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3 years ago

Solve this equation.

Mathematics

2 answers:

tensa zangetsu [6.8K]3 years ago

8 0

Answer:

The value of y is 2.

Step-by-step explanation:

1/3 ( y + 7 ) = 3 ( y - 1 ) Distribute

1/3y + 7/3 = 3y - 3 Subtract 1/3y on both sides

7/3 = 2 2/3y - 3 Add 3 on both sides

5 1/3 = 2 2/3 y Divide 2 2/3 on both sides

2 = y

<h3>Please rate this and please give brainliest. Thanks!!! </h3><h3>Appreciate it! : ) </h3><h3>And always, </h3><h3>SIMPLIFY BANANAS : )</h3>

baherus [9]3 years ago

7 0

Answer:

y=2

Step-by-step explanation:

(y+7)÷3=3y-3

y+7=9y-9

7+9=9y-y

16=8y

2=y

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2 years ago

Which equation is equivalent to log Subscript 3 Baseline (x + 5) = 2?

adell [148]

Answer:

The correct option is right-bracket squared 3 squared =x+5

Step-by-step explanation:

The equation is \log _{3}(x+5)=2

Option a: \log _{3}(x+5)=3^{2}

This is not possible because using logarithmic rule, if \log _{a} b=c then b=a^{c}

Hence, option a is not equivalent to \log _{3}(x+5)=2

Option b: \log _{3}(x+5)=2^{3}

This is not possible because using logarithmic rule, if \log _{a} b=c then b=a^{c}

Hence, option b is not equivalent to \log _{3}(x+5)=2

Option c: x+5=3^{2}

This is possible because using logarithmic rule, if \log _{a} b=c then b=a^{c}

Hence, option c is equivalent to \log _{3}(x+5)=2

Option b: x+5=2^{3}

This is not possible because using logarithmic rule, if \log _{a} b=c then b=a^{c}

Hence, option b is not equivalent to \log _{3}(x+5)=2

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Hence, the equation x+5=3^{2} is equivalent to \log _{3}(x+5)=2

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3 years ago

In a set of 3 numbers the first is a positive integer, the second is 3 more than the first and the 3rd is a square of the second

Molodets [167]

Given:

In a set of 3 numbers the first is a positive integer, the second is 3 more than the first and the 3rd is a square of the second.

To find:

The equation for the given situation if the sum of the numbers is 77.

Solution:

a. Let the first number in the set is x.

The second is 3 more than the first. So, the second number is $(x+3)$ .

The 3rd number is a square of the second. So, the third number is $(x+3)^2$ .

Therefore, the first, second and third numbers are $x,(x+3),(x+3)^2$ respectively.

b. The sum of the numbers is 77.

First number + Second number + Third number = 77

c. So, the equation in terms of x is:

$x+(x+3)+(x+3)^2=77$

Therefore, the required equation is $x+(x+3)+(x+3)^2=77$ .

d. On simplification, we get

$x+x+3+x^2+6x+9=77$ $[\because (a+b)^2=a^2+2ab+b^2]$

$x^2+(6x+x+x)+(3+9)=77$

$x^2+8x+12=77$

Subtract 77 from both sides.

$x^2+8x+12-77=77-77$

$x^2+8x-65=0$

Therefore, the simplified form of the required equation is $x^2+8x-65=0$ .

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