What are the answers to these questions Please help

Mathematics

1 answer:

svetlana [45]3 years ago

3 0

Toby ...

Look at the drawing I made.
It's attached to my answer.
You can count the blocks.

Also ... stay inside tonight.
There's still going to be more rain around here.
Good luck.

You might be interested in

Please please please help!

lys-0071 [83]

Answer:

6 pounds

Step-by-step explanation:

every 16 ounces =1 pound

so 96/16 = 6

6 *1 = 6

therefour 6 pounds

4 0

3 years ago

Water is flowing into a large spherical tank at a constant rate. Let V (t) be the volume of water in the tank at time t, and h(t

aleksley [76]

Answer:

See solutions for detail.

Step-by-step explanation:

a. $\frac{dV}{dt}$ is the instantaneous rate of change of volume given with respect to time, t.

The volume's rate of change is written as a function of time.

- $\frac{dh}{dt}$ is the rate of change in the height of water in the tank with respect to time, t.

b. $\frac{dV}{dt}-$ is the only constant. Water flows into the constant at a constant rate, say $6cm^3$ per minute.

c. $\frac{dV}{dt}$ is positive. Volume water in the take is increasing from time to time.

-The volume at time t=1 is greater than the volume at t=0, hence, it's a positive rate of change.

d. $\frac{dh}{dt}$ is a positive rate. The initial height of water in the tank is zero.

-The final height at time t is 0.25h. The height is increasing with time.

Hence, it is positive.

8 0

3 years ago

If the second number is subtracted from the sum of the first number and 2 times the third number, the result is 1. The thrid num

weeeeeb [17]

Answer:

Step-by-step explanation:

$x,\ y,\ z-\text{three numbers}\\\\\left\{\begin{array}{ccc}(x+2z)-y=1&(1)\\z+2x=3&(2)\\x+3y+z=18&(3)\end{array}\right\\\\(2)\\z+2x=3\qquad\text{subtract}\ 2x\ \text{from both sides}\\z=3-2x\qquad(*)\\\\\text{Substitute}\ (*)\ \text{to (1) and (3)}\\\\\left\{\begin{array}{ccc}x+2(3-2x)-y=1&\text{use the distributive property}\\x+3y+(3-2x)=18\end{array}\right$

$\left\{\begin{array}{ccc}x+(2)(3)+(2)(-2x)-y=1\\x+3y+3-2x=18&\text{subtract 3 from both sides}\end{array}\right\\\left\{\begin{array}{ccc}x+6-4x-y=1&\text{subtract 6 from both sides}\\(x-2x)+3y=15\end{array}\right\\\left\{\begin{array}{ccc}(x-4x)-y=-5\\-x+3y=15\end{array}\right\\\left\{\begin{array}{ccc}-3x-y=-5&\text{multiply both sides by 3}\\-x+3y=15\end{array}\right$

$\underline{+\left\{\begin{array}{ccc}-9x-3y=-15\\-x+3y=15\end{array}\right}\qquad\text{add all sides of the equations}\\.\qquad-10x=0\qquad\text{divide both sides by (-10)}\\.\qquad\boxed{x=0}\\\\\text{Put it to the second equation:}\\-0+3y=15\\3y=15\qquad\text{divide both sides by 3}\\\boxed{y=5}\\\\\text{Put the value of}\ x\ \text{to}\ (*):\\\\z=3-2(0)\\\boxed{z=3}$