Find the slope between the points (8, 10) and (-3, 10).

PLZ HELP MARKIN BRAINIEST!!!

Dahasolnce [82]

From the graph, when x = 1, y = 57,000.

Replace x with 1 in the equations and see if any of the Y 's equal 57,000 :

y = -2610.82(1) + 47860.82 = 45,250

y = 219(1)^2 - 6,506.78(1) + 59,385 = 219 - 6506.78 + 59385 = 53,097.22

y = 54041.5(0.9)^1 = 48,637.35

y = 10,504.6 (1.1)^1 = 11,555.06

The second equation is the closest. so try another x value to see if it is close to the Y value:

Let's try x = 14:

y = 219(14)^2 - 6506.78(14) + 59,385 = 42924 - 91094.92 + 59385 = 11,214.08

This is close to Y = 12,00 shown on the graph

SO the closest equitation is y = 219x^2 - 6506.78x + 59385

8 0

2 years ago

Plz help me with this i will give u brainliest

Darya [45]

Answer:

2 tons are 4000 pounds

......

3 0

2 years ago

The population of two different villages is modeled by the equations shown

vlada-n [284]

Answer:

Year = 1995 and population = 315

Year = 2015 and population = 715

Step-by-step explanation:

It is given that the population of two different villages is modeled by the given equations:

$y=x^2-30x+540$

$y=20x+15$

The population of both villages are same after x years after 1980 if

$x^2-30x+540=20x+15$

$x^2-30x+540-20x-15=0$

$x^2-50x+525=0$

Splitting the middle term, we get

$x^2-15x-35x+525=0$

$x(x-15)-35(x-15)=0$

$(x-15)(x-35)=0$

$x=15,35$

It means, after 15 or 35 years, the population will same.

For x=15, years is $1980+15=1995$ and population is

$y=20(15)+15=315$

For x=35, years is $1980+35=2015$ and population is

$y=20(35)+15=715$ .

Therefore, population are equation in Year = 1995 and population = 315 or Year = 2015 and population = 715.

7 0

2 years ago

(c). It is well known that the rate of flow can be found by measuring the volume of blood that flows past a point in a given tim

aleksklad [387]

(i) Given that

$V(R) = \displaystyle \int_0^R 2\pi K(R^2r-r^3) \, dr$

when R = 0.30 cm and v = (0.30 - 3.33r²) cm/s (which additionally tells us to take K = 1), then

$V(0.30) = \displaystyle \int_0^{0.30} 2\pi \left(0.30-3.33r^2\right)r \, dr \approx \boxed{0.0425}$

and this is a volume so it must be reported with units of cm³.

In Mathematica, you can first define the velocity function with

v[r_] := 0.30 - 3.33r^2

and additionally define the volume function with

V[R_] := Integrate[2 Pi v[r] r, {r, 0, R}]

Then get the desired volume by running V[0.30].

(ii) In full, the volume function is

$\displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr$

Compute the integral:

$V(R) = \displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr$

$V(R) = \displaystyle 2\pi K \int_0^R (R^2r-r^3) \, dr$

$V(R) = \displaystyle 2\pi K \left(\frac12 R^2r^2 - \frac14 r^4\right)\bigg_0^R$

$V(R) = \displaystyle 2\pi K \left(\frac{R^4}2- \frac{R^4}4\right)$

$V(R) = \displaystyle \boxed{\frac{\pi KR^4}2}$

In M, redefine the velocity function as

v[r_] := k*(R^2 - r^2)

(you can't use capital K because it's reserved for a built-in function)

Then run

Integrate[2 Pi v[r] r, {r, 0, R}]

This may take a little longer to compute than expected because M tries to generate a result to cover all cases (it doesn't automatically know that R is a real number, for instance). You can make it run faster by including the Assumptions option, as with

Integrate[2 Pi v[r] r, {r, 0, R}, Assumptions -> R > 0]

which ensures that R is positive, and moreover a real number.

5 0

2 years ago

A cylindrical container with a radius of 5 cm and a height of 14 cm is completely filled with liquid. Some of the liquid from th

Flura [38]

Answer:

The remaining liquid has a volume of 345.4 cm³.

Step-by-step explanation:

To solve this question we first need to calculate the volume of each container.

Cylindrical:

Volume = h*pi*r² = 14*3.14*(5)² = 1099 cm³

Cone:

Volume = (1/3)*h*pi*r² = (1/3)*20*3.14*(6)² = 753.6 cm³

Since the liquid from the cylinder was transfered to the cone until the latter was full, then what remains in the cylinder is it's original volume minus the cone volume. So we have:

remaining liquid = 1099 - 753.6 = 345.4 cm³

4 0

2 years ago