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3 years ago

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January 30, 2019 was the coldest day in Minneapolis-St paul i’m over 20 years the lowest temperature recorded that day was -28f

the warmest temperature was 15 higher

2 answers:

marissa [1.9K]3 years ago

8 0

Answer:

Whats the question?????

d1i1m1o1n [39]3 years ago

5 0

I think that's true..?

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SIMPLIFY (x + 5)(2x - 6)<br> A) 2x2 - 30 <br> B) 2x2 - 6x <br> C) 2x2 + 7x - 6 <br> D) 2x2 + 4x - 30

alina1380 [7]

Answer:

D) 2x²+4x-30

Step-by-step explanation:

You would multiply x and 2x; x and -6; 5 and 2x; 5 and -6

(x)(2x) = 2x²

(x)(-6) = -6x

(5)(2x) = 10x

(5)(-6) = -30

Now just put it back in the order that it was listed in the equation starting with (x)(2x)

This would give you 2x²-6x+10x-30

Simplify even further to get 2x²+4x-30

4 0

3 years ago

Read 2 more answers

Simplify 1/9 - 1/x over 1/81 - 1/x^2

cricket20 [7]

Answer:

9x over x+9

Step-by-step explanation:

7 0

2 years ago

A tank contains 60 kg of salt and 1000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain

MissTica

Answer:

(a) 60 kg; (b) 21.6 kg; (c) 0 kg/L

Step-by-step explanation:

(a) Initial amount of salt in tank

The tank initially contains 60 kg of salt.

(b) Amount of salt after 4.5 h

$\text{Let A = mass of salt after t min}\\\text{and }r_{i} = \text{rate of salt coming into tank}\\\text{and }r_{0} =\text{rate of salt going out of tank}$

(i) Set up an expression for the rate of change of salt concentration.

$\dfrac{\text{d}A}{\text{d}t} = r_{i} - r_{o}\\\\\text{The fresh water is entering with no salt, so}\\ r_{i} = 0\\r_{o} = \dfrac{\text{3 L}}{\text{1 min}} \times \dfrac {A\text{ kg}}{\text{1000 L}} =\dfrac{3A}{1000}\text{ kg/min}\\\\\dfrac{\text{d}A}{\text{d}t} = -0.003A \text{ kg/min}$

(ii) Integrate the expression

$\dfrac{\text{d}A}{\text{d}t} = -0.003A\\\\\dfrac{\text{d}A}{A} = -0.003\text{d}t\\\\\int \dfrac{\text{d}A}{A} = -\int 0.003\text{d}t\\\\\ln A = -0.003t + C$

(iii) Find the constant of integration

$\ln A = -0.003t + C\\\text{At t = 0, A = 60 kg/1000 L = 0.060 kg/L} \\\ln (0.060) = -0.003\times0 + C\\C = \ln(0.060)$

(iv) Solve for A as a function of time.

$\text{The integrated rate expression is}\\\ln A = -0.003t + \ln(0.060)\\\text{Solve for } A\\A = 0.060e^{-0.003t}$

(v) Calculate the amount of salt after 4.5 h

a. Convert hours to minutes

$\text{Time} = \text{4.5 h} \times \dfrac{\text{60 min}}{\text{1h}} = \text{270 min}$

b.Calculate the concentration

$A = 0.060e^{-0.003t} = 0.060e^{-0.003\times270} = 0.060e^{-0.81} = 0.060 \times 0.445 = \text{0.0267 kg/L}$

c. Calculate the volume

The tank has been filling at 6 L/min and draining at 3 L/min, so it is filling at a net rate of 3 L/min.

The volume added in 4.5 h is

$\text{Volume added} = \text{270 min} \times \dfrac{\text{3 L}}{\text{1 min}} = \text{810 L}$

Total volume in tank = 1000 L + 810 L = 1810 L

d. Calculate the mass of salt in the tank

$\text{Mass of salt in tank } = \text{1810 L} \times \dfrac{\text{0.0267 kg}}{\text{1 L}} = \textbf{21.6 kg}$

(c) Concentration at infinite time

$\text{As t $\longrightarrow \, -\infty,\, e^{-\infty} \longrightarrow \, 0$, so A $\longrightarrow \, 0$.}$

This makes sense, because the salt is continuously being flushed out by the fresh water coming in.

The graph below shows how the concentration of salt varies with time.

3 0

3 years ago

What’s 18x3+12-1x34+13-2x56?

Alina [70]

18 x 3 + 12 - 1 x 34 + 13 - 2 x 56 = 67

<=Work=>

18 x 3 = 54

1 x 34 = 34

2 x 56 = 112

(54) + 12 - (34) + 13 - (112)

66 - 34 + 13 - 112

32 + 13 - 112

45 - 112

= 67

8 0

3 years ago

Read 2 more answers

the measure of the angle is 88 degrees greater than its complement. what is the measure of the complement?

vazorg [7]

Let the angle be x and it's complement be c.

Then, x = c + 88 and x + c = 90

Substitute the first equation in the second.

(c+88) + c = 90
2c+88 = 90
2c = 2
c =1

Compliment = 1
Angle = 89

6 0

3 years ago