2) When we divide ______ by 8 gives a quotient of 75 and a remainder of 3

Mathematics

1 answer:

Pani-rosa [81]2 years ago

4 0

Answer:

603

Step-by-step explanation:

603/8

600/8 + 3/8 = 75, 3/8

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Parker offers to pay for 5 of his friends to play laser tag, but 3 of them don’t want to play. Parker offers to pay for 5 of his

gtnhenbr [62]

Answer: B

Step-by-step explanation:

First, you have to find how many friends are playing. This can be found in the equation 5-3. Now you multiply by six to find the cost. If you chose C, that would be incorrect, because the () you would mltilply first (30) and then subtract 3 (27). You can also think of it as finding the cost of how much all of them want to play by using the distributive property, leaving you with 30 - 18, and subtract from there to get 12.

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2 years ago

Higher-order thinking What two multiplication tables can help you find 3x 9? How could you use 3 x 9 to find 9 x 3?

babunello [35]

Answer:

Its simple if you will multiple 3 by 9 the answer will be 27 and if we multiple 9 by 3 so the answer will be same 27.

Step-by-step explanation:

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2 years ago

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

In-s [12.5K]

Answer:

Step-by-step explanation:

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

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

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

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

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