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TiliK225 [7]
3 years ago
8

What is the slope slope of a line that is perpendicular to the line represented by the equation y-5x=5 ?

Mathematics
2 answers:
Oliga [24]3 years ago
8 0

Answer: m=\dfrac{-1}{5}

Step-by-step explanation:

We know that the general intercept form of equation of line is given by :-

y=mx+c, where m is the slope of the line.

The given equation of line : y-5x=5

Subtract 5x both sides, we get

y=5x+5

On comparing the equation to the general intercept form of equation of line , we get

Slope =5

Since the product of slopes of perpendicular lines is -1.

Then the slope of line that is perpendicular to the given equation will be :-

m=\dfrac{-1}{5}

Rzqust [24]3 years ago
6 0
You have to divide the 5 and 5x by 5 add your variables together in alpha order=1
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faust18 [17]
1. Divide 95 million by the amount of days in a year

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Multiply it by 9.

So, 260273.97 x 9 = $2342465.75

$2342465.75 is made in 9 days.

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3 years ago
the admission fee to an amusement park is $27.it cost an additional p dollars to rent a locker to hold your belongings. the tota
Mademuasel [1]

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A=244in2<br> H=<br> B=12.6in<br> (1in=2.54cm)
pshichka [43]
Although there is no picture, I will assume this is a triangle we are talking about since the terms base and height are being used. If that is the case, the height is roughly 38.72in. 

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6 0
3 years ago
HURRY PLEASE
Delicious77 [7]

Answer:

Infinite solutions

Step-by-step explanation:

2x−7y=12 (1)

−x+3.5y=−6 (2)

(2) x = 3.5y + 6

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8 0
2 years ago
Read 2 more answers
The daily average temperature in Santiago, Chile, varies over time in a periodic way that can be modeled approximately by a trig
natali 33 [55]

Answer:

a) the trigonometric function is;

y = 7.5 sin ( \frac{2 \pi}{365}t + \frac{337 \pi}{730})+ 21.5

b) y = 28.36^0 \ C    ( to two decimal places)

Step-by-step explanation:

This data can be represented by the sinusoidal function of the form :

\mathbf{y = A sin (Bt -C)+D}

where A = amplitude and which can be determined via the formula:

A = \dfrac{largest \ temperature -  lowest \ temperature}{2}

A = \dfrac{29-14}{2}

A = \dfrac{15}{2}

A = 7.5° C

where B = the frequency;

Since the data covers a period of 3 days ; then \dfrac{2 \pi}{B } =365

B = \dfrac{2 \pi}{365}   ( where 365 is the time period )

The vertical shift is found by the equation D;

D =  \frac{largest \ temperature + lowest \ temperature}{2}

D = \frac{29+14}{2}

D = 21.5

Replacing the values of A ; B and D into the above sinusoidal function; we have :

y = 7.5 sin (\frac{2 \pi}{365}t -C) + 21.5

From the question; when it is 7th of the year ( i.e January 7);

t =  7 and the temperature (y) = 29° C

replacing that too into the above equation; we have:

29= 7.5 sin (\frac{2 \pi}{365}*7 -C) + 21.5

29= 7.5 sin (\frac{14 \pi}{365} -C) + 21.5

\frac{29-21.5}{7.5}=  sin (\frac{14 \pi}{365} -C)

1=  sin (\frac{14 \pi}{365} -C)

sin^{-1}(1)=   (\frac{14 \pi}{365} -C)

\frac{\pi}{2}=   (\frac{14 \pi}{365} -C)

C=   (\frac{14 \pi}{365} -\frac{\pi}{2})

C=   (\frac{28 \pi- 365 \pi}{730} )

C=  \frac{-337 \pi}{730}

Thus; the trigonometric function is;

y = 7.5 sin ( \frac{2 \pi}{365}t + \frac{337 \pi}{730})+ 21.5

Similarly; to determine the temperature o Jan 31; i.e when t= 31 ; we have :

y = 7.5 sin ( \frac{2 \pi}{365}*31+ \frac{337 \pi}{730})+ 21.5

y = 7.5 sin ( \frac{62 \pi}{365}+ \frac{337 \pi}{730})+ 21.5

y = 7.5 sin ( \frac{124 \pi+ 337 \pi }{730})+ 21.5

y = 7.5 sin ( \frac{461 \pi }{730})+ 21.5

y = 7.5 *( 0.915)+ 21.5

y = 6.8689+ 21.5

y = 28.36^0 \ C    ( to two decimal places)

7 0
3 years ago
Read 2 more answers
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