Ask question

Ask question

All categories

2 years ago

13

A pole that is 3.2m tall casts a shadow that is 1.54m long. At the same time, a nearby building casts a shadow that is 50.75m lo

ng. How tall is the building? Round your answer to the nearest meter.

1 answer:

Nastasia [14]2 years ago

6 0

3.2m casts 1.54m shadow.

3.2÷1.54= 160/77 (2.077922077922...)

50.75 × 160/77= 105.454545...

105m to the nearest meter.

You might be interested in

The differenceof a minus 3 divided by 3

Vika [28.1K]

Answer:

The difference of a minus 3 divided by 3 is - 1.

Step-by-step explanation:

-3/3= - 1

Hope it's helpful.

4 0

3 years ago

Please help with these Math question's ASP.

vodomira [7]

Answer:

See below in bold.

Step-by-step explanation:

1. x/8 = 15/24

Cross multiplying:

24x = 8*15

x = (8*15)/24

x = 5.

2. 5/6 = 35/x

5x = 6*35

5x = 210

x = 210/5

x = 42.

3.<em> </em>18/42

Divide top and bottom by 6:

18 / 6 / 42/6

= 3/7.

4. 15/75

Divide top and bottom by 15:

= 1/5.

5 0

3 years ago

Can anyone help me? i need to provide work!!! 50 points and brainliest answer to whoever helps the most!

mylen [45]

Answer:

Maybe if you look the name of your worksheet and the answers to it, you can find the answers. That has helped me before...

Step-by-step explanation:

3 0

3 years ago

The table below shows data from a survey about the amount of time students spend doing homework each week. The students were in

nadezda [96]

Answer:

The correct option is;

Both spreads are best described by the standard deviation

Step-by-step explanation:

The given information are;

, College High School

High, 20 20

Low, 6 3

Q₁, 8 5.5

Q₃, 18 16

IQR, 10 10.5

Median, 14 11

Mean, 13.3 11

σ, 5.2 5.4

Checking for outliers, we have

College

Q₁ - 1.5×IQR gives 8 - 1.5×10 = -7

Q₃ + 1.5×IQR gives 18 + 1.5×10 = 33

For high school

Q₁ - 1.5×IQR gives 5.5 - 1.5×10.5 = -10.25

Q₃ + 1.5×IQR gives 16 + 1.5×10.5 = 31.75

Therefore, there are no outliers and the data is representative of the population

From the data, for the college students, it is observed that the difference between the mean, 13.3 and Q₁, 8, and between Q₃, 18 and the mean,13.3 is approximately the standard deviation, σ, 5.2

The difference between the low and the high is also approximately 3 standard deviations

Therefore the college spread is best described by the standard deviation

Similarly for the high school students, the IQR is approximately two standard deviations, the difference between the mean, 11 and Q₁, 5.5, and between Q₃, 16 and the mean,11 is approximately the standard deviation, σ, 5.4

Therefore the high school spread is also best described by the standard deviation.

4 0

3 years ago

Let S be the solid beneath z = 12xy^2 and above z = 0, over the rectangle [0, 1] × [0, 1]. Find the value of m > 1 so that th

jonny [76]

Answer:

The answer is $\sqrt{\frac{6}{5}}$

Step-by-step explanation:

To calculate the volumen of the solid we solve the next double integral:

$\int\limits^1_0\int\limits^1_0 {12xy^{2} } \, dxdy$

Solving:

$\int\limits^1_0 {12x} \, dx \int\limits^1_0 {y^{2} } \, dy$

$[6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.$

Replacing the limits:

$6*\frac{1}{3} =2$

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.

Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤ $\frac{1}{m}$

Solving the double integral with these new limits we have:

$\int\limits^\frac{1}{m} _0\int\limits^{1}_{mx} {12xy^{2} } \, dxdy$

This part is a little bit tricky so let's solve the integral first for dy:

$\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx$

Replacing the limits:

$\int\limits^\frac{1}{m}_0 {4x(1-(mx)^{3} )\, dx =\int\limits^\frac{1}{m}_0 {4x-4x(m^{3} x^{3} )\, dx =\int\limits^\frac{1}{m}_0 ({4x-4m^{3} x^{4}) \, dx$

Solving now for dx:

$[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.$

Replacing the limits:

$\frac{2}{m^{2} }-\frac{4m^{3}\frac{1}{m^{5}}}{5} =\frac{2}{m^{2} }-\frac{4\frac{1}{m^{2}}}{5} \\ \frac{2}{m^{2} }-\frac{4}{5m^{2} }=\frac{10m^{2}-4m^{2} }{5m^{4}} \\ \frac{6m^{2} }{5m^{4}} =\frac{6}{5m^{2}}$

As I mentioned before, this volume is equal to 1, hence:

$\frac{6}{5m^{2}}=1\\m^{2} =\frac{6}{5} \\m=\sqrt{\frac{6}{5} }$

3 0

2 years ago