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3 years ago

Can you please help me i really need help you get 15 points if you answer all please just say a or b or c or d for answers

Mathematics

1 answer:

Hoochie [10]3 years ago

3 0

For the brownie one it’s C
For the train one it’s B
For the flour one it’s C

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A function, F(x), is shown below.

Maksim231197 [3]

Answer:

range of f(x) = [-4, -2) ∪ [2, 8)

a+b+c+d = -4

Step-by-step explanation:

The graph is attached. The range is the vertical extent of the function. It is defined at f(0) = -4 and f(2) = 2.

The limits f(2-) and f(4-) are -2 and 8, respectively, so the graph has open circles there. These are the ends of the two half-open intervals that make up the range of the function.

The portion of the graph in the domain [4, 7) is included in the range [2, 8), so no special treatment is needed for that piece of the function.

7 0

3 years ago

Marco needs to buy cat food. At the nearest store, 3 bags of cat food cost $15.75. How much would Marco spend on 5 bags of cat f

Ber [7]

Answer:

$29 dollars and 60 cents

Step-by-step explanation:

15.75 divided by 3 is 5.92 if rounded.

then 5.92x 5= 29.60

5 0

3 years ago

Read 2 more answers

Solve for X ...........

musickatia [10]

Answer:

x = 46

Step-by-step explanation:

Given 2 secants from an external point to the circle, then

The product of the external part and the whole of one secant is equal to the product of the external part and the whole of the other secant, that is

2(2 + x) = 6(6 + 10)

2(2 + x) = 6 × 16 = 96 ( divide both sides by 2 )

2 + x = 48 ( subtract 2 from both sides )

x = 46

5 0

3 years ago

If 3 bean bags were tossed onto the above board

omeli [17]

Answer:

I dont understand the question if you're talking about the game then the points are 0

Step-by-step explanation:

4 0

3 years ago

A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

7 0

2 years ago