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sergeinik [125]
3 years ago
6

Simplify the square root of 27 over 12

Mathematics
1 answer:
Hunter-Best [27]3 years ago
5 0
\sqrt{ \frac{27}{12} }

{ \frac{ \sqrt{27} {} }{ \sqrt{12}}

\frac{ \sqrt{9 * 3} }{ \sqrt{4 * 3} }

\frac{ \sqrt{9}  \sqrt{3} }{ \sqrt{4}  \sqrt{3} }

\frac{3 \sqrt{3} }{2 \sqrt{3} }

\frac{3}{2}

1  \frac{1}{2}
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Measure of Angle 1 is 5x.<br> Measure of Angle 3 is x-24.<br><br> What is the measure of Angle 1?
olga55 [171]

Answer:

The answer to your question is:  30°

Step-by-step explanation:

Data

m∠ 1 = 5x

m∠ 3 = x - 24

Process

m ∠ 1 = m ∠ 3 because they are vertical angles

            5x = x - 24

            5x - x = 24

            4x = 24

            x = 24 / 4

            x = 6

Angle 1 = 5(6)

            = 30°

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3 years ago
What is 0.15% of 43 i suck at math it will help me so much
butalik [34]
.15% translates to
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.0645
7 0
3 years ago
The perimeter of a rectangle is 50 inches the length of the rectangle is 17 inches He cuts out another rectangle that is the sam
Brums [2.3K]

Step-by-step explanation:

A rectangle: P=2l+2w or P=2(l+w) (same formula written differently)

50=2x17+2w

50=34+2w

2w=50-34

2w=16

w=16÷2=8

Since it says twice as wide then we will multiply w with 2

2xW

2x8=16 which is the width of the 2nd rectangle

Hope this helps :)

6 0
2 years ago
You have $44 to spend at the music store. Each cassette tape costs $10 and each CD costs $12. Write a linear inequality that rep
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Read 2 more answers
A fair coin is tossed three times and the events A, B, and C are defined as follows: A:{ At least one head is observed } B:{ At
kicyunya [14]

Answer:

(a) 1/2

(b) 1/2

(c) 1/8

Step-by-step explanation:

Since, when a fair coin is tossed three times,

The the total number of possible outcomes

n(S) = 2 × 2 × 2

= 8 { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT },

Here, B : { At least two heads are observed } ,

⇒ B = {HHH, HHT, HTH, THH},

⇒ n(B) = 4,

Since,

\text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}

(a) So, the probability of B,

P(B) =\frac{n(B)}{n(S)}=\frac{4}{8}=\frac{1}{2}

(b) A : { At least one head is observed },

⇒ A = {HHH, HHT, HTH, THH, HTT, THT, TTH},

∵ A ∩ B = {HHH, HHT, HTH, THH},

n(A∩ B) = 4,

\implies P(A\cap B) = \frac{n(A\cap B)}{n(S)} = \frac{4}{8}=\frac{1}{2}

(c) C: { The number of heads observed is odd },

⇒ C = { HHH, HTT, THT, TTH},

∵ A ∩ B ∩ C = {HHH},

⇒ n(A ∩ B ∩ C) = 1,

\implies P(A\cap B\cap C)=\frac{1}{8}

7 0
2 years ago
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