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Anon25 [30]
3 years ago
15

In a scale drawing the length of a rectangular room is 6 inches and the width is 3 inches and the actual length of the room is 1

8 feet. What is the scale of the drawing
Mathematics
1 answer:
sashaice [31]3 years ago
7 0

Answer:

1:36

Step-by-step explanation:

Length:

6 inches : 18 feet

1 ft = 12 inches

6 inches = 6/12 = 0.5 ft

Length:

0.5 ft : 18 ft

then. rhe scale is:

1: 36

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What is the point of symmetry for the circle (x + 2)2 + (y â 4)2 = 25?
irina [24]
It's its center:

(-2, 4)

Can't read y-4 or y+4, change the sign.

y-4 ---> SOLUTION: (-2,4)

y+4 ---> SOLUTION: (-2, -4)
3 0
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Step-by-step explanation:

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3 0
3 years ago
What would the sum of -18-(-12) be?
Mademuasel [1]
-18-(-12)=-6 
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7 0
3 years ago
Read 2 more answers
Sin tita= 0.6892.find the value Of tita correct to two decimal places<br><br>​
Anvisha [2.4K]

Answer:

\theta \approx 6.28n + 2.38,  \quad  n \in \mathbb{Z}

or

\theta \approx 6.28n + 0.76, \quad n \in \mathbb{Z}

Considering \theta \in (0, 2\pi]

\theta \approx 2.38

or

\theta \approx 0.76

Step-by-step explanation:

\sin(\theta)=0.6892

We have:

\sin (x)=a \Longrightarrow x=\arcsin (a)+2\pi n \text{ or } x=\pi -\arcsin (a)+2\pi n \text{ as } n\in \mathbb{Z}

Therefore,

\theta= \arcsin (0.6892)+2\pi n, \quad n \in \mathbb{Z}

or

\theta = \pi -\arcsin (0.6892)+2\pi n, \quad  n\in \mathbb{Z}

---------------------------------

\theta \approx 6.28n + 2.38,  \quad  n \in \mathbb{Z}

or

\theta \approx 6.28n + 0.76, \quad n \in \mathbb{Z}

4 0
3 years ago
Use the distributive property to express 15+45
Nuetrik [128]
<u><em>Answer:</em></u>
15(1+3)

<u><em>Explanation:</em></u>
<u>The distributive property can be generally expressed as follows:</u>
ab + ac = a(b+c)

<u>The given expression is:</u>
15 + 45

<u>We know that:</u>
15 = 1*15
45 = 3*15

<u>Therefore, the given expression can be written as:</u>
1*15 + 3*15

<u>Taking 15 as a common factor and applying the above rule, we will reach the following expression:</u>
15(1+3)

Hope this helps :)
5 0
3 years ago
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