What is the solution to the equation 2(x + 3)2 = 98?

Mathematics

1 answer:

Basile [38]3 years ago

8 0

Answer: if you're solving by x it would be x=43/2

Step-by-step explanation:

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Simplify: −z^3+5k^6−(−z^3+10k^6)

Georgia [21]

-z³ + 5k^6 + z³ -10k^6

(-z³ cancels out with z³)

5k^6 -10k^6

(then subtract)

Answer is -5k^6

6 0

3 years ago

I need help with math thanks

Viktor [21]

The Last on = 45x^2 - 2x - 8

Reason being is because you would group and multiple:
(9x - 4) • (5x +2)

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3 years ago

It takes you 1 minute 40 seconds to walk 550 feet. What is your average speed?

erica [24]

Answer:

5.5 feet per second or 330 feet per minute

Step-by-step explanation:

550/100 seconds = 5.5 feet per second

5.5 x 60 = 330 feet per minute

3 0

3 years ago

Read 2 more answers

Log(x-2)+log2=2logy<br>log(x-3y+3)=0

In-s [12.5K]

Answer:

Step-by-step explanation:

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

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

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

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

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