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andrey2020 [161]
3 years ago
5

Two siblings, Andy and Alexa, bake 24 cookies. Alexa eats some number of cookies that is a positive multiple of the number of co

okies eaten by her brother. If the siblings finish all 24 cookies, then what is the maximum number of cookies that the brother, Andy, could have eaten?
Mathematics
2 answers:
kolbaska11 [484]3 years ago
4 0

Answer:

12.

Step-by-step explanation:

The number of cookies eaten by Andy depends on the number eaten by his sister: if Andy eats more, then Alexa eats fewer, and the total always adds up to 24. We want to maximize the number eaten by the brother, so we want to minimize the number eaten by the sister. The smallest positive multiple of the number eaten by Andy is one times that number, which is the number itself. Alexa must eat the same number as Andy, so each sibling eats half the cookies.

Gemiola [76]3 years ago
3 0
Solution?????????????????
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Need help with another problem
Verdich [7]
Find the midpoint:

m= x1+x2/2; y1+y2/2

m= 9+-1/2; 8+-2/2

m= 8/2; 6/2

m= (4,3)

(4,3) is your answer.

I hope this helps!
~kaikers
4 0
3 years ago
A box contains 35 coins either nickels or dimes all worth $2.90. How many dimes are in the box?
tino4ka555 [31]

Answer:

There are 23 dimes in the box

Step-by-step explanation:

Let nickels=n

Let dimes=(35-n)

Equation: 0.05n+0.10(35-n)=2.90

Multiply both sides by 100: 5n+10(35-n)=290

-5n+350=290

-5n=-60

n=12

35-12=23

Therefore, there are 23 dimes in the box

3 0
3 years ago
The braking distance, in feet of a car a Travling at v miles per hour is given.
irakobra [83]

The braking distance is the distance the car travels before coming to a stop after the brakes are applied

a. The braking distances are as follows;

  • The braking distance at 25 mph, is approximately <u>63.7 ft.</u>
  • The braking distance at 55 mph,  is approximately <u>298.35 ft.</u>
  • The braking distance at 85 mph,  is approximately <u>708.92 ft.</u>

b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s

Reason:

The given function for the braking distance is D = 2.6 + v²/22

a. The braking distance if the car is going 25 mph is therefore;

25 mph = 36.66339 ft./s

D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.

At 25 mph, the braking distance is approximately <u>63.7 ft.</u>

At 55 mph, the braking distance is given as follows;

55 mph = 80.65945  ft.s

D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.

At 55 mph, the braking distance is approximately <u>298.35 ft.</u>

At 85 mph, the braking distance is given as follows;

85 mph = 124.6555 ft.s

D = 2.6 + \dfrac{124.6555^2}{22} \approx 708.92 \ ft.

At 85 mph, the braking distance is approximately <u>708.92 ft.</u>

b. The speed of the car when the braking distance is 450 feet is given as follows;

450 = 2.6 + \dfrac{v^2}{22}

v² = (450 - 2.6) × 22 = 9842.8

v = √(9842.2) ≈ 98.211 ft./s

The car was moving at v ≈ <u>98.211 ft./s</u>

Learn more here:

brainly.com/question/18591940

8 0
2 years ago
Step 3a: Calculating the average rate
Alex73 [517]

9514 1404 393

Answer:

  a) average rate = (total distance)/(total time)

  b) Rave = 2·R1·R2/(R1 +R2)

  c) cheetah's average rate ≈ 50.91 mph

Step-by-step explanation:

a) Let AB represent the distance from A to B. Let t1 and t2 represent the travel times (in hours) on leg1 and leg2 of the trip, respectively. Then the distances traveled are...

  First leg distance: AB = 70·t1   ⇒   t1 = AB/70

  Second leg distance: AB = 40·t2   ⇒   t2 = AB/40

The average rate is the ratio of total distance to total time:

  average rate = (AB +AB)/(t1 +t2)

  average rate = 2AB/(AB/70 +AB/40) = 2/(1/70 +1/40) = 2(40)(70)/(70+40)

  average rate = 560/11 = 50 10/11 . . . mph

__

No equations are given, so we cannot compare what we wrote with the given equations. In each step of the solution, we have used the rules of algebra and equality.

b) For two rates over the same distance (as above), the average is their harmonic mean:

  average rate = 2r1·r2/(r1+r2)

__

c) The cheetah's average rate was 50 10/11 mph ≈ 50.91 mph.

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3 years ago
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