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2 years ago

Helpppppppppppppppppppppppppppppppppppppppppppppppppppppppppp

Mathematics

2 answers:

Roman55 [17]2 years ago

6 0

Answer:

Your answer is 1.

Step-by-step explanation:

So to find the slope take the x and y values and subract them from eachother this looks like this:

-4 - 3 7 (two negatives make a positive)

---------- = -------

2 - 9 7(It doesn't matter the order you put the number 9-2 or 2-9)

This gives you a whole number of 1.

Hope this helped!

Oliga [24]2 years ago

5 0

Answer: the answer is A the slop is 1

Step-by-step explanation:

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A 400 gallon tank initially contains 100 gal of brine containing 50 pounds of salt. Brine containing 1 pound of salt per gallon

posledela

Answer:

The amount of salt in the tank when it is full of brine is 393.75 pounds.

Step-by-step explanation:

This is a mixing problem. In these problems we will start with a substance that is dissolved in a liquid. Liquid will be entering and leaving a holding tank. The liquid entering the tank may or may not contain more of the substance dissolved in it. Liquid leaving the tank will of course contain the substance dissolved in it. If Q(t) gives the amount of the substance dissolved in the liquid in the tank at any time t we want to develop a differential equation that, when solved, will give us an expression for Q(t).

The main equation that we’ll be using to model this situation is:

Rate of change of Q(t) = Rate at which Q(t) enters the tank – Rate at which Q(t) exits the tank

where,

Rate at which Q(t) enters the tank = (flow rate of liquid entering) x

(concentration of substance in liquid entering)

Rate at which Q(t) exits the tank = (flow rate of liquid exiting) x

(concentration of substance in liquid exiting)

Let y(t) be the amount of salt (in pounds) in the tank at time t (in seconds). Then we can represent the situation with the below picture.

Then the differential equation we’re after is

$\frac{dy}{dt} = (Rate \:in)- (Rate \:out)\\\\\frac{dy}{dt} = 5 \:\frac{gal}{s} \cdot 1 \:\frac{pound}{gal}-3 \:\frac{gal}{s}\cdot \frac{y(t)}{V(t)} \:\frac{pound}{gal}\\\\\frac{dy}{dt} =5\:\frac{pound}{s}-3 \frac{y(t)}{V(t)} \:\frac{pound}{s}$

$V(t)$ is the volume of brine in the tank at time t. To find it we know that at time 0 there were 100 gallons, 5 gallons are added and 3 are drained, and the net increase is 2 gallons per second. So,

$V(t)=100 + 2t$

We can then write the initial value problem:

$\frac{dy}{dt} =5-\frac{3y}{100+2t} , \quad y(0)=50$

We have a linear differential equation. A first-order linear differential equation is one that can be put into the form

$\frac{dy}{dx}+P(x)y =Q(x)$

where P and Q are continuous functions on a given interval.

In our case, we have that

$\frac{dy}{dt}+\frac{3y}{100+2t} =5 , \quad y(0)=50$

The solution process for a first order linear differential equation is as follows.

Step 1: Find the integrating factor, $\mu \left( x \right)$ , using $\mu \left( x \right) = \,{{\bf{e}}^{\int{{P\left( x \right)\,dx}}}$

$\mu \left( t \right) = \,{{e}}^{\int{{\frac{3}{100+2t}\,dt}}}\\\int \frac{3}{100+2t}dt=\frac{3}{2}\ln \left|100+2t\right|\\\\\mu \left( t \right) =e^{\frac{3}{2}\ln \left|100+2t\right|}\\\\\mu \left( t \right) =(100+2t)^{\frac{3}{2}$

Step 2: Multiply everything in the differential equation by $\mu \left( x \right)$ and verify that the left side becomes the product rule $\left( {\mu \left( t \right)y\left( t \right)} \right)'$ and write it as such.

$\frac{dy}{dt}\cdot \left(100+2t\right)^{\frac{3}{2}}+\frac{3y}{100+2t}\cdot \left(100+2t\right)^{\frac{3}{2}}=5 \left(100+2t\right)^{\frac{3}{2}}\\\\\frac{dy}{dt}\cdot \left(100+2t\right)^{\frac{3}{2}}+3y\cdot \left(100+2t\right)^{\frac{1}{2}}=5 \left(100+2t\right)^{\frac{3}{2}}\\\\\frac{dy}{dt}(y \left(100+2t\right)^{\frac{3}{2}})=5\left(100+2t\right)^{\frac{3}{2}}$

Step 3: Integrate both sides.

$\int \frac{dy}{dt}(y \left(100+2t\right)^{\frac{3}{2}})dt=\int 5\left(100+2t\right)^{\frac{3}{2}}dt\\\\y \left(100+2t\right)^{\frac{3}{2}}=(100+2t)^{\frac{5}{2} }+ C$

Step 4: Find the value of the constant and solve for the solution y(t).

$50 \left(100+2(0)\right)^{\frac{3}{2}}=(100+2(0))^{\frac{5}{2} }+ C\\\\100000+C=50000\\\\C=-50000$

$y \left(100+2t\right)^{\frac{3}{2}}=(100+2t)^{\frac{5}{2} }-50000\\\\y(t)=100+2t-\frac{50000}{\left(100+2t\right)^{\frac{3}{2}}}$

Now, the tank is full of brine when:

$V(t) = 400\\100+2t=400\\t=150$

The amount of salt in the tank when it is full of brine is

$y(150)=100+2(150)-\frac{50000}{\left(100+2(150)\right)^{\frac{3}{2}}}\\\\y(150)=393.75$

6 0

2 years ago

In triangle ABC, m∠ABC = (4x – 12)° and m∠ACB = (2x + 26)°. Yin says that if x = 19, the triangle must be equilateral. Is he cor

spin [16.1K]

He is not correct. If this is to be an equilateral triangle then all the angles must be the same measure. That means that 180/3 = 60°. They all have to equal 60°. If one of the angles measures 4x-12, then 4x-12=60. Solving for x, we will add 12 to both sides to get 4x=72. x here is 18. For the other angle measuring 2x+26, we would do the same. 2x+26=60. Subtract 26 from both sides to get 2x= 34 and x = 17. He was way off.

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2 years ago

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The table lists recommended amounts of food to order for 25 party guests. Amanda and Syndey are hosting a graduation party for 4

yan [13]

Answer:

Answer: 25 guests is 2.5 times more than 10

Multiply each order by 2.5

Chicken = 16 x 2.5 = 40

Lasagna = 7 x 2.5 = 17.5 , round to 18

Deli meat = 1.8 x 2.5 = 4.5, round to 5

Sliced cheese = 15 x 2.5 = 37.5, round to 38

Buns = 1 x 2.5 = 2.5, round to 3

Potato salad = 2 x 2.5 = 5

Step-by-step explanation:

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2 years ago

Which decimal is less than 3.19 A. 3.179 B. 3.2 C. 3.291 D. 3.1911

astra-53 [7]

Answer:

Step-by-step explanation:

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1 year ago

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A savings account deposit of $300 is to earn 5.8% interest. After how many years will the investment be worth $900?

KatRina [158]

900=300(1+0.058)^t divide each side by 300
3=1.059^t
Log(3)=t*log(1.058)
T=log(3)/log(1.058)
T==19.5 years

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2 years ago