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Viefleur [7K]
4 years ago
7

What’s the 10th term of the geometric sequence 3,15,75

Mathematics
2 answers:
mina [271]4 years ago
6 0

Answer:

45 dagree

Step-by-step explanation:

Sauron [17]4 years ago
5 0

Answer:

5859375

Step-by-step explanation:

an=(3)(5)^n-1

a10=(3)(5)^9

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Answer if you want to
prohojiy [21]

Answer:

B: 24(0.3)

Step-by-step explanation:

24 * 0.3 = 7.2

So Mandy spent $7.20 on arts and crafts.

24 - 7.2 = $16.80 remaining.

So this is the final answer is 24 * 0.3

5 0
3 years ago
Someone help me with this ASAP
Archy [21]

Answer:

1:6

3/5×2=6/10

2:14

21÷3=7

2×7=14

8 0
3 years ago
Read 2 more answers
Explain how a division problem is like an unknown factor?
steposvetlana [31]
A division problem is unsolved, so it is an unknown factor until ithe problem is solved.
7 0
4 years ago
A football is thrown from the top of the stands, 50 feet above the ground at an initial velocity of 62 ft/sec and at an angle of
Anvisha [2.4K]

a. i. The parametric equation for the horizontal movement is x = 43.84t

ii. The parametric equation for the vertical movement is y = 50 + 43.84t

b. the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

<h2>a. Parametric equations</h2>

A parametric equation is an equation that defines a set of quantities a functions of one or more independent variables called parameters.

<h3>i. Parametric equation for the horizontal movement</h3>

The parametric equation for the horizontal movement is x = 43.84t

Since

  • the angle of elevation is Ф = 45° and
  • the initial velocity, v = 62 ft/s,

the horizontal component of the velocity is v' = vcosФ.

So, the horizontal distance the football moves in time, t is x = vcosФt

= vtcosФ

= 62tcos45°

= 62t × 0.7071

= 43.84t

So, the parametric equation for the horizontal movement is x = 43.84t

<h3>ii Parametric equation for the vertical movement</h3>

The parametric equation for the vertical movement is y = 50 + 43.84t

Also, since

  • the angle of elevation is Ф = 45° and
  • the initial velocity, v = 62 ft/s,

the vertical component of the velocity is v" = vsinФ.

Since the football is initially at a height of h = 50 feet, the vertical distance the football moves in time, t relative to the ground is y = 50 + vsinФt

= 50 + vtcosФ

= 50 + 62tsin45°

= 50 + 62t × 0.7071

= 50 + 43.84t

<h3>b. Location of football at maximum height relative to starting point</h3>

The location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Since the football reaches maximum height at t = 1.37 s

The x coordinate of its location at maximum height is gotten by substituting t = 1.37 into x = 48.84t

So, x = 43.84t

x = 43.84 × 1.37

x = 60.0608

x ≅ 60.1 ft

The y coordinate of the football's location at maximum height relative to the ground is y = 50 + 48.84t

The y coordinate of the football's location at maximum height relative to the starting point is y - 50 = 48.84t

So,  y - 50 = 48.84t

y - 50 = 43.84 × 1.37

y - 50 = 60.0608

y - 50 ≅ 60.1 ft

So, the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Learn ore about parametric equations here:

brainly.com/question/8674159

5 0
2 years ago
Cellus<br> The figure below is a rhombus.<br> 630<br> x<br> y = <br> Enter
sleet_krkn [62]

Answer:

y=27

Step-by-step explanation:

6 0
3 years ago
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