since both y's are already by itself.

If ONE of the equations is solved for x or y and the other equation is not, then the-substitution-? method is best.

If BOTH equations are lined up in standard form & the coefficients of x or y are opposites then the BEST method is definitely the-elimination--? method.

If BOTH equations are lined up in standard form the elimination method would be best. But if the coefficient of x or y is 1, then the-substitution--? method is also effective.

5 0

2 years ago

Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite

zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt be a gram/L

Let the amount salt in the tank at any time t be Q(t).

$\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}$

Incoming rate = (a g/L)×(1 L/min)

=a g/min

The concentration of salt in the tank at any time t is = $\frac{Q(t)}{10}$ g/L

Outgoing rate = $(\frac{Q(t)}{10} g/L)(1 L/ min)$ $\frac{Q(t)}{10} g/min$

$\frac{dQ}{dt} = a- \frac{Q(t)}{10}$

$\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt$

Integrating both sides

$\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt$

$\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c$ [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

$\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c$

$\Rightarrow -log|10a-15|-2=c$

Therefore ,

$-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2$ .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0 in equation (1) we get

$- log|10a|= -log|10a-15| -2$

$\Rightarrow- log|10a|+log|10a-15|= -2$

$\Rightarrow log|\frac{10a-15}{10a}|= -2$

$\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}$

$\Rightarrow 1-\frac{15}{10a} =e^{-2}$

$\Rightarrow \frac{15}{10a} =1-e^{-2}$

$\Rightarrow \frac{3}{2a} =1-e^{-2}$

$\Rightarrow2a= \frac{3}{1-e^{-2}}$

$\Rightarrow a = 1.73$

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0

3 years ago

12. A carpenter made two tables. The first table is 4 feet

Arte-miy333 [17]

Answer:

a. True

b. True

c. False

d. False

Step-by-step explanation:

<u>First Table</u>

Length=10 feet, Width=4 feet

Area =10 X 4=40 Square Feet

<u>Second Table</u>

The second table is half as long as the first table.

Length =0.5 X 10 =5 feet

Area=5 X Width= 5w

The area of the second table is one fourth the area of the first table.

Area of Second Table = $\frac{1}{4}$ X Area of first table

5w= $\frac{1}{4}$ X 40

5w=10

Width of the Second table, w=2 feet

The following are true.

(a)The width of the second table is 2 feet.

(b) The area of the second table is 10 square feet.

6 0

2 years ago

A regular 10-sided polygon is called a decagon. What is the area of the decagon that has sides that measure 4.5 inches and an ap

scoray [572]

Answer:

94.5 square inches :)

Step-by-step explanation:

3 0

3 years ago