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2 years ago

9

Fill in the blank with the correct factor, dividend, divisor or product to complete the number sentence correctly. ____ divided

by 1.5 = (-6)

2 answers:

hoa [83]2 years ago

7 0

You can solve this equation easily if you understand that to find the answer you have to do the opposite sign of the question. Multiply 1.5 by -6 and that should give you -9. From there plug in -9/1.5= -6 and you should get -6

slavikrds [6]2 years ago

5 0

Answer:

the answer is negative nine

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0.9<2x-0.7<1.5<br><img src="https://tex.z-dn.net/?f=0.9%20%3C%202x%20-%200.7%20%3C%201.5" id="TexFormula1" title="0.9 <

Nitella [24]

Answer:

0.8<x<1.1

Step-by-step explanation:

Take 0.9<2x-0.7<1.5 add everything by 0.7 to get 1.6<2x<2.2 then divide everything by 2 to get 0.8<x<1.1

3 0

3 years ago

Write an equation for the line parallel to y= -2x+1 that contains (-2,5)

grin007 [14]

Answer:

y = - 2x + 1

Step-by-step explanation:

Parallels line have the same slope, when an equation is in the form y= mx + b, m is the slope. In this problem slope = - 2

Now with the slope what is missing is the y-intercept, the problem says that the line contains the point (-2, 5), replacing that point in the equation you can solve it to find the y-intercept

y = mx + b

5 = (-2)(-2) + b

5 = 4 + b

1 + b

y = - 2x +1

5 0

3 years ago

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows

4 0

3 years ago

Translate each sentence into an equation

Mazyrski [523]

Answer:

1. X/2 = 4

2. 3x - 2 = 11

3. 2x - 5 = 15

4. 2x-1=12

5. 9x = 2(x+5)

Step-by-step explanation:

8 0

3 years ago

Why do prices increase when demand for a product is high?

andrew11 [14]

Answer:When demand exceeds supply, prices tend to rise. ... If there is a decrease in supply of goods and services while demand remains the same, prices tend to rise to a higher equilibrium price and a lower quantity of goods and services. The same inverse relationship holds for the demand for goods and services.

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers