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Ilya [14]
3 years ago
12

HELP!! I DON't HAVE MUCH TIME

Mathematics
1 answer:
babunello [35]3 years ago
5 0
The answer is 153 you’re welcome !!
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The function h(n) describes the total amount of money a movie theater receives for n tickets sold.
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Answer:

Let x be the amount of money

and n be the number of tickets:

h(n) = x \times n

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What is the rule for calculating |x| for any value of x?
padilas [110]
Anything with || around it is a whole value. It’s always a positive. For example, |-56|=56. Now, if the negative was on the outside, it would be a negative like so; -|10| =-10
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Match the irrational numbers with their estimated positions on the number line.
IrinaVladis [17]
The sqrt of 19 is between 4.3 and 4.4.
The sqrt of 18 is between 4.2 and 4.3.
The sqrt of 22 is between 4.6 and 4.7.
And finally, the sqrt of 21 is between 4.5 and 4.6.
3 0
3 years ago
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A soccer store can ship 7 soccer balls in each box. How many boxes are needed to ship 81 soccer balls?
yanalaym [24]
The 81 balls need to be split into some number of boxes. To figure out how many boxes, we can divide the 81 balls by the 7 ball capacity of a single box.

81/7 = 11.571

This means that we will need 11 full boxes, but there will be a remainder of balls which did not fit. Since we can't have a fraction of a box, we need to round up to 12 boxes to hold all 81 balls.

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6 0
3 years ago
Angle α lies in quadrant II , and tanα=−125 . Angle β lies in quadrant IV , and cosβ=35 .
Artist 52 [7]

Answer:

cos(\alpha+\beta)=\frac{33}{65}

Step-by-step explanation:

step 1

Find cos α

we know that

tan^2(\alpha)+1=sec^2(\alpha)

we have

tan(\alpha)=-\frac{12}{5}

substitute

(-\frac{12}{5})^2+1=sec^2(\alpha)

sec^2(\alpha)=\frac{144}{25}+1

sec^2(\alpha)=\frac{169}{25}

sec(\alpha)=\pm\frac{13}{5}

Remember that Angle α lies in quadrant II

so

sec α is negative

sec(\alpha)=-\frac{13}{5}

Find the value of cos α

cos)\alpha)=\frac{1}{sec(\alpha)}

so

cos(\alpha)=-\frac{5}{13}

step 2

Find sin α

we know that

tan(\alpha)=\frac{sin(\alpha)}{cos(\alpha)}

sin(\alpha)=tan(\alpha)cos(\alpha)

we have

tan(\alpha)=-\frac{12}{5}

cos(\alpha)=-\frac{5}{13}

substitute

sin(\alpha)=(-\frac{12}{5})(-\frac{5}{13})

sin(\alpha)=\frac{12}{13}

step 3

Find sin β

we know that

sin^2(\beta)+cos^2(\beta)=1

we have

cos(\beta)=\frac{3}{5}

substitute

sin^2(\beta)+(\frac{3}{5})^2=1

sin^2(\beta)=1-(\frac{3}{5})^2

sin^2(\beta)=1-\frac{9}{25}

sin^2(\beta)=\frac{16}{25}

sin(\beta)=\pm\frac{4}{5}

Remember that

Angle β lies in quadrant IV

so

sin β is negative

sin(\beta)=-\frac{4}{5}

step 4

Find cos(α−β)

we know that

cos(\alpha+\beta)=cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta)

we have

cos(\alpha)=-\frac{5}{13}

cos(\beta)=\frac{3}{5}

sin(\alpha)=\frac{12}{13}

sin(\beta)=-\frac{4}{5}

substitute the given values

cos(\alpha+\beta)=(-\frac{5}{13})(\frac{3}{5})-(\frac{12}{13})(-\frac{4}{5})

cos(\alpha+\beta)=(-\frac{15}{65})+(\frac{48}{65})

cos(\alpha+\beta)=\frac{33}{65}

7 0
4 years ago
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