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Anton [14]
3 years ago
11

Isaac has a piece of rope that 5 yards long. Into how many 1/2 - yard pieces of rope isaac can cut?

Mathematics
2 answers:
Trava [24]3 years ago
7 0

Answer:

10

Step-by-step explanation:

So there are 5 pieces of rope, if you cut them all in half then that would double the number.

alexira [117]3 years ago
5 0

Answer:

10

Step-by-step explanation:

5 divided by 1/2 is 10

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What is the equation represented​
ZanzabumX [31]

Answer:

y=-8x-6, y=-4x, y=-2x+3, x=6

Step-by-step explanation:

y=mx+b

use that formula to make you're equations but remember m represents slope and b what is going up by

8 0
4 years ago
HELP I DONT UNDERSTAND
klemol [59]

Answer:

A -4,4, A 2,-1,    B -4 1/2, 1 1/2.   C 2 1/2, 2 1/2, 1,1.

Step-by-step explanation:

count the caculations

4 0
3 years ago
Help out for these two questions please
attashe74 [19]

Answer:

23 - A

24 - x = -3 and x = 1

Step-by-step explanation:

Hey There!

So if you didnt know the " zeroes" are the coordinates of where x=0

For the first one the zeroes would be 1 because that's where the line is at x=0

For the second one the zeroes would be x=-3 and x=1 because that's where x=0 if that makes sense

8 0
3 years ago
(a) Use the Quotient Rule to differentiate the function f(x)=tan(x)-1/sec(x). f'(x)=
yKpoI14uk [10]

Answer:

(a) sin(x) + cos(x)

(b) sin(x) + cos(x)

(c) Both answers are equivalent

Step-by-step explanation:

(a) The given function is:

f(x) = \frac{tan(x)-1}{sec(x)}

According to the quotient rule:

f(x) = \frac{g(x)}{h(x)}\\f'(x)= \frac{g'(x)h(x)-g(x)h'(x)}{h(x)^2}

Applying the quotient rule:

f(x) = \frac{tan(x)-1}{sec(x)}\\f'(x)=\frac{sec^2(x)*sec(x)-(tan(x)-1)*sec(x)tan(x)}{sec(x)^2}\\f'(x)=\frac{sec^3(x)-sec(x)tan^2(x)+sec(x)tan(x)}{sec(x)^2}\\f'(x)=sec(x)+\frac{tan(x)-tan^2(x)}{sec(x)}\\ \frac{tan(x)}{sec(x)}=sin(x) \\f'(x)=sec(x)+sin(x)-sin(x)tan(x)\\

This can be simplified to:

f'(x)=sec(x)+sin(x)-sin(x)tan(x)\\f'(x) = \frac{1}{cos(x)}+sin(x)-\frac{sin^2(x)}{cos(x)}\\f'(x)=\frac{1+sin(x)cos(x)-sin^2(x)}{cos(x)}\\f'(x)=\frac{sin^2(x)+cos^2(x)+sin(x)cos(x)-sin^2(x)}{cos(x)}\\f'(x)=\frac{cos^2(x)+sin(x)cos(x)}{cos(x)}\\ f'(x)=sin(x) +cos(x)

(b) Simplifying in terms of sin(x) and cos(x):

f(x) = \frac{tan(x)-1}{sec(x)}\\f(x)=\frac{\frac{sin(x)}{cos(x)}-1 }{\frac{1}{cos(x)} } \\f(x)=sin(x)-cos(x)\\f'(x) = cos(x)+sin(x)

(c) As proven above, both answers are equivalent.

4 0
3 years ago
224 divided by 7 in distributive property
zhenek [66]
(56/7)+(168/7)=8+24=32. Your answer is: 32
8 0
4 years ago
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