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OleMash [197]
3 years ago
11

An artist makes a scale drawing of a parallelogram-shaped sculpture. The scale is 10 cm on the drawing for every 8 meters on the

sculpture. What is the area of the scale drawing? Show your work.

Mathematics
1 answer:
8_murik_8 [283]3 years ago
3 0

Answer:

Area of drawing= 0.315 m²

Step-by-step explanation:

Base = 6m

Height = 4.2 m

Area = height* base

Area= 4.2*6

Area= 25.2 m²

The scale is 10 cm on the drawing for every 8 meters on the sculpture.

Which is 10 cm to 800 cm

Or

0.1m to 8m

Arra of sculpture= 25.2 m²

area of drawing= (25.2*0.1)/8

Area of drawing= 0.315 m²

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Answer:

she will make $270 and each friend buys 9 CD's

Step-by-step explanation:

To find the money she makes take the CDs and multiply it times the money each one costs. To find the amount each friend buys divide the cds by th amount of friends

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3 years ago
Exspress the ratio from the simplest form 72:36
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First, find the GCF. For these two numbers, it is 36. (Because 36 times 1 is 36 and 36 times 2 is 72)
Divide both sides by this, which gives you a
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3 years ago
Read 2 more answers
A traffic helicopter descends 102 meters to be 331 meters above the ground, as illustrated in the diagram to the right. Write an
lana66690 [7]

Answer:

  • h -102 = 331
  • h = 433

Step-by-step explanation:

Using the problem statement as the basis for the equation, we can write ...

  h -102 = 331

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Using the diagram as the basis for the equation, we could write it as ...

  h = 331 +102

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The original height of the helicopter was 433 m.

3 0
3 years ago
Which of the following is not one of the 8th roots of unity?
Anika [276]

Answer:

1+i

Step-by-step explanation:

To find the 8th roots of unity, you have to find the trigonometric form of unity.

1.  Since z=1=1+0\cdot i, then

Rez=1,\\ \\Im z=0

and

|z|=\sqrt{1^2+0^2}=1,\\ \\\\\cos\varphi =\dfrac{Rez}{|z|}=\dfrac{1}{1}=1,\\ \\\sin\varphi =\dfrac{Imz}{|z|}=\dfrac{0}{1}=0.

This gives you \varphi=0.

Thus,

z=1\cdot(\cos 0+i\sin 0).

2. The 8th roots can be calculated using following formula:

\sqrt[8]{z}=\{\sqrt[8]{|z|} (\cos\dfrac{\varphi+2\pi k}{8}+i\sin \dfrac{\varphi+2\pi k}{8}), k=0,\ 1,\dots,7\}.

Now

at k=0,  z_0=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 0}{8}+i\sin \dfrac{0+2\pi \cdot 0}{8})=1\cdot (1+0\cdot i)=1;

at k=1,  z_1=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 1}{8}+i\sin \dfrac{0+2\pi \cdot 1}{8})=1\cdot (\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};

at k=2,  z_2=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 2}{8}+i\sin \dfrac{0+2\pi \cdot 2}{8})=1\cdot (0+1\cdot i)=i;

at k=3,  z_3=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 3}{8}+i\sin \dfrac{0+2\pi \cdot 3}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};

at k=4,  z_4=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 4}{8}+i\sin \dfrac{0+2\pi \cdot 4}{8})=1\cdot (-1+0\cdot i)=-1;

at k=5,  z_5=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 5}{8}+i\sin \dfrac{0+2\pi \cdot 5}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};

at k=6,  z_6=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 6}{8}+i\sin \dfrac{0+2\pi \cdot 6}{8})=1\cdot (0-1\cdot i)=-i;

at k=7,  z_7=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 7}{8}+i\sin \dfrac{0+2\pi \cdot 7}{8})=1\cdot (\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};

The 8th roots are

\{1,\ \dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ i, -\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ -1, -\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2},\ -i,\ \dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2}\}.

Option C is icncorrect.

5 0
3 years ago
What is the slope of a line that is parallel to the line y = 3/4 + 2?
Elodia [21]
3/4 because it’s parallel so it has the same slope. It just doesn’t pass through the same line.
7 0
3 years ago
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