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Sav [38]
3 years ago
7

Please help with this question it is 60% of my grade.

Mathematics
2 answers:
zubka84 [21]3 years ago
7 0

Answer:

B

Step-by-step explanation:

After the first minute it should be at 21, because it would be reduced by 3. Second, the same. ETC

Vsevolod [243]3 years ago
3 0

Answer: b

Step-by-step explanation:

you start with 0 min- equals 24

1 min- 21

2 min- 18

3 min- 15

4 min- 12

5 min- 9

and so on- you look for the table that best describes that. x represents the minutes, and y represent the amount of water left in the sink.

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The class president wants to survey the students at the high school about the selection of food offered by the school cafeteria.
lesantik [10]

Answer:

C . Sample : 6 boy and 7 girls

Step-by-step explanation:

Data :

  Boys = 150

  Girls = 175

So ratio between boys and girls

     175/150  = 1.67

So when boy is 6

   6 ×1.667 = 7

5 0
2 years ago
Given the graph below, locate the points of the reflection over the x-axis.
Zinaida [17]

Answer:

A: (1,3)

A': (1,-3)

---

B: (5,1)

B': (5,-1)

---

C: (-2,2)

C': (-2,-2)

---

D: (-5,4)

D': (-5,-4)

Hope it helps, sorry I was late; just got around to this lesson.

6 0
2 years ago
Express 5/8 in forieths plese show work
fenix001 [56]

<u><em>25/40</em></u>

Work:

5/8 x 5 = 25/40

7 0
3 years ago
Peyton is going to invest $71,000 and leave it in an account for 18 years. Assuming the interest is compounded continuously, wha
rewona [7]

Answer:

2.23

Step-by-step explanation:

4 0
3 years ago
Attached as picture. Please read fully
Troyanec [42]

a. The velocity t = v = Ce_{n} (\frac{mo}{mo - kt} ) - gt

b. v60 = 7164

<h3>How to solve for the velocity</h3>

mdv/dt = ck - mg

dv/dt = ck/m - mg/m

= ck/m - g

dv = (\frac{ck}{Mo-Kt} -g)dv

Integrate the two sides of the equation to get

v -\frac{ck}{k} e_{n} (Mo- kt)-gt+c

v = Ce_{n} (\frac{mo}{mo - kt} ) - gt

b. fuel accounts for 55% of the mass

So final mass after fuel is burned out is = 0.45

c=2500

g=9.8

t=60

v = -2500ln0.45 - 9.8 x 60

= 7752 - 588

= 7164

<h3>Complete question</h3>

A rocket, fired from rest at time t = 0, has an initial mass of m0 (including its fuel). Assuming that the fuel is consumed at a constant rate k, the mass m of the rocket, while fuel is being burned, will be given by m0 - kt. It can be shown that if air resistance is neglected and the fuel gases are expelled at a constant speed c relative to the rocket, then the velocity of the rocket will satisfy the equation where g is the acceleration due to gravity.

dv dt m =ck - mg

(a) Find v(t) keeping in mind that the mass m is a function of t.

v(t) =

m/sec

(b) Suppose that the fuel accounts for 55% of the initial mass of the rocket and that all of the fuel is consumed at 60 s. Find the velocity of the rocket in meters per second at the instant the fuel is exhausted. [Note: Take g = 9.8 m/s² and c = 2500 m/s.]

v(60) =

m/sec [Round to nearest whole number]

Raed more on velocity here

brainly.com/question/25749514

#SPJ1

8 0
1 year ago
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