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Vlada [557]
2 years ago
7

What is the slope of a line perpendicular to the line 6x-y=2?

Mathematics
1 answer:
DENIUS [597]2 years ago
5 0

Answer:

C) -1/6

Step-by-step explanation:

Slope of perpendicular line = -1/m

6x - y = 2

Slope - y-intercept form:

           -y = -6x + 2

Divide the equation by (-1)

    y = 6x - 2

m = 6

Slope of perpendicular line = -1/m = -1/6

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F' : (-5,5)

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H' : (0,9)

Step-by-step explanation:

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4 0
3 years ago
Read 2 more answers
An invoice is dated January 17 with terms of 8/10 , 6/20 , 4/30 , n/40 . find the three final discount dates and the net payment
IRISSAK [1]

Invoice Date = January 17

• First terms:

8/10 and n/40

This means the final discount can be availed "10" days from Invoice Date (Jan17)

So, it will be:

Jan 17 + 10 days = January 27

Final Discount Date = Jan 27

The net payment date will be "n/40", which is 40 days from invoice date, thus

Net payment date will be:

Jan 17 - Jan 31 = 14 days

Feb 1 - Feb 26 = 26 days

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

40 days

Payment Date = Feb 26

• Second terms:

6/20 and n/40

This means the final discount can be availed "20" days from Invoice Date (Jan17)

So, it will be:

Jan 17 + 20 days = Feb 6

Final Discount Date = Feb 6

The net payment date will be "n/40", which is 40 days from invoice date, thus, same as before,

Payment Date = Feb 26

• Third terms:

4/30 and n/40

This means the final discount can be availed "30" days from Invoice Date (Jan17)

So, it will be:

Jan 17 + 20 days = Feb 16

Final Discount Date = Feb 16

The net payment date will be "n/40", which is 40 days from invoice date, thus, same as before,

Payment Date = Feb 26

6 0
1 year ago
How can you find the area of a polygon that is not one for which you know an area formula
Anon25 [30]
You would divide it into shapes that you already know, like squares, rectangles and triangles. Find the area of each shape and add them together
8 0
3 years ago
Verify the trigonometric identities
snow_lady [41]
1)

here, we do the left-hand-side

\bf [sin(x)+cos(x)]^2+[sin(x)-cos(x)]^2=2&#10;\\\\\\\&#10;[sin^2(x)+2sin(x)cos(x)+cos^2(x)]\\\\+~ [sin^2(x)-2sin(x)cos(x)+cos^2(x)]&#10;\\\\\\&#10;2sin^2(x)+2cos^2(x)\implies 2[sin^2(x)+cos^2(x)]\implies 2[1]\implies 2

2)

here we also do the left-hand-side

\bf \cfrac{2-cos^2(x)}{sin(x)}=csc(x)+sin(x)&#10;\\\\\\&#10;\cfrac{2-[1-sin^2(x)]}{sin(x)}\implies \cfrac{2-1+sin^2(x)}{sin(x)}\implies \cfrac{1+sin^2(x)}{sin(x)}&#10;\\\\\\&#10;\cfrac{1}{sin(x)}+\cfrac{sin^2(x)}{sin(x)}\implies csc(x)+sin(x)

3)

here, we do the right-hand-side

\bf \cfrac{cos(x)-sin^2(x)}{sin(x)+cos^2(x)}=\cfrac{csc(x)-tan(x)}{sec(x)+cot(x)}&#10;\\\\\\&#10;\cfrac{csc(x)-tan(x)}{sec(x)+cot(x)}\implies \cfrac{\frac{1}{sin(x)}-\frac{sin(x)}{cos(x)}}{\frac{1}{cos(x)}-\frac{cos(x)}{sin(x)}}\implies \cfrac{\frac{cos(x)-sin^2(x)}{sin(x)cos(x)}}{\frac{sin(x)+cos^2(x)}{sin(x)cos(x)}}&#10;\\\\\\&#10;\cfrac{cos(x)-sin^2(x)}{\underline{sin(x)cos(x)}}\cdot \cfrac{\underline{sin(x)cos(x)}}{sin(x)+cos^2(x)}\implies \cfrac{cos(x)-sin^2(x)}{sin(x)+cos^2(x)}
8 0
3 years ago
Find the area A of a circle with a radius of 2.4yd. Round the answer to the nearest hundredth. Use 3.14
dezoksy [38]

Answer: 18.09yd

Step-by-step explanation:

A=πr^2

A= 3.14*2.4^2

A= 18.09yd

5 0
3 years ago
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