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Pepsi [2]
3 years ago
9

Why is 4/5 closer to 1 than 5/4

Mathematics
2 answers:
pshichka [43]3 years ago
7 0
It isn’t. They are both 1/5 away from being 1.
Sergio [31]3 years ago
4 0

4/5=0.8             1-0.8= 0.2

5/4= 1.25          1.25-1= 0.25

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Solve the given equation over the interval [0, 2π): csc^2 x + 2 csc x = 0.
Anastaziya [24]
csc^2x+2cscx=0
Factor out the GCF which is csc x
cscx(cscx+2)=0
Set each factor equal to zero.
cscx=0  or cscx+2=0
Using the reciprocal sine function we can replace csc in the equations.
\frac{1}{sinx} =0  or \frac{1}{sinx} +2=0
The first equation yields 1 = 0 which is not true. However the second equation does offer 2 possible answers.
When solved for sinx you get 
sinx=- \frac{1}{2}
The sine function is negative in the 3rd and 4th quadrants.
You can use the unit circle or the 
sin^-^1 feature on your calculator  sin^-^1(1/2)
210^o  or 330^o  alter radian answer which is probably what you want is \frac{7 \pi }{6}    or   \frac{11 \pi }{6}

8 0
3 years ago
Which of these terms does not describe polygon A'B'C'D?
chubhunter [2.5K]
I can’t answer if you don’t show the multiple choice...
5 0
3 years ago
Read 2 more answers
What is the value of 4
Stells [14]

The value of 4 is in the ones place

8 0
3 years ago
Does the point (6,3) satisfy the equation y = 4(x – 5)?
Darina [25.2K]

Answer:

No

Step-by-step explanation:

plug (6,3) into the given equation to find out

3 = 4(6-5)

3 = 4(1)

3 = 4

since 3 does not equal 4, the point (6,3) does not satisfy the equation y=4(x-5)

5 0
4 years ago
when the polynomial f(x) is divided by (x-2) the remainder is 4, and when it is divided by (x-3) the remainder is 7. Given that
natima [27]

f(x)=(x-2)(x-3)Q(x)+ax+b

Recall the polynomial remainder theorem: the remainder upon dividing a polynomial p(x) by x-c is equal to p(c). This means that f(2)=4 and f(3)=7, which tell us

4=2a+b

7=3a+b

From here we can solve for a,b:

4=2a+b\implies b=4-2a

7=3a+b=3a+(4-2a)\implies a=3\implies b=-2

so that

f(x)=(x-2)(x-3)Q(x)+3x-2

Now,

\dfrac{f(x)}{(x-2)(x-3)}=Q(x)+\dfrac{3x-2}{(x-2)(x-3)}

so the remainder upon dividing f(x) by (x-2)(x-3) is 3x-2.

Next, if f is a cubic function, then Q(x) is a linear polynomial that can be written as Q(x)=cx-d. The coefficient of x^3 in f(x) is 1 (unity), so that expanding f(x) gives us

f(x)=(x-2)(x-3)(cx-d)+3x-2

f(x)=(cx^3-(5c+d)x^2+(6c+5d)x-6d)+3x-2

f(x)=cx^3-(5c+d)x^2+(6c+5d+3)x-(6d+2)

\implies c=1

and we also have that f(1)=1, so that

1=1-(5+d)+(6+5d+3)-(6d+2)

\implies2d=2\implies d=1

so that

Q(x)=x-1

4 0
3 years ago
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