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Marrrta [24]
3 years ago
12

Petco makes a profit of $9 per bag on its line of natural cat food. If the store wants to make a profit of at least

Mathematics
1 answer:
Nata [24]3 years ago
8 0

Answer:

x ≥ 767

The number of bags sold must be greater than or equal to 767 to make a minimum profit of $6900.

Step-by-step explanation:

If they want a profit of at least $6900, then the total amount earned would have to be ≥ 6900. You can always earn more than $6900, but the lowest should be equal to $6900.

So if we use x as the number of bags sold, and we know that each bag will earn $9, then x • 9 ≥ 6900.

If we divide both sides by 9, then the number of bags sold must be ≥ 767.

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What is x? -3(52)=-3(x)+-3(2)
krek1111 [17]

Answer:

x = 50

Step-by-step explanation:

-3(52)=-3(x)+-3(2)

Divide both sides by -3

52 = x + 2

x = 50

8 0
3 years ago
What is 50×10 plz help me
AveGali [126]
The answer is for the problem is 500
6 0
3 years ago
Read 2 more answers
A brick is thrown upward from the top of a building at an angle of 250 above the horizontal, and with an initial speed of 15 m/s
mojhsa [17]

Answer:

(a) 25.08 m

(b) 2.05 m

Step-by-step explanation:

Let the height of the building be 'h'.

Given:

Angle of projection is, \theta=25°

Initial speed is, u=15\ m/s

Time of flight is, t=3.0\ s

(a)

Consider the vertical motion of the brick.

Vertical component of initial velocity is given as:

u_y=u\sin\theta\\\\u_y=15\sin(25)=6.34\ m/s

Vertical displacement of the brick is equal to the height of the building.

So, vertical displacement = -h (Negative sign implies downward motion)

Acceleration is due to gravity in the downward direction. So,

Acceleration is, g=-9.8\ m/s^2

Now, using the following equation of motion;

-h=u_yt+\frac{1}{2}gt^2\\-h=6.34(3)-\frac{9.8}{2}(3)^2\\\\-h=19.02-44.1\\\\-h=-25.08\\\\h=25.08\ m

Therefore, the building is 25.08 m tall.

(b)

Let the maximum height be 'H'.

At maximum height, the vertical component of velocity is 0 as the brick stops temporarily in the vertical direction.

So, v_y=0\ m/s

Now, using the following equation of motion, we have:

v_y^2=u_y^2+2gH\\\\0=(6.34)^2-2\times 9.8\times H\\\\19.6H=40.2\\\\H=\frac{40.2}{19.6}=2.05\ m

Therefore, the maximum height of the brick is 2.05 m.

5 0
3 years ago
PLEASE PLEASE HELP ME PLEASE I AM SO DUM I DONT KNOW THE ANSWER PLEASE JUST HELP ME
pogonyaev

Answer:

Cougar population was 790 at the beginning

830-790=40

3 0
3 years ago
Lm=6x+3 mn=4x-17 ln=x+14 (find the range of possible values for x, help a sister out)
katen-ka-za [31]

Answer:

Step-by-step explanation:

lm=6x+3

mn=4x-17

ln=x+14

lm+mn>ln

6x+3+4x-17>x+14

10x-14>x+14

9x>28

x>28/9

mn+ln>lm

4x-17+x+14>6x+3

5x-3>6x+3

-x>6

x<-6

lm+ln>mn

6x+3+x+14>4x-17

7x+17>4x-17

3x>-34

x>-34/3

combining x<-6 or x>28/9

5 0
3 years ago
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