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ale4655 [162]
3 years ago
10

How to change 3 7/8 into an improper fraction

Mathematics
2 answers:
Anon25 [30]3 years ago
8 0
3\dfrac{7}{8}=\dfrac{3\cdot8+7}{8}=\dfrac{31}{8}
VladimirAG [237]3 years ago
6 0
3 7/8

So we can use this a \frac{b}{c} = \frac{(a \times c) + b}{c}

So we get

3 \frac{7}{8} = \frac{(3 \times 8) + 7}{8}

= \frac{24+7}{8} = \frac{31}{8}
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Solve the equation (x-2)^2/3=9
statuscvo [17]

Answer:

x = 2±3sqrt(3)

Step-by-step explanation:

(x-2)^2/3=9

Multiply each side by 3

(x-2)^2/3 *3=9*3

(x-2)^2=27

Take the square root of each side

sqrt( (x-2)^2)=±sqrt(27)

x-2 = ±3sqrt(3)

Add 2 to each side

x = 2±3sqrt(3)

6 0
3 years ago
<img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7Bx-5%7D%7B2x%5E%7B2%7D-5x-3%20%7D" id="TexFormula1" title="f(x) = \f
Vedmedyk [2.9K]

i) The given function is

f(x)=\frac{x-5}{2x^2-5x-3}

The factored form is

f(x)=\frac{x-5}{(x-3)(2x+1)}

The domain are the values of  x for which the function is defined.

(x-3)(2x+1)\ne 0

(x-3)\ne0,(2x+1)\ne 0

x\ne3,x\ne-\frac{1}{2}

ii) To find the vertical asymptotes, equate the denominator to zero.

(x-3)(2x+1)=0

(x-3)=\ne0,(2x+1)=0

x=3,x=-\frac{1}{2}

iii) To find the roots, equate the numerator to zero.

x-5=0

The root is x=5

iv) To find the y-intercept, put x=0 into the function.

f(0)=\frac{0-5}{(0-3)(2(0)+1)}

f(0)=\frac{-5}{(-3)(1)}

f(0)=\frac{5}{3}

The y-intercept is \frac{5}{3}

v) The horizontal asymptote is given by;

lim_{x\to \infty}\frac{x-5}{2x^2-5x-3}=0

The horizontal asymptote is y=0

vi) The function is not reducible. There are no holes.

vii) The given function is a proper rational function.

Proper rational functions do not have oblique asymptotes.

3 0
3 years ago
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SCORPION-xisa [38]

Answer:

Step-by-step explanation:

No that’s wrong

4 0
3 years ago
What is the answer to 2h/15=20?
GarryVolchara [31]
2h/15=20 so the answer to h would be h=150
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Which property of equality can be used to solve the equation n plus nine equals negative three?
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aAnswer:

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Step-by-step explanation:

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