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sergeinik [125]

3 years ago

8

Samuel recorded the number of different types of pets his friends have in the table shown below: Cats 5 Dogs 3 Rabbits 1 Guinea

Pigs 3

1 answer:

rjkz [21]3 years ago

3 0

4 different pets is the an

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1. 8y^2-4y+1 divided by 2y-1

____ [38]

Q1. The answer is $4y+ \frac{1}{2y-1}$

$\frac{8y^{2}-4y+1 }{2y-1} = \frac{4y*2y-4y+1}{2y-1} = \frac{4y(2y-1)+1}{2y-1} = \frac{4y(2y-1)}{2y-1}+ \frac{1}{2y-1} =4y+ \frac{1}{2y-1}$

Q2. The answer is $2a+3+ \frac{6}{a-1}$

$\frac{2 a^{2}+a+3 }{a-1} = \frac{a*2a-2a+3a+3}{a-1} = \frac{2a(a-1)+3a+3}{a-1}= \frac{2a(a-1)}{a-1}+ \frac{3a+3}{a-1} = \\ \\ =2a+ \frac{3a+3}{a-1}=2a+ \frac{3a-3+3+3}{a-1}=2a+ \frac{3(a-1)+6}{a-1} =2a+ \frac{3(a-1)}{a-1} + \frac{6}{a-1}= \\ \\ =2a+3+ \frac{6}{a-1}$
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Q3. The answer is </span> $2 x^{2} +5x+2$ <span>

</span> $\frac{6 x^{3} +11 x^{2} -4x-4}{3x-2} = \frac{3x*2 x^{2}-2 x^{2} *2+15 x^{2} -4x-4 }{3x-2} = \frac{2 x^{2} (3x-2)+15 x^{2} -4x-4}{3x-2}= \\ \\ = \frac{2 x^{2} (3x-2)}{3x-2} + \frac{15 x^{2} -4x-4}{3x-2} =2 x^{2} +\frac{15 x^{2} -4x-4}{3x-2}=2 x^{2} + \frac{15 x^{2} -10x+6x-4}{3x-2}= \\ \\ =2 x^{2} + \frac{5x*3x-5x*2+6x-4}{3x-2} =2 x^{2} + \frac{5x(3x-2)+3x*2-2*2}{3x-2} = \\ \\ =2 x^{2} + \frac{5x(3x-2)}{3x-2} + \frac{3x*2-2*2}{3x-2} =2 x^{2} +5x+ \frac{2(3x-2)}{3x-2} =2 x^{2} +5x+2$
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Q4. The answer is 2x + 7

</span> $\frac{6 x^{2} +11x-35}{3x-5} = \frac{6 x^{2} -10x+21x-35 }{3x-5} = \frac{3 x *2x-5*2x+7*3x-7*5 }{3x-5} = \\ \\ = \frac{2x(3x-5)+7(3x-5)}{3x-5}= = \frac{(3x-5)(2x+7)}{3x-5} =2x+7$
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Q5. The answer is </span> $x+1- \frac{3}{x-1}$ <span>

</span> $\frac{ x^{2} -4}{x-1} = \frac{ x^{2} -x+x-1-3 }{x-1} = \frac{x*x-x+x-1-3}{x-1} = \frac{x(x-1)+(x-1)-3}{x-1} = \\ \\ \frac{(x+1)(x-1)-3}{x-1} = \frac{(x+1)(x-1)}{x-1} -\frac{3}{x-1} =x+1- \frac{3}{x-1}$

Q6. The answer is $y^{2} -2y+3$

$\frac{ y^{3}-4 y^{2}+7y-6 }{y-2} = \frac{y* y^{2} -2y^{2}-2 y^{2} +7y-6 }{y-2} = \frac{y^{2}(y-2)-2 y^{2} +7y-6}{y-2}= \\ \\ = \frac{y^{2}(y-2)}{y-2}+ \frac{-2 y^{2} +7y-6}{y-2} = y^{2} + \frac{-2 y^{2} +4y + 3y-6}{y-2} = \\ \\ =y^{2} + \frac{-2y*y-2y(-2)+3y-3*2}{y-2} = y^{2} + \frac{(-2y)(y-2)+3(y-2)}{y-2} = \\ \\ = y^{2} + \frac{(-2y+3)(y-2)}{y-2} = y^{2} +(-2y+3) =y^{2} -2y+3$
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Q7. The answer is </span> $x^{2} +xy+ y^{2}}{x-y}$ <span>

</span> $\frac{ x^{3} - \frac{x}{y} y^{3} }{x-y} = \frac{(x-y)( x^{2} +xy+ y^{2}) }{x-y} = \frac{ x^{2} +xy+ y^{2}}{x-y}$
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Q8. The answer is </span> $(a^{2} +2ab+2b^{2})$ <span>

</span> $\frac{a^{4} +4b^{4} }{a^{2}-2ab+2 b^{2} } = \frac{ (a^{2})^{2} +(2b)^{2}}{a^{2}-2ab+2 b^{2}} = \frac{(a^{2} -2ab+2b^{2})(a^{2} +2ab+2b^{2}) }{(a^{2} -2ab+2b^{2})} =(a^{2} +2ab+2b^{2})$ <span>

Q9. The answer is a^{n-8} - a^{-14}

</span> $\frac{ (a^{2}) ^{n} - a^{n-6} }{ a^{n+8} }= \frac{(a^{2}) ^{n} }{a^{n+8}}- \frac{ a^{n-6} }{a^{n+8}} \\ \\ (x^{y}) ^{z}= x^{y*z} \\ \\ \frac{ x^{y} }{ x^{z} } = x^{y-z} \\ \\ 
\frac{(a^{2}) ^{n} }{a^{n+8}}- \frac{ a^{n-6} }{a^{n+8}} =\frac{a^{2n} }{a^{n+8}}- \frac{ a^{n-6} }{a^{n+8}} = a^{2n-(n+8)} - a^{n-6-(n+8)} = \\ \\ =a^{2n-n-8} - a^{n-6-n-8} = a^{n-8} - a^{-14}$

5 0

3 years ago

6m + 15n- 12p/n for m = -3.5, n=4, and p=3

Alexus [3.1K]

Answer: B. 30

Step-by-step explanation:

First, you re-write the expression

6(-3.5) + 15(4) - 12(3/4)

Then multiply all the parts correctly.

-21 + 60 - 9

-21 + 60 = 39

-9

= 30

8 0

3 years ago

Read 2 more answers

Valentin [98]

Answer:

4.229

Step-by-step explanation:

75 = 4(2)^x

75/4 = 2^x

Apply ln both sides

ln(75/4) = ln(2^x)

ln(75/4) = x ln(2)

x = ln(75/4) ÷ ln(2)

x = 4.22881869

3 0

3 years ago

2,5, 25/2, ... <br> find the 10th term

notka56 [123]

Answer:

The $10^{th}$ term of the given sequence

$t_{10} = \frac{5^{9} }{2^{8} }$

Step-by-step explanation:

<u>Step(i):-</u>

Given sequence 2 , 5, $\frac{25}{2}$

First term a = 2

The difference of given geometric sequence

$d = \frac{r_{2} }{r_{1} } = \frac{5}{2}$

<u><em>Step(ii):-</em></u>

The $n^{th}$ term of the given sequence

$t_{n} = ar^{n-1}$

The $10^{th}$ term of the given sequence

$t_{10} = (2)(\frac{5}{2} )^{10-1}$

$t_{10} = (2)(\frac{5}{2} )^{9}= \frac{5^{9} }{2^{8} }$

8 0

3 years ago

If 4x^2 - 100 = 0, the roots of the equation are

ololo11 [35]

$4x^2-100=0 \ \ \ \ \ \ |+100 \\
4x^2=100 \ \ \ \ \ \ \ \ \ \ \ |\div 4 \\
x^2=25 \\
x=\pm \sqrt{25} \\
x=\pm 5 \\
x=-5 \hbox{ or } x=5$

The answer is (3) -5 and 5.

5 0

3 years ago