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attashe74 [19]
3 years ago
6

What's 5 2/5 as a mixed number??​

Mathematics
2 answers:
frosja888 [35]3 years ago
8 0
Yeah 27/5 and it’s improper because the numerator bigger
Dmitrij [34]3 years ago
5 0
5 2/5 is a mixed number, as an improper fraction it’s 27/5
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What is the value of x in the equation
Nastasia [14]

Answer:

54

Step-by-step explanation:

\frac{1}{3x}  -  \frac{1}{2}  = 18 \frac{1}{2}  \\  \\  \frac{1}{3x}  -  \frac{1}{2}  =  \frac{37}{2}   \\  \\  \frac{1}{3x}  =  \frac{37}{2}  -  \frac{1}{2}  \\  \\  \frac{1}{3x}  =  \frac{37 - 1}{2}  \\  \\  \frac{1}{3x}  =  \frac{36}{2}  \\  \\  \frac{1}{3x}  = 18 \\  \\ 1 = 54x

8 0
3 years ago
Solve 2sin2(x) – 5sin(x) – 3 = 0. Let u = sin(x). Which of the following is equivalent to the given equation?
Orlov [11]

Answer:

x=(-1)^{k+1}\cdot \dfrac{\pi}{6}+\pi k,\ k\in Z

Step-by-step explanation:

Consider the equation

2\sin^2x-5\sin x-3=0

Substitute

u=\sin x

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2u^2-5u-3=0

This is quadratic equation, where

D=(-5)^2-4\cdot 2\cdot (-3)=25+24=49\\ \\\sqrt{D}=7

Hence,

u_1=\dfrac{-(-5)-\sqrt{D}}{2\cdot 2}=\dfrac{5-7}{4}=-\dfrac{1}{2}\\ \\u_2=\dfrac{-(-5)+\sqrt{D}}{2\cdot 2}=\dfrac{5+7}{4}=3

Now, use the substitution again:

1. When u_1=-\dfrac{1}{2}, then

\sin x=-\dfrac{1}{2}\\ \\x=(-1)^k\sin^{-1}\left(-\dfrac{1}{2}\right)+\pi k,\ k\in Z\\ \\x=(-1)^k\cdot \left(-\dfrac{\pi}{6}\right)+\pi k,\ k\in Z\\ \\x=(-1)^{k+1}\cdot \dfrac{\pi}{6}+\pi k,\ k\in Z

2. When u_2=3, then

\sin x=3

This equation has no solutions, because the range of \sin x is [-1,1] and 3>1.

4 0
4 years ago
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anastassius [24]

Answer:

y=-3/2x+4

Step-by-step explanation:

In order to solve this you need to graph the points and then draw a line threw them. The place where the line crosses through the y-intercept is the value for b in y=mx+b.

6 0
3 years ago
Fractions in simplest form that have denominators of 2,
Montano1993 [528]

One of the fractions could be 1/2

Hoped this helped at all :)

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Which property would allow you to use mental computation to simplify the problem 23 + (7 + 9)?
katrin2010 [14]
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3 years ago
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