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2 years ago

What is 80 times 30 = plz help me

Mathematics

2 answers:

goblinko [34]2 years ago

6 0

Answer:

80 x 30 = 2,400

Step-by-step explanation:

Zarrin [17]2 years ago

3 0

80*30=2,400

I would recomend getting a calculator or searching one up online.

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11. A roofer calculates his bid price using the formula P = 1.85s + 4.2f, where s is the area of the roof in square feet and f i

storchak [24]

Replace f with 190, replace P with 4148 and solve for s:

4148 = 1.85s + 4.2(190)

Simplify:

4148 = 1.85 + 798

Subtract 798 from both sides:

3350 = 1.85s

Divide both sides by 1.85:

s = 1,810.81

Rounded to nearest square foot = 1811 square feet.

8 0

3 years ago

Read 2 more answers

Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g

AysviL [449]

Answer:

The quantity of salt at time t is $m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$ , where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

$\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}$

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

$(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}$

Where:

$c$ - The salt concentration in the tank, as well at the exit of the tank, measured in $\frac{pd}{gal}$ .

$\frac{dc}{dt}$ - Concentration rate of change in the tank, measured in $\frac{pd}{min}$ .

$V$ - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

$V \cdot \frac{dc}{dt} + 6\cdot c = 3$

$60\cdot \frac{dc}{dt} + 6\cdot c = 3$

$\frac{dc}{dt} + \frac{1}{10}\cdot c = 3$

This equation is solved as follows:

$e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }$

$\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }$

$e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt$

$e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C$

$c = 30 + C\cdot e^{-\frac{t}{10} }$

The initial concentration in the tank is:

$c_{o} = \frac{10\,pd}{60\,gal}$

$c_{o} = 0.167\,\frac{pd}{gal}$

Now, the integration constant is:

$0.167 = 30 + C$

$C = -29.833$

The solution of the differential equation is:

$c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }$

Now, the quantity of salt at time t is:

$m_{salt} = V_{tank}\cdot c(t)$

$m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$

Where t is measured in minutes.

7 0

3 years ago

Which equation represents the line that passes through the points (6,-3) and (-4,-9)?

Citrus2011 [14]

I will give in y=mx+b and ax+by=c form

using point slope form
y-y1=m(x-x1)
where the slope is m and a point on that line is (x1,y1)

slope between 2 points (x1,y1) and (x2,y2) is (y2-y1)/(x1-x1)

given
(6,-3) and (-4,-9)
slope=(-9-(-3))/(-4-6)=(-9+3)/(-10)=-6/-10=3/5
a point is (6,-3)

y-(-3)=3/5(x-6)
y+3=3/5(x-6)
y+3=3/5x-18/5
y=3/5x-33/5
-3/5x+y=-33/5
3x-5y=33

equation is
y=3/5x-33 or
3x-5y=33 or
y+3=3/5(x-6) or
y+9=3/5(x+4)

7 0

3 years ago

I dont understand this I need to show work to. Please help me.

ra1l [238]

Answer:

a. m = Vd

b. 6.012g

Step-by-step explanation:

a. you need to rearrange the equation so that you're solving for m. you can do so by multiplying both sides by V so you end up with

Vd = m or m = Vd (i prefer writing it this was purely cause it looks better)

b. now, you can plug in each value into the new equation so you get

5.01g/cm3 x 1.2 cm3 = m

from here you just multiply as normal

6 0

2 years ago

What is 2(x+6) in mathematics

Tanya [424]

Answer:

2x+12, the answer was derived by multiplying the variable in the bracket by the two

Step-by-step explanation:

3 0

2 years ago