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oksano4ka [1.4K]

3 years ago

5

The book fair sold 60% of the books

1 answer:

natka813 [3]3 years ago

6 0

Answer:

120 Books

Step-by-step explanation:

60% of 120 is 72

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you are selling pandesal to raise money for a school field trip. Pandesal with cheese cost 8.00 pesos and ube pandesal cost 9.00

Lera25 [3.4K]

Answer:

Let's define:

C = number of pandesal with cheese that you sell

U = number of pandesal with ube that you sell.

If you sell these numbers of each, the total profit you get is:

C*8.00 pesos + U*9.00 pesos.

And yo want to get at least 180 pesos, then:

C*8.00 pesos + U*9.00 pesos ≥ 180 pesos.

And you also want to sell two of each, then:

C ≥ 2

U ≥2

So the system of inequalities is:

C*8.00 pesos + U*9.00 pesos ≥ 180 pesos.

C ≥ 2

U ≥2

If U is on the x-axis, and C is on the y-axis, then the graph is: (where the region at the right of the vertical line should be shaded)

6 0

3 years ago

HI could some one please help me I will give you BRAINLEST!!!!

tatuchka [14]

Answer:

16

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Which inequality corresponds<br>to the following graph?

alexira [117]

The first one x is greater than and equal to 34

4 0

3 years ago

Find all of the equilibrium solutions. Enter your answer as a list of ordered pairs (R,W), where R is the number of rabbits and

zloy xaker [14]

Answer:

$(0,0)$ $(4000,0)$ and $(500,79)$

Step-by-step explanation:

Given

See attachment for complete question

Required

Determine the equilibrium solutions

We have:

$\frac{dR}{dt} = 0.09R(1 - 0.00025R) - 0.001RW$

$\frac{dW}{dt} = -0.02W + 0.00004RW$

To solve this, we first equate $\frac{dR}{dt}$ and $\frac{dW}{dt}$ to 0.

So, we have:

$0.09R(1 - 0.00025R) - 0.001RW = 0$

$-0.02W + 0.00004RW = 0$

Factor out R in $0.09R(1 - 0.00025R) - 0.001RW = 0$

$R(0.09(1 - 0.00025R) - 0.001W) = 0$

Split

$R = 0$ or $0.09(1 - 0.00025R) - 0.001W = 0$

$R = 0$ or $0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

Factor out W in $-0.02W + 0.00004RW = 0$

$W(-0.02 + 0.00004R) = 0$

Split

$W = 0$ or $-0.02 + 0.00004R = 0$

Solve for R

$-0.02 + 0.00004R = 0$

$0.00004R = 0.02$

Make R the subject

$R = \frac{0.02}{0.00004}$

$R = 500$

When $R = 500$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 -2.25 * 10^{-5} * 500 - 0.001W = 0$

$0.09 -0.01125 - 0.001W = 0$

$0.07875 - 0.001W = 0$

Collect like terms

$- 0.001W = -0.07875$

Solve for W

$W = \frac{-0.07875}{ - 0.001}$

$W = 78.75$

$W \approx 79$

$(R,W) \to (500,79)$

When $W = 0$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 - 2.25 * 10^{-5}R - 0.001*0 = 0$

$0.09 - 2.25 * 10^{-5}R = 0$

Collect like terms

$- 2.25 * 10^{-5}R = -0.09$

Solve for R

$R = \frac{-0.09}{- 2.25 * 10^{-5}}$

$R = 4000$

So, we have:

$(R,W) \to (4000,0)$

When $R =0$ , we have:

$-0.02W + 0.00004RW = 0$

$-0.02W + 0.00004W*0 = 0$

$-0.02W + 0 = 0$

$-0.02W = 0$

$W=0$

So, we have:

$(R,W) \to (0,0)$

Hence, the points of equilibrium are:

$(0,0)$ $(4000,0)$ and $(500,79)$

4 0

3 years ago

1) what is the answer to<br><br> (5²-7) ₃ 3

valentinak56 [21]

Answer: 54

Explanation step by step: according to bidmas you must open the bracket first before any other thing. so the solution is 5×5-7×3

8 0

3 years ago

Read 2 more answers