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3 years ago

9

PLS HELP ME ASAP FOR ALL 4! (SHOW WORK!!) + LOTS OF POINTS!! :)

2 answers:

frez [133]3 years ago

6 0

1 is B this should help

butalik [34]3 years ago

5 0

80n+20; You wanna know?
Well, since you only have to pay for every night and not everyday for the dog, 'a' is the correct answer.

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Least common denumertor of 11/3 9/10

Alisiya [41]

30 since that’s the lowest number both 3 and 10 can fit in

8 0

2 years ago

Solve the following equations.<br> log2(x^2 − 16) − log^2(x − 4) = 1

Alenkasestr [34]

Answer:

$x=\frac{4*(2+e)}{e-2}$

Step-by-step explanation:

Let's rewrite the left side keeping in mind the next propierties:

$log(\frac{1}{x} )=-log(x)$

$log(x*y)=log(x)+log(y)$

Therefore:

$log(2*(x^{2} -16))+log(\frac{1}{(x-4)^{2} })=1\\ log(\frac{2*(x^{2} -16)}{(x-4)^{2}})=1$

Now, cancel logarithms by taking exp of both sides:

$e^{log(\frac{2*(x^{2} -16)}{(x-4)^{2}})} =e^{1} \\\frac{2*(x^{2} -16)}{(x-4)^{2}}=e$

Multiply both sides by $(x-4)^{2}$ and using distributive propierty:

$2x^{2} -32=16e-8ex+ex^{2}$

Substract $16e-8ex+ex^{2}$ from both sides and factoring:

$-(x-4)*(-8-4e-2x+ex)=0$

Multiply both sides by -1:

$(x-4)*(-8-4e-2x+ex)=0$

Split into two equations:

$x-4=0\hspace{3}or\hspace{3}-8-4e-2x+ex=0$

Solving for $x-4=0$

Add 4 to both sides:

$x=4$

Solving for $-8-4e-2x+ex=0$

Collect in terms of x and add $4e+8$ to both sides:

$x(e-2)=4e+8$

Divide both sides by e-2:

$x=\frac{4*(2+e)}{e-2}$

The solutions are:

$x=4\hspace{3}or\hspace{3}x=\frac{4*(2+e)}{e-2}$

If we evaluate x=4 in the original equation:

$log(0)-log(0)=1$

This is an absurd because log (x) is undefined for $x\leq 0$

If we evaluate $x=\frac{4*(2+e)}{e-2}$ in the original equation:

$log(2*((\frac{4e+8}{e-2})^2-16))-log((\frac{4e+8}{e-2}-4)^2)=1$

Which is correct, therefore the solution is:

$x=\frac{4*(2+e)}{e-2}$

6 0

3 years ago

Help please! Will give brainlist for correct answer with steps. If you’re a troll I will report!

atroni [7]

Answer:

x²-10x+34=0. here u go I'm 12

8 0

2 years ago

I need to know what to do with the 12

vekshin1

Try doing 1/6x + 1/2y > 1...that should be asier to graph....all I did was divide 2x by 12 and 6y by 12

6 0

3 years ago

Can i get help with this 8th grade algebra question plz?

grandymaker [24]

The answer is -2 to your 8th grade algebra question hope it is right

4 0

2 years ago