What is the slope of the line represented by the equation y=-1/2x+1/4

In the set of real numbers, the first number less than -3 is -4. True False

mariarad [96]

True, -4 seems bigger because 4 is bigger than 3 but it's negative and is further away from 0

6 0

3 years ago

georges playground is shown in the diagram he wants to buy timber to put all around it so that mulch does not spread how much ti

Xelga [282]

He needs 84 ft.
Step by step explanation:
(24*2)+(18*2)
48+36
84

7 0

3 years ago

Suppose that the quantity supplied S and quantity demanded D of T-shirts at a concert are given by the following functions wher

harkovskaia [24]

Answer:

Step-by-step explanation:

The demand function is expressed as

D(p )=1150-50p

The supply function is expressed as S(p)=−200 + 40p

Where

p represents the price

A) The equilibrium price is the price at which the quantity supplied and the quantity demanded would be equal. Therefore

1150 - 50p = - 200 + 40p

40p + 50p = 1150 + 200

90p = 1350

p = 1350/90

p = 15

The equilibrium price is $15

b) For quantity demanded to be greater than quantity supplied, the price would be

1150 - 50p > - 200 + 40p

40p + 50p > 1150 + 200

90p > 1350

p > 1350/90

p > 15

The price would be greater than 15

c) if the quantity demanded is greater than the quantity supplied, the prices of the T shirt would increase.

7 0

3 years ago

Solve the initial value problem y'=2 cos 2x/(3+2y),y(0)=−1 and determine where the solution attains its maximum value.

zloy xaker [14]

Answer:

$y=\frac{-3\pm\sqrt{4sin2x+1}}{2}$

$x={\pi}{4}$

Step-by-step explanation:

We are given that

$y'=\frac{2cos2x}{3+2y}$

y(0)=-1

$\frac{dy}{dx}=\frac{2cos2x}{3+2y}$

$(3+2y)dy=2cos2x dx$

Taking integration on both sides then we get

$\int (3+2y)dy=2\int cos 2xdx$

$3y+y^2=sin2x+C$

Using formula

$\int x^n=\frac{x^{n+1}}{n+1}+C$

$\int cosx dx=sinx$

Substitute x=0 and y=-1

$-3+1=sin0+C$

$-2=C$

$sin0=0$

Substitute the value of C

$y^2+3y=sin2x-2$

$y^2+3y-sin 2x+2=0$

$y=\frac{-3\pm\sqrt{(3)^2-4(1)(-sin2x+2)}}{2}$

By using quadratic formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$y=\frac{-3\pm\sqrt{9+4sin2x-8}}{2}=\frac{-3\pm\sqrt{4sin2x+1}}{2}$

Hence, the solution $y=\frac{-3\pm\sqrt{4sin2x+1}}{2}$

When the solution is maximum then y'=0

$\frac{2cos2x}{3+2y}=0$

$2cos2x=0$

$cos2x=0$

$cos2x=cos\frac{\pi}{2}$

$cos\frac{\pi}{2}=0$

$2x=\frac{\pi}{2}$

$x=\frac{\pi}{4}$

3 0

3 years ago

A runner runs around a circular track. He completes one lap at a time of t = 269 s at a constant speed of v = 4.6 m/s. What is t

AURORKA [14]

Answer:

$\boxed{\text{197 m}}$

Step-by-step explanation:

The formula relating distance (d), speed (s), and time (t) is

d = st

1. Calculate the distance

d = 269 s × 4.6 m·s⁻¹ = 1240 m

2.Calculate the track radius

The distance travelled is the circumference of a circle

$\begin{array}{rcl}C & = & 2 \pi r\\1240 & = & 2 \pi r\\\\r & = & \dfrac{1240}{2 \pi }\\\\& = & 197\\\end{array}\\\text{The radius of the track is }\boxed{\textbf{197 m}}$

5 0

4 years ago