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laila [671]
3 years ago
6

QUESTION ONE: Solve for h.

Mathematics
2 answers:
Setler79 [48]3 years ago
8 0

Answer:

h = - 105

Step-by-step explanation:

Given

\frac{h}{14} = - 7.5 ( multiply both sides by 14 to clear the fraction )

h = - 105

kykrilka [37]3 years ago
8 0

Answer:1.86

Step-by-step explanation:

You might be interested in
Describe the correct answer to this problem including units what is the volume of the figure
Andru [333]

Step-by-step explanation:

height = √6.5² - 2.5² = 6ft

8 0
3 years ago
A 9,000-lb load is suspended from the roof in a shopping mall with a 16-ft-long solid aluminum rod. The modulus of elasticity of
Westkost [7]

Answer:

Step-by-step explanation:

Given:

elongation, x = 0.50 in

Force, f = 9000 lb

Young modulus, E = 10,000,000 psi

Maximum Stress, Sm = 30000 psi

Length, L = 16 ft

Converting ft to in,

12 in = 1 ft

=16 × 12 = 192 in

Young modulus, E = stress/strain

Stress = force/area, A

Strain = elongation, x/Length, L

E = f × L/A × E

1 × 10^7 = stress/(0.5/16)

= 26041.7 psi

Minimum stress = 26041.7 psi

Maximum stress = 30,000 psi

Stress = force/area

Area = 9000/26041.7

= 0.3456 in^2

Stress = force/area

Area = 9000/30000

= 0.3 in^2

Using minimum area of 0.3 in^2,

A = (pi/4)(d^2)

0.3 in^2 = (pi/4)(d^2)

d = 0.618 inches

diameter, d = 0.618 inches

7 0
3 years ago
You are trying to bulk and gain muscle. Your trainer says your daily calorie intake should increases based on hours you spend ex
s2008m [1.1K]

a) T - Calories, 2000 - <em>Minimum</em> calories, 500 - Additional calories per exercise hour, b) We should consume 3500 calories to spend three hours exercising.

<h3>How to analyse a linear function of calorie gain in terms of the exercise time</h3>

In this question we have a <em>linear</em> function where the <em>independent</em> variable is the exercise time (h), in hours, and the <em>dependent</em> variable is the calorie gain (T), in calories. The constant "2000" is the <em>minimum</em> calorie consumption and the constant "500" represents the amount of <em>additional</em> calories per each hour of exercise.

Now we proceed to respond the questions of the part a):

(i) T - Amount of calories needed, (ii) 2000 - Minimum calorie consumption, (iii) 500 - Amount of additional calories per hour of exercise.

b) And lastly we calculate the amount of calories by evaluating the function:

T = 2000 + 500 · 3

T = 2000 + 1500

T = 3500

We should consume 3500 calories to spend three hours exercising.

<h3>Remark</h3>

The statement is incomplete, the complete form is shown below:

You are trying to bulk up and gain muscle. Your trainer says that your daily calorie intake should increase based on the hours you spend exercising and can be represented by the following equation: T = 2000 +  500 · h

a) What does each part of this expression represent? (i) T = , (ii) 2000, (iii) 500, (iv) h

b) If you spend 3 hours exercising, how many calories should you consume?

To learn more on linear functions: brainly.com/question/21107621

#SPJ1

4 0
2 years ago
xand y are light house. y being 20km due east of x and from a ship due south of x the bearing of y was 055°. what is the distanc
leonid [27]

Answer:

I) |xz| ≈ 28.6 km

II) |yz| ≈ 34.8 km

Step-by-step explanation:

Let's assume that the position of ship due south of x is z (aà pictor representation of the question is attached)

|xy| = 20 km, |xz| = ?, |yz| = ?, θ(y) = 55°

Using Trigonometric ratio - SOHCAHTOA

I) Tan θ = |xz| ÷ |xy| ⇒ Tan 55° = |xz| ÷ 20

|xz| = 20 * Tan 55 = 20 * 1.428

|xz| = 28.56 km

|xz| ≈ <u>28.6 km</u>

<u />

II) Cos θ = |xy| ÷ |yz| ⇒ Cos 55° = 20 ÷ |yz|

|yz| * Cos 55° = 20 ⇒ |yz| = 20 ÷ Cos 55°

|yz| = 20 ÷ 0.574 = 34.84 km

|yz| ≈ <u>34.8 km</u>

4 0
3 years ago
The probability that a professor arrives on time is 0.8 and the probability that a student arrives on time is 0.6. Assuming thes
saul85 [17]

Answer:

a)0.08  , b)0.4  , C) i)0.84  , ii)0.56

Step-by-step explanation:

Given data

P(A) =  professor arrives on time

P(A) = 0.8

P(B) =  Student aarive on time

P(B) = 0.6

According to the question A & B are Independent  

P(A∩B) = P(A) . P(B)

Therefore  

{A}' & {B}' is also independent

{A}' = 1-0.8 = 0.2

{B}' = 1-0.6 = 0.4

part a)

Probability of both student and the professor are late

P(A'∩B') = P(A') . P(B')  (only for independent cases)

= 0.2 x 0.4

= 0.08

Part b)

The probability that the student is late given that the professor is on time

P(\frac{B'}{A}) = \frac{P(B'\cap A)}{P(A)} = \frac{0.4\times 0.8}{0.8} = 0.4

Part c)

Assume the events are not independent

Given Data

P(\frac{{A}'}{{B}'}) = 0.4

=\frac{P({A}'\cap {B}')}{P({B}')} = 0.4

P({A}'\cap {B}') = 0.4 x P({B}')

= 0.4 x 0.4 = 0.16

P({A}'\cap {B}') = 0.16

i)

The probability that at least one of them is on time

P(A\cup B) = 1- P({A}'\cap {B}')  

=  1 - 0.16 = 0.84

ii)The probability that they are both on time

P(A\cap  B) = 1 - P({A}'\cup {B}') = 1 - [P({A}')+P({B}') - P({A}'\cap {B}')]

= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56

6 0
3 years ago
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