Eighty is 0.4 of what number

Blababa [14]

Answer:

73.72 ft Squared

Step-by-step explanation:

First you need to find the area of the rectangle the area of rectangle is 8x10=80. Then You take that 80=3.14x2. then you find the area by dividing it. so...

1.) 80=3.14x2

3.14x2=6.28 80=6.28 (Then you divide 6.28 by 80)

80 80

3 0

3 years ago

you are asked to size the diameter of a bolt loaded in shear to the nearest 1/8 inch with a factor of safety of 1.5. you find th

kupik [55]

Answer:

2.375 inches

Step-by-step explanation:

Main Steps

1. Identifying important parts of the question

2. Solving the problem

Step 1. - Identifying important parts of the question

In the question, there is some extraneous information. For instance, the question introduces the terms "diameter" and "loaded in shear" and "with a safety factor of 1.5". However, given the information in the problem, and the question asked, it isn't necessary to know that it is "loaded in shear" means, or to know what a diameter is (since everything stays with diameters, and doesn't involve another dimension), and the problem gives the result that will satisfy "a safety factor of 1.5".

Sifting through the problem/question, these are the important parts:

- "to the nearest 1/8 inch" (meaning round to the nearest 1/8th of a unit)

- with "units of inches to 3 decimal places"

- The number to round is "2.4183 inches

Step 2. - Solving the Problem

To round to the nearest 1/8 inch, we can divide the number 2.4183 by 1/8.

$2.4183 \div(1/8)$

Recall that dividing by a fraction is the same as multiplying by the reciprocal

$2.4183 * 8$

19.3464

So, there are approximately 19, one-eighth-inches in the measurement. (Note that 19.3464 is closer to 19 than it is to 20). To find the final rounded measurement, multiply 19 by 1/8

$19 * (1/8)$

2.375 inches

So, the diameter of the bolt that should be used (if rounded to the nearest 1/8 inch, to allow for a factor of safety of 1.5) is 2.375 inches

5 0

1 year ago

An object moving along a curve is at position P(2cost, sint) where t denotes time in seconds. Find the speed of the object at po

shepuryov [24]

We have 2 equations to specify the location of the object and we desire the velocity. In order to get that, we simply need to calculate the first derivative of each location equation. So: X = 2 cos(t) X' = 2 (-sin(t)) X' = -2 sin(t) Y = sin(t) Y' = cos(t) So the velocity vector at time t is (-2sin(t), cos(t)). But you want the velocity. So using the Pythagorean theorem we can get that by calculating the square root of the sum of the squares. So: V = sqrt((-2sin(t))^2 + cos^2(t)) V = sqrt(4sin^2(t) + cos^2(t)) Speed at t = 1, is V = sqrt(4sin^2(1) + cos^2(1)) V = sqrt(2.832293673 + 0.291926582) V = sqrt(3.124220255) V = 1.767546394 And t=3: V = sqrt(4sin^2(3) + cos^2(3)) V = sqrt(0.079659427 + 0.980085143) V = sqrt(1.05974457) V = 1.029438959 Now asking for velocity as a function of P, we have a bit of a complication. As shown above, it's trivial to calculate velocity as a function of t. But if all you're given is the X and Y coordinates of the object, we have a bit more work to do. The below equations will be using the trigonometric identity of cos^2(a) + sin^2(a) = 1 for any angle a. X = 2 cos(t) X' = -2 sin(t) We want to get from X which is 2cos(t) to X'^2 which is 4sin^2(t). So: X/2; We now have cos(t) (X/2)^2: We now have cos^2(t) 1-(X/2)^2: We now have sin^2(t) 4(1-(X/2)^2): We now have 4sin^2(t) which is what we want. Time to simplify 4(1 - (X/2)^2) 4(1 - (X^2/4)) 4 - 4(X^2/4) 4 - X^2 Now we need to get from Y to Y'^2. Will do the same as for X to X'^2, but without all the comments. Y = sin(t) Y' = cos(t) Y'^2 = 1 - Y^2 So the equation for the velocity as a function of X,Y we get V = sqrt(4 - X^2 + 1 - Y^2) V = sqrt(5 - X^2 - Y^2) In summary: Position at time t = (2cos(t), sin(t)) Velocity vector at time t = (-2 sin(t), cos(t)) Velocity as function of t is: V = sqrt(4sin^2(t) + cos^2(t)) Velocity as function of P is: V = sqrt(5 - X^2 - Y^2) Is object traveling at constant speed? NO Velocity at t = 1 is: V = 1.767546394 Velocity at t = 2 is: V = 1.029438959

7 0

4 years ago

The value of the 5 in the ones place in the number 5,555 is how many times as much as the value of the 5 in the hundred place?

Radda [10]

100 times, because 5 times 100 equals 500

6 0

4 years ago

The water diet requires you to drink two cups of water every half hour from the time you get up until you go to bed, but otherwi

Elan Coil [88]

Answer:

$7-3.182 \frac{13.638}{\sqrt{4}}= -14.698$

$7+3.182 \frac{13.638}{\sqrt{4}}= 28.698$

Step-by-step explanation:

For this case we have the following info given:

Weight before diet 180 125 240 150

Weight after diet 170 130 215 152

We define the random variable $D = before-after$ and we can calculate the inidividual values:

D: 10, -5, 25, -2

And we can calculate the mean with this formula:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

And the deviation with:

$s= \sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n-1}}$

And after replace we got:

$\bar D= 7, s_d = 13.638$

And the confidence interval for this case would be given by:

$\bar D \pm t_{\alpha/2} \frac{s_d}{\sqrt{n}}$

The degrees of freedom are given by:

$df = n-1= 4-1=3$

For the 95% of confidence the value for the significance is $\alpha=0.05$ and the critical value would be $t_{\alpha/2}= 3.182$ . And replacing we got:

$7-3.182 \frac{13.638}{\sqrt{4}}= -14.698$

$7+3.182 \frac{13.638}{\sqrt{4}}= 28.698$

6 0

3 years ago