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valina [46]
3 years ago
8

(12) Which equation has irrational solutions? Group of answer choices

Mathematics
1 answer:
Drupady [299]3 years ago
8 0

Answer:

9(x+3)²=27

Step-by-step explanation:

hello :

9(x+3)²=27   means : (x+3)²=27/9

(x+3)²=3  because 3 is not the perfect square

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A girl ran from her home to a store at the rate of 6 mph, and she walked back home at the rate of 4 mph. if it took 15 minutes f
viva [34]

Home is 0.6 miles away from the store

How to solve such time related questions?

The key concept for solving such questions is the relationship between speed, distance and time.

The relationship between them is Distance = Speed x Time

Given:-

Total time taken by girl to complete the trip = 15 minutes = \frac{1}{4} hours

Speed while riding going = 6 miles per hour

Speed while coming back = 4 miles per hour

Total time taken by girl  =

Time taken while going +   Time taken while coming back      - (1)

  Let x be distance from home to store

       \frac{x}{6} =  Time taken while going

\frac{x}{4} = Time taken while coming back

    Substituting in (1) we get

      \frac{1}{4} =  \frac{x}{6} + \frac{x}{4}

      \frac{1}{4} =  \frac{2x + 3x}{12}

       \frac{1}{4} =  \frac{5x}{12}

      x= \frac{12}{20}

    x = 0.6 miles

Thus home is 0.6 miles away from the store

Learn more about time related  here :

brainly.com/question/12759408

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6 0
2 years ago
Translate into an expression the sum of 4 and m
jek_recluse [69]

Answer:

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Step-by-step explanation:

8 0
3 years ago
Log(x-2)+log2=2logy<br>log(x-3y+3)=0​
In-s [12.5K]

Answer:

<h2>x = 4 and y = 2 or x = 10 and y = 4</h2>

Step-by-step explanation:

\left\{\begin{array}{ccc}\log(x-2)+\log2=2\log y&(1)\\\log(x-3y+3)=0&(2)\end{array}\right

===========================\\\text{DOMAIN:}\\\\x-2>0\to x>2\\y>0\\x-3y+3>0\to x-3y>-3\\=====================\\\\\text{We know:}\ \log_ab=c\iff a^c=b,\\\\\text{therefore}\ \log_ab=0\iff a^0=b\to b=1\\\\\text{From this we have}\\\\\log(x-3y+3)=0\iff x-3y+3=1\qquad\text{add}\ 3y\ \text{to both sides}\\\\x+3=3y+1\qquad\text{subtract 3 from both isdes}\\\\\boxed{x=3y-2}\qquad(*)\\\\\text{Substitute it to (1)}

\log(3y-2-2)+\log2=2\log(y)\qquad\text{use}\ \log_ab^n=n\log_ab\\\\\log(3y-4)+\log2=\log(y^2)\qquad\text{subtract}\ \log(y^2)\ \text{from both sides}\\\\\log(3y-4)+\log2-\log(y^2)=0\\\\\text{Use}\ \log_ab+\log_ac=\log_a(bc)\ \text{and}\ \log_ab-\log_ac=\log_a\dfrac{b}{c}\\\\\log\dfrac{(3y-4)(2)}{y^2}=0\qquad\text{use}\ \log_ab=0\Rightarrow b=1\\\\\dfrac{(3y-4)(2)}{y^2}=1\iff y^2=(3y-4)(2)\\\\\text{Use the distributive property}\ a(b+c)=ab+ac

y^2=(3y)(2)+(-4)(2)\\\\y^2=6y-8\qquad\text{subtract}\ 6y\ \text{from both sides}\\\\y^2-6y=-8\qquad\text{add 8 to both sides}\\\\y^2-6y+8=0\\\\y^2-2y-4y+8=0\\\\y(y-2)-4(y-2)=0\\\\(y-2)(y-4)=0\iff y-2=0\ \vee\ y-4=0\\\\\boxed{y=2\ \vee\ y=4}\in D

\text{Put the values of}\ y\ \text{to }\ (*):\\\\\text{for}\ y=2:\\\\x=3(2)-2\\\\x=6-2\\\\\boxed{x=4}\in D\\\\\text{for}\ y=4:\\\\x=3(4)-2\\\\x=12-2\\\\\boxed{x=10}\in D

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