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ASHA 777 [7]
2 years ago
8

Tia made a scale drawing of the White House for her history project. The actual length of the building is 168 feet, and its widt

h (with porticoes) is 152 feet. If the scaled length of the building in the drawing is 21 inches and the width is 19 inches, what scale did Tia use to make the drawing?
Mathematics
2 answers:
Delicious77 [7]2 years ago
6 0

Answer:

One IN to 9 FT!


Step-by-step explanation:


Murrr4er [49]2 years ago
4 0
The answer is 1 inch to 9 feet
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A theater group made appearances in two cities. The hotel charge before tax in the second city was $1000 lower than in the first
Bess [88]

Answer:

  • $5000 in the first city
  • $4000 in the second city

Step-by-step explanation:

Let x represent the pre-tax hotel charge in the first city in dollars. Then the hotel charge in the second city is x-1000 dollars, and the total tax is ...

... 7.5%·x +5%·(x -1000) = 575

... 12.5%·x = 625 . . . . simplify, add 50

... x = 5000 . . . . . . . . divide by 12.5% (=0.125)

The first-city charge was $5000; the second-city charge was $4000.

8 0
2 years ago
Square ABCD is shown below with line EF passing through the center: If Square ABCD is dilated by a scale factor of two about the
dlinn [17]
Line E'F' would also double.


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2 years ago
Use the formula A=bh , where A is the area, b is the base length, and h is the height of the parallelogram, to solve this proble
Jlenok [28]
33 in because 1056 divided by 32 is 33
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3 0
2 years ago
Read 2 more answers
The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillat
Greeley [361]

Answer:

1) L \propto T^2

Using the condition given:

2.205 m = K (3)^2

K = 0.245 \approx \frac{g}{4\pi^2}

So then if we want to create an equation we need to do this:

L = K T^2

With K a constant. For this case the period of a pendulumn is given by this general expression:

T = 2\pi \sqrt{\frac{L}{g}}

Where L is the length in m and g the gravity g = 9.8 \frac{m}{s^2}.

2) T = 2\pi \sqrt{\frac{L}{g}}

If we square both sides of the equation we got:

T^2 = 4 \pi^2 \frac{L}{g}

And solving for L we got:

L = \frac{g T^2}{4 \pi^2}

Replacing we got:

L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m

3) T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s

Step-by-step explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

L \propto T^2

Using the condition given:

2.205 m = K (3)^2

K = 0.245 \approx \frac{g}{4\pi^2}

So then if we want to create an equation we need to do this:

L = K T^2

With K a constant. For this case the period of a pendulumn is given by this general expression:

T = 2\pi \sqrt{\frac{L}{g}}

Where L is the length in m and g the gravity g = 9.8 \frac{m}{s^2}.

Part 2

For this case using the function in part a we got:

T = 2\pi \sqrt{\frac{L}{g}}

If we square both sides of the equation we got:

T^2 = 4 \pi^2 \frac{L}{g}

And solving for L we got:

L = \frac{g T^2}{4 \pi^2}

Replacing we got:

L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m

Part 3

For this case using the function in part a we got:

T = 2\pi \sqrt{\frac{L}{g}}

Replacing we got:

T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s

8 0
2 years ago
. Given f(x) = e 2x e 2x + 3e x + 2 : (a) Make the substitution u = e x to convert Z f(x) dx into an integral in u (HINT: The ea
MrRa [10]

Answer:

Step-by-step explanation:

Given;

f(x)=\frac{e^{2x}}{e^{2x}+3e^x+2}

a)

substitute u=e^x\\du=e^x dx\\\\\int\frac{e^{2x}}{e^{2x}+3e^x+2}dx=\int\frac{e^x\dot e^x}{e^x^{2x}+3e^x+2}dx\\\\=\int\frac{udu}{u^2+3u+2}

b)

Apply partial fraction in (a), we get;

\frac{u}{u^2+3u+2}=\frac{2}{u+2}-\frac{1}{u+1}\\\\\therefore u^2+3u+2\\=u^2+2u+u+2\\=u(u+2)+1(u+1)\\=(u+2)(u+1)\\\\Now\,\int\frac{u}{u^2+3u+2}\,du=\int\frac{2}{u+2}du-\int\frac{1}{u+1}du\\\\=2ln|u+2|-ln|u+1|+c\\=2ln|e^x+2|-lm|e^x+1|+c

where C is an arbitrary constant

8 0
2 years ago
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