What is the answer? Please! ill mark brainliest!

88+88+88+88+88+88+88+88 times 88=_________

Nastasia [14]

Answer:

It's your friend at school. 8360 is your answer.

Step-by-step explanation:

88 + 88

= 176 + 88

= 264 + 88

= 352 + 88

= 440 + 88

= 528 + 88

= 616 + 88 =

704 x 88

= 8360

3 0

3 years ago

Read 2 more answers

A patio has an area of 1420 square feet. If the length is 45 1/2, what is the width

Anni [7]

[ Answer ]

$\boxed{Width \ = \ 31\frac{19}{91} \ Ft^{2}}$

[ Explanation ]

Area = 1420

Width = $45\frac{1}{2}$

Area For Rectangle: Length · Width

Use division to find width:

1420 ÷ $45\frac{1}{2}$

Convert mixed numbers to fractions:

1420 = $\frac{1420}{1}$

$45\frac{1}{2}$ = $\frac{91}{2}$

$\frac{1420}{1}$ ÷ $\frac{91}{2}$

Use mixed multiplication:

$\frac{1420*2}{1*91}$

= $\frac{2140}{91}$

Simplify:

= $31\frac{19}{91}$

$\boxed{[ \ Eclipsed \ ]}$

7 0

3 years ago

Helpp I’ll Mark you brainiest if your correct

patriot [66]

The answer is 586.17 cm^2

6 0

2 years ago

Solve the inequality. x – 1 ≤ – 9

ohaa [14]

i think the answer is x ≤ -8

7 0

3 years ago

A swimming pool with a volume of 30,000 liters originally contains water that is 0.01% chlorine (i.e. it contains 0.1 mL of chlo

SpyIntel [72]

Answer:

$R_{in}=0.2\dfrac{mL}{min}$

$C(t)=\dfrac{A(t)}{30000}$

$R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

$A(t)=300+2700e^{-\dfrac{t}{1500}},$ A(0)=3000$

Step-by-step explanation:

The volume of the swimming pool = 30,000 liters

(a) Amount of chlorine initially in the tank.

It originally contains water that is 0.01% chlorine.

0.01% of 30000=3000 mL of chlorine per liter

A(0)= 3000 mL of chlorine per liter

(b) Rate at which the chlorine is entering the pool.

City water containing 0.001%(0.01 mL of chlorine per liter) chlorine is pumped into the pool at a rate of 20 liters/min.

$R_{in}=$ (concentration of chlorine in inflow)(input rate of the water)

$=(0.01\dfrac{mL}{liter}) (20\dfrac{liter}{min})\\R_{in}=0.2\dfrac{mL}{min}$

(c) Concentration of chlorine in the pool at time t

Volume of the pool =30,000 Liter

$Concentration, C(t)= \dfrac{Amount}{Volume}\\C(t)=\dfrac{A(t)}{30000}$

(d) Rate at which the chlorine is leaving the pool

$R_{out}=$ (concentration of chlorine in outflow)(output rate of the water)

$= (\dfrac{A(t)}{30000})(20\dfrac{liter}{min})\\R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

(e) Differential equation representing the rate at which the amount of sugar in the tank is changing at time t.

$\dfrac{dA}{dt}=R_{in}-R_{out}\\\dfrac{dA}{dt}=0.2- \dfrac{A(t)}{1500}$

We then solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{1500}=0.2\\$The integrating factor: e^{\int \frac{1}{1500}dt} =e^{\frac{t}{1500}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{1500}}+\dfrac{A}{1500}e^{\frac{t}{1500}}=0.2e^{\frac{t}{1500}}\\(Ae^{\frac{t}{1500}})'=0.2e^{\frac{t}{1500}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{1500}})'=\int 0.2e^{\frac{t}{1500}} dt\\Ae^{\frac{t}{1500}}=0.2*1500e^{\frac{t}{1500}}+C, $(C a constant of integration)\\Ae^{\frac{t}{1500}}=300e^{\frac{t}{1500}}+C\\$Divide all through by e^{\frac{t}{1500}}\\A(t)=300+Ce^{-\frac{t}{1500}}$

Recall that when t=0, A(t)=3000 (our initial condition)

$3000=300+Ce^{0}\\C=2700\\$Therefore:\\A(t)=300+2700e^{-\dfrac{t}{1500}}$

3 0

3 years ago