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OLEGan [10]
3 years ago
10

Jennifer wants to purchase a book but only has $62.10 to spend. What is the maximum number of pages she can have in her book if

P= (20+0.5x) + 0.15 (20+ 0.5x)?
Mathematics
1 answer:
nevsk [136]3 years ago
7 0
Its not a full questions...... send the full question
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So you're initial equation is (2/6)*24=m. Well since 2/6=1/3, plug in 24 for the 1 in 1/3. So 24/3=m, and 24/3=8 so m=8.
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Which expressions are equivalent to 2(2x + 4y + x − 2y)?
Greeley [361]
C should be your answer:))))
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What is the probability that a randomly dropped marker will fall in the unshaded region?
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The correct answer is 14 by 15
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Read 2 more answers
If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by
Vitek1552 [10]

Answer:

a) Average velocity at 0.1 s is 696 ft/s.

b) Average velocity at 0.01 s is 7536 ft/s.

c) Average velocity at 0.001 s is 75936 ft/s.

Step-by-step explanation:

Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by y = 70t-16t^2.

To find : The average velocity for the time period beginning when t = 2 and lasting.  a. 0.1 s. , b. 0.01 s. , c. 0.001 s.

Solution :    

a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.

(\text{Average velocity})_{0.1\ s}=\frac{\text{Change in height}}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{h_{2.1}-h_2}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{(70(2.1)-16(2.1)^2)-(70(0.1)-16(0.1)^2)}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{69.6}{0.1}

(\text{Average velocity})_{0.1\ s}=696\ ft/s

b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.

(\text{Average velocity})_{0.01\ s}=\frac{\text{Change in height}}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{h_{2.01}-h_2}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{(70(2.01)-16(2.01)^2)-(70(0.01)-16(0.01)^2)}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{75.36}{0.01}

(\text{Average velocity})_{0.01\ s}=7536\ ft/s

c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.

(\text{Average velocity})_{0.001\ s}=\frac{\text{Change in height}}{0.001}  

(\text{Average velocity})_{0.001\ s}=\frac{h_{2.001}-h_2}{0.001}

(\text{Average velocity})_{0.001\ s}=\frac{(70(2.001)-16(2.001)^2)-(70(0.001)-16(0.001)^2)}{0.001}

(\text{Average velocity})_{0.001\ s}=\frac{75.936}{0.001}

(\text{Average velocity})_{0.001\ s}=75936\ ft/s

5 0
3 years ago
A spherical hot air balloon has a diameter of 55 feet when the balloon is inflated the radius increases at a rate of 1.5 feet pe
Nuetrik [128]

Answer: 46.90mins

Step-by-step explanation:

The given data:

The diameter of the balloon = 55 feet

The rate of increase of the radius of the balloon when inflated = 1.5 feet/min.

Solution:

dr/dt = 1.5 feet per minute = 1.5 ft/min

V = 4/3·π·r³

The maximum volume of the balloon

= 4/3 × 3.14 × 55³

= 696556.67 ft³

When the volume 2/3 the maximum volume

= 2/3 × 696556.67 ft³

= 464371.11 ft³

The radius, r₂ at the point is

= 4/3·π·r₂³

= 464371.11 ft³

r₂³ = 464371.11 ft³ × 3/4

= 348278.33 ft³

348278.333333

r₂ = ∛(348278.33 ft³) ≈ 70.36 ft

The time for the radius to increase to the above length = Length/(Rate of increase of length of the radius)

The time for the radius to increase to the

above length

Time taken for the radius to increase the length.

= is 70.369 ft/(1.5 ft/min)

= 46.90 minutes

46.90mins is the time taken to inflate the balloon.

5 0
2 years ago
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