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1 year ago

A city determines that a planned community must have at least 4 acres of developed and open space, and the difference

Mathematics

1 answer:

FinnZ [79.3K]1 year ago

3 0

Answer:

The graph in the attached figure

Step-by-step explanation:

Let

x -----> the number of open acres

y ----> the number of developed acres

we know that

$x+y \geq 4$ -----> inequality A

$x\leq 1$ ----> inequality B

Remember that

if the number of open acres, x, can be no more than 1. then the number of open acres, x must be less than or equal to 1

Solve the system of equations by graphing

The solution is the shaded area (Note The number of acres cannot be a negative number)

The graph in the attached figure

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A pure acid measuring x liters is added to 300 liters of a 20% acidic solution. The concentration of acid, f(x), in the new subs

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Answer How many liters of a 20% acid solution should be mixed with 30 liters of 50% acid solution in order to obtain a 40% solution. ... x=15 liters 15*(.20 pure acid)=3 liters 30*(.50 pure acid)=15 liters That is 18 liters pure acid That is 45 liters solution *0.45 pure acid=18 liters.

Step-by-step explanation:

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1 year ago

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In the diagram below AB is parallel to CD what is the value of y

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y=60 due to adjacent angle

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How do you write 7.75 in expanded form?

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2 years ago

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Given: The coordinates of triangle PQR are P(0, 0), Q(2a, 0), and R(2b, 2c).

masha68 [24]

Answer:

The line containing the midpoints of two sides of a triangle is parallel to the third side ⇒ proved down

Step-by-step explanation:

* Lets revise the rules of the midpoint and the slope to prove the

problem

- The slope of a line whose endpoints are (x1 , y1) and (x2 , y2) is

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

- The mid-point of a line whose endpoints are (x1 , y1) and (x2 , y2) is

$(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

* Lets solve the problem

- PQR is a triangle of vertices P (0 , 0) , Q (2a , 0) , R (2b , 2c)

- Lets find the mid-poits of PQ called A

∵ Point P is (x1 , y1) and point Q is (x2 , y2)

∴ x1 = 0 , x2 = 2a and y1 = 0 , y2 = 0

∵ A is the mid-point of PQ

∴ $A=(\frac{0+2a}{2},\frac{0+0}{2})=(\frac{2a}{2},\frac{0}{2})=(a,0)$

- Lets find the mid-poits of PR which called B

∵ Point P is (x1 , y1) and point R is (x2 , y2)

∴ x1 = 0 , x2 = 2b and y1 = 0 , y2 = 2c

∵ B is the mid-point of PR

∴ $B=(\frac{0+2b}{2},\frac{0+2c}{2})=(\frac{2b}{2},\frac{2c}{2})=(b,c)$

- The parallel line have equal slopes, so lets find the slopes of AB and

QR to prove that they have same slopes then they are parallel

# Slope of AB

∵ Point A is (x1 , y1) and point B is (x2 , y2)

∵ Point A = (a , 0) and point B = (b , c)

∴ x1 = a , x2 = b and y1 = 0 and y2 = c

∴ The slope of AB is $m=\frac{c-0}{b-a}=\frac{c}{b-a}$

# Slope of QR

∵ Point Q is (x1 , y1) and point R is (x2 , y2)

∵ Point Q = (2a , 0) and point R = (2b , 2c)

∴ x1 = 2a , x2 = 2b and y1 = 0 and y2 = 2c

∴ The slope of AB is $m=\frac{2c-0}{2b-2a}=\frac{2c}{2(b-c)}=\frac{c}{b-a}$

∵ The slopes of AB and QR are equal

∴ AB // QR

∵ AB is the line containing the midpoints of PQ and PR of Δ PQR

∵ QR is the third side of the triangle

∴ The line containing the midpoints of two sides of a triangle is parallel

to the third side

6 0

1 year ago

The graphs of the quadratic functions f(x) = 6 – 10x2 and g(x) = 8 – (x – 2)2 are provided below. Observe there are TWO lines si

natta225 [31]

Answer:

a) y = 7.74*x + 7.5

b) y = 1.148*x + 6.036

Step-by-step explanation:

Given:

f(x) = 6 - 10*x^2

g(x) = 8 - (x-2)^2

Find:

(a) The line simultaneously tangent to both graphs having the LARGEST slope has equation

(b) The other line simultaneously tangent to both graphs has equation,

Solution:

- Find the derivatives of the two functions given:

f'(x) = -20*x

g'(x) = -2*(x-2)

- Since, the derivative of both function depends on the x coordinate. We will choose a point x_o which is common for both the functions f(x) and g(x). Point: ( x_o , g(x_o)) Hence,

g'(x_o) = -2*(x_o -2)

- Now compute the gradient of a line tangent to both graphs at point (x_o , g(x_o) ) on g(x) graph and point ( x , f(x) ) on function f(x):

m = (g(x_o) - f(x)) / (x_o - x)

m = (8 - (x_o-2)^2 - 6 + 10*x^2) / (x_o - x)

m = (8 - (x_o^2 - 4*x_o + 4) - 6 + 10*x^2)/(x_o - x)

m = ( 8 - x_o^2 + 4*x_o -4 -6 +10*x^2) /(x_o - x)

m = ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x)

- Now the gradient of the line computed from a point on each graph m must be equal to the derivatives computed earlier for each function:

m = f'(x) = g'(x_o)

- We will develop the first expression:

m = f'(x)

( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

Eq 1. (-2 - x_o^2 + 4*x_o + 10*x^2) = -20*x*x_o + 20*x^2

And,

m = g'(x_o)

( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

-2 - x_o^2 + 4*x_o + 10*x^2 = -2(x_o - 2)(x_o - x)

Eq 2 -2 - x_o^2 + 4*x_o+ 10*x^2 = -2(x_o^2 - x_o*(x + 2) + 2*x)

- Now subtract the two equations (Eq 1 - Eq 2):

-20*x*x_o + 20*x^2 + 2*x_o^2 - 2*x_o*(x + 2) + 4*x = 0

-22*x*x_o + 20*x^2 + 2*x_o^2 - 4*x_o + 4*x = 0

- Form factors: 20*x^2 - 20*x*x_o - 2*x*x_o + 2*x_o^2 - 4*x_o + 4*x = 0

20*x*(x - x_o) - 2*x_o*(x - x_o) + 4*(x - x_o) = 0

(x - x_o)(20*x - 2*x_o + 4) = 0

x = x_o , x_o = 10x + 2

- For x_o = 10x + 2 ,

(g(10*x + 2) - f(x))/(10*x + 2 - x) = -20*x

(8 - 100*x^2 - 6 + 10*x^2)/(9*x + 2) = -20*x

(-90*x^2 + 2) = -180*x^2 - 40*x

90*x^2 + 40*x + 2 = 0

- Solve the quadratic equation above:

x = -0.0574, -0.387

- Largest slope is at x = -0.387 where equation of line is:

y - 4.502 = -20*(-0.387)*(x + 0.387)

y = 7.74*x + 7.5

- Other tangent line:

y - 5.97 = 1.148*(x + 0.0574)

y = 1.148*x + 6.036

6 0

1 year ago