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2 years ago

(6x -8)(2x -3) B) 12x2 + 24 U) 70x L) 8x2 - 21x + 11 E) 12x2 -34x + 24

Mathematics

2 answers:

deff fn [24]2 years ago

5 0

Answer is E!
12x^2-34x+24

ss7ja [257]2 years ago

4 0

Answer:

E) 12x^2 - 34x +24

Step-by-step explanation:

6x times 2x Is 12x^2

6x times -3 = -18x

-8 times 2x = -16x

-8 times -3 = 24

combine terms and you get the answer

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The admission fee at a movie theater is $5 for children and $9 for adults. If 3200 people go to the movies and $24000 is collect

Usimov [2.4K]

For this problem, let x be the number of children and y for adults. Formulate the equations: 1st equation, x + y = 3,200 and 2nd equation 5x + 9y = 24,000. Re-arrange 1st equation into x = 3200 - y. Then, substitute into 2nd equation, 5(3,200-y) + 9y = 24,000. Then, solve for y. The 16,000 - 5y + 9y = 24000. Final answer is, y = 2000 adults went to watch the movie.

6 0

2 years ago

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

Alondra is nearly 6 feet tall. Her normal stride is 24 inches long. She wants to walk 5 miles each day for exercise. The table b

Alenkinab [10]

There are 253,440 inches in 4 miles. If you divide this quantity into the 24 inch strides she takes, you get your answer of 10,560 strides.

Hope this helps! <3

8 0

2 years ago

this past Sunday, the giants scored 9 less than twice the cowboys. the packers scored 14 points more than the giants. if the tea

Grace [21]

Let the score of cowboys is x
and giants make score 9 which is twice less than the cowboys score so
giants score will be = 2x -9
and packers scored 14 more than giants that is (2x - 9) + 14
now sum of their scores is equal to 81 it means:
x + (2x - 9) + (2x -9) + 14 = 81
x + 2x - 9 + 2x - 9 + 14 = 81
5x = 81 + 4
5x = 85
x = 17
packers scored = (2x - 9) + 14
= 2 (17) -9 + 14
=38 + 5 = 43 points

5 0

2 years ago

Ted has a credit card that uses the average daily balance method. For the first 9 days of one of his billing cycles, his balance

slega [8]

Your question doesn't say what are the options, but we can make some reasoning.

The average daily balance method is based, obviously, on the average daily balance, which is the average balance for every day of the billing cycle. Therefore, in order to calculate the average daily balance, you need to sum the balance of every day and then divide it by the days of the billing cycle.
In your case:
ADB = (9×2030 + 21×1450) / 30 = 1624 $

Now, in order to calculate the interest, you should first calculate the daily rate, since APR is usually defined yearly, and therefore:
rate = 0.23 ÷ 365 = 0.00063

Finally, the expression to calculate the interest could be:
interest = ADB × rate × days in the billing cycle
or else:
interest = ADB × APR ÷ 365 × days in the billing cycle

In your case:
interest = 1624 × 0.23 ÷ 365 × 30
= 30.70 $

4 0

2 years ago

Read 2 more answers