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maw [93]
3 years ago
14

What is indicated by arrowheads on a line

Mathematics
2 answers:
Molodets [167]3 years ago
7 0
Arrowheads indicate that the line will continue traveling on a number line infinitely. In other words the line never ends. 
spayn [35]3 years ago
4 0

Answer:

the line will continue traveling on a number line infinitely. In other words the line never ends.

Step-by-step explanation:

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Can anyone help me solve a trigonomic identity problem and also help me how to do it step by step?
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\\\\\\
sin^2(\theta)+cos^2(\theta)=1\\\\
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\bf \cfrac{cos(\theta )cot(\theta )}{1-sin(\theta )}-1=csc(\theta )\\\\
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\cfrac{cos(\theta )\cdot \frac{cos(\theta )}{sin(\theta )}}{1-sin(\theta )}-1\implies \cfrac{\frac{cos^2(\theta )}{sin(\theta )}}{\frac{1-sin(\theta )}{1}}-1\implies 
\cfrac{cos^2(\theta )}{sin(\theta )}\cdot \cfrac{1}{1-sin(\theta )}-1
\\\\\\
\cfrac{cos^2(\theta )}{sin(\theta )[1-sin(\theta )]}-1\implies 
\cfrac{cos^2(\theta )-1[sin(\theta )[1-sin(\theta )]]}{sin(\theta )[1-sin(\theta )]}

\bf \cfrac{cos^2(\theta )-1[sin(\theta )-sin^2(\theta )]}{sin(\theta )[1-sin(\theta )]}\implies \cfrac{cos^2(\theta )-sin(\theta )+sin^2(\theta )}{sin(\theta )[1-sin(\theta )]}
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\cfrac{cos^2(\theta )+sin^2(\theta )-sin(\theta )}{sin(\theta )[1-sin(\theta )]}\implies \cfrac{\underline{1-sin(\theta )}}{sin(\theta )\underline{[1-sin(\theta )]}}
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7 0
3 years ago
How do I do this please help!
jeka94

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6 0
3 years ago
QUIZ<br> Find the volume of the square pyramid.<br> 9 cm<br> 4 cm
kari74 [83]

Answer:

SEE BELOW

Step-by-step explanation:

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5 0
2 years ago
Samples of emissions from three suppliers are classified for conformance to air-quality specifications. The results from 100 sam
tigry1 [53]

Answer:

P(A) = \frac{30}{100}

P(B) = \frac{77}{100}

P(A\ n\ B) = \frac{22}{100}

P(A\ u\ B) = \frac{85}{100}

Step-by-step explanation:

Given

See attachment for proper format of table

n = 100 --- Sample

A = Supplier 1

B = Conforms to specification

Solving (a): P(A)

Here, we only consider data in sample 1 row.

In this row:

Yes = 22 and No = 8

So, we have:

n(A) = Yes + No

n(A) = 22 + 8

n(A) = 30

P(A) is then calculated as:

P(A) = \frac{n(A)}{Sample}

P(A) = \frac{30}{100}

Solving (b): P(B)

Here, we only consider data in the Yes column.

In this column:

(1) = 22    (2) = 25 and (3) = 30

So, we have:

n(B) = (1) + (2) + (3)

n(B) = 22 + 25 + 30

n(B) = 77

P(B) is then calculated as:

P(B) = \frac{n(B)}{Sample}

P(B) = \frac{77}{100}

Solving (c): P(A n B)

Here, we only consider the similar cell in the yes column and sample 1 row.

This cell is: [Supplier 1][Yes]

And it is represented with; n(A n B)

So, we have:

n(A\ n\ B) = 22

The probability is then calculated as:

P(A\ n\ B) = \frac{n(A\ n\ B)}{Sample}

P(A\ n\ B) = \frac{22}{100}

Solving (d): P(A u B)

This is calculated as:

P(A\ u\ B) = P(A) + P(B) - P(A\ n\ B)

This gives:

P(A\ u\ B) = \frac{30}{100} + \frac{77}{100} - \frac{22}{100}

Take LCM

P(A\ u\ B) = \frac{30+77-22}{100}

P(A\ u\ B) = \frac{85}{100}

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How to find the missing factor for binomials
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Answer: pizza

Step-by-step explanation:

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3 years ago
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