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vazorg [7]
2 years ago
10

Please help i’m sorry!!! 15 points again!

Mathematics
1 answer:
morpeh [17]2 years ago
7 0

Answer:

Guess I'll agree with the other person.

Step-by-step explanation:

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Please help me I don't get it
noname [10]
D is 90 and e is 130
6 0
3 years ago
A scale for a scale drawing is 10 cm:1 mm. Which is larger, the actual object or scale drawing? Explain.
Marysya12 [62]
The <u>correct answer</u> is:

The scale drawing is larger.

Explanation:

A scale for a scale drawing is written as a ratio in the form
(scale size):(actual size).  

For example, on a map you may have a scale that says 1 in: 2 mi.  This means that 1 inch on the scale is equal to 2 miles on the map.

The scale for this problem is 10 cm: 1 mm.  This means that 10 cm on the scale drawing represents 1 mm on the actual object.

10 cm is larger than 1 mm, so the scale drawing is larger.
4 0
3 years ago
Read 2 more answers
Factorize 2√2a∧3 +8b∧3 -27c∧3+18√2abc
rosijanka [135]

Answer:

{sqrt(2) 2 sqrt(a), 8 b + 3, 3 - 27 c, 18 sqrt(2) a b c + 3}

Step-by-step explanation:

7 0
3 years ago
Anybody know the answer ?
algol [13]
C=2πr
 We have 3/4 of circumference
 (3π/2)*2π=3/4.
 and it is equal 5π/2.
(3/4)*C=(3/4)*2πr
5π/2 = 6πr/4
5/2=6r/4
r=(5/2)*(4/6)=(5*4)/(2*6)=5/3
r=5/3
6 0
3 years ago
Convert from rectangular to polar coordinates: note: choose rr and θθ such that rr is nonnegative and 0≤θ&lt;2π0≤θ&lt;2π (a)(9,0
DochEvi [55]

Answer:

Step-by-step explanation:

Convert rectangle (x , y) to polar coordinates ( r , θ)

x=r  \cos \theta, y= r \sin \theta

r=\sqrt{x^2+y^2} , \theta =tan^-^1 (\frac{y}{x} )

a) converts (9, 0) to polar coordinates  ( r , θ)

r=\sqrt{x^2+y^2} \\\\=\sqrt{9^2+0} \\\\=9

\theta= \tan^-^1 (\frac{0}{9} )\\\\=0

b) Convert (18,\frac{18}{\sqrt{3} } ) to polar coordinates ( r, θ)

r = \sqrt{18^2+(\frac{18}{\sqrt{3} })^2 } \\\\=\sqrt{324+108} \\\\=\sqrt{432}

\frac{x}{y} \theta = \tan^-^1(\frac{\frac{18}{\sqrt{3} } }{18} )\\\\= \tan ^-^1(\frac{1}{\sqrt{3} } )\\\\= \frac{\pi}{6}

c)  converts (-5, 5) to polar coordinates  ( r , θ)

r =\sqrt{(-5)^2+(5)^2} \\\\=\sqrt{50} \\\\=5\sqrt{2}

\theta=\tan^-^1(\frac{5}{-5} )\\\\= \tan^-^1(-1)\\\\=\frac{3\pi}{4}

d)  converts (-1, √3) to polar coordinates  ( r , θ)

r=\sqrt{(-1)^2+(\sqrt{3})^2 } \\\\= \sqrt{4} \\\\=2

\theta=\tan^-^1(\frac{\sqrt{3} }{-1} )\\\=\tan^-^1(-\sqrt{3} )\\\\=\frac{2\pi}{3}

= \frac{2\pi}{\sqrt{3} }

6 0
3 years ago
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