Answer: 35 t shirts
Step-by-step explanation:
Let number of t shirts be x
Let profit made be y
On main street, the store costs $650,
Selling the t shirt at $32 per 1
He would make a revenue of 32x
Profit = revenue - cost accrued
y1 = 32x - 650
On Broad street, the store costs $440,
Selling the t shirt at $26 per 1
He would make a revenue of 26x
Profit = revenue - cost accrued
y2 = 26x - 440
To make same profit on either location
y1 = y2
32x - 650 =26x - 440
32x -26x = -440+650
6x = 210
x = 210/6
= 35 t shirts
Answer:
act as if the ratio is a fraction 4:3 would be 4/3..
Step-by-step explanation:
Answer:
Pecan cheesecake = $11
Strawberry cheesecake= $19
Step-by-step explanation:
To solve this, we will need to set up a system of equations:
Let's call "Pecan cheesecakes" 'x', and
"Strawberry cheesecakes" 'y'.
We can set up the equations once we have our variables:
276= 13x + 7y
295 = 13x +8y
We will need to subtract these equations from each other to get rid of the 'x' variable. This gives us:
295 = 13x+ 8y
-
276 = 13x+ 7y
19 = y
Therefore, Strawberry cheesecakes cost $19. Plug this into an equation to get the value of 'x':
276= 13x + 7(19)
276 = 13x + 133
143 = 13x
x= $11.
Therefore, Pecan cheesecakes cost $11.
The sum of any two rational numbers is a rational number, so it's true.
Answer:
Step-by-step explanation:
The position of an object moving horizontally after t seconds is given by the function
s = 3t - t³
a) The object is stationary when there is no external force acting on the body. When the body is at rest, the body remains in a position and here is no distance covered by the object i.e s = 0
b) velocity is the change in displacement of a body with respect to time.
v = ds/dt
S = 3t - t³
V = ds/dt = 3-3t²
at t = 2
Velocity = 3-3(2)²
Velocity = 3-12
Velocity = -9m/s
c) acceleration is the change in velocity of a body with respect to time.
acceleration = dv/dt
If v = 3-3t²
a = dv/dt = -6t
When v = 0
0 = 3-3t²
-3 = -3t²
t² = 1
t = ±√1
t = 1sec
The acceleration of the object at v = 0 occurs at t = 1sec and -1sec
a = -6(1)
a = -6m/s²
d) Given the speed of the body v modelled by the function
v = 3-3t²
The speed is decreasing when it is less than zero as shown:
3-3t²< 0
3<3t²
1<t²
±1<t
1<t and -1<t
t>±1
t >1 and t>-1
The speed is decreasing when
3-3t²<0 and t>1 or t>-1
