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Stella [2.4K]
1 year ago
8

Marika is training for a track race. She start sprinting by 100 yards. She gradually increases her distance, adding 4 yards a da

y for 21 days. the explicit formula that models this situation is an=100+(n-1)4. how far does seh sprint on day 21
Mathematics
1 answer:
Hunter-Best [27]1 year ago
8 0

Answer:

  180 yards

Step-by-step explanation:

The number of yards she sprints on day 21 can be found by evaluating the formula with n=21.

<h3>Solution</h3>

  a21 = 100 +(21 -1)4 = 100 +(20)(4) = 100 +80 . . . . . use 21 in place of n

  a21 = 180

Marika sprints 180 yards on day 21.

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Answer: 1:8

Step-by-step explanation:

First you add all of the candies together (so 18) then you see how many there are grape flavored (2), put them in a ratio (2:18) and simplify (1:8)

hope this helps :)

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2 years ago
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If your income is $1240
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Answer: 9/20 for rent = $558

1/2 for food implies= $620

Step-by-step explanation:

9/20 ×$1240= $558

While

1/2 ×$1240= $620

8 0
2 years ago
Will award brainliest
Ostrovityanka [42]
Infinite solution Bc both sides are equal
8 0
3 years ago
1. On a train, it takes 3 days, 5 hours, and 15 minutes to get to from Denver to Seattle by train. How many hours does it take t
jarptica [38.1K]
For 
1. its 
D=Days
H=Hours
M=Minutes
3d+5h+15m
3d=72h+5h
72h+5h=77h
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5 0
3 years ago
A hoop, a uniform solid cylinder, a spherical shell, and a uniform solid sphere are released from rest at the top of an incline.
Serjik [45]

Answer:

Step-by-step explanation:

Given

Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height

Suppose they have same mass and radius

time Period is given by

t=\sqrt{\frac{2h}{a}} ,where h=height of release

a=acceleration

a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}

Where I=moment of inertia

a for hoop

a=\frac{g\sin \theta }{1+\frac{mr^2}{mr^2}}

a=\frac{g\sin \theta }{2}

a for Uniform solid cylinder

a=\frac{g\sin \theta }{1+\frac{mr^2}{2mr^2}}

a=\frac{2g\sin \theta }{3}

a for spherical shell

a=\frac{g\sin \theta }{1+\frac{2mr^2}{3mr^2}}

a=\frac{3g\sin \theta }{5}

a for Uniform Solid

a=\frac{g\sin \theta }{1+\frac{2mr^2}{5mr^2}}

a=\frac{5g\sin \theta }{7}

time taken will be inversely proportional to the square root of acceleration

t_1=k\sqrt{2}=1.414k

t_2=k\sqrt{\frac{3}{2}}=1.224k

t_3=k\sqrt{\frac{5}{3}}=1.2909k

t_4=k\sqrt{\frac{7}{5}}=1.183k

thus first one to reach is Solid Sphere

second is Uniform solid cylinder

third is Spherical Shell

Fourth is hoop

3 0
3 years ago
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