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Thanks,

I love that fireball 

You are supposed to work out some things for yourself.

If your calculator did everything for you then your school subject would be called  "calculators" not "mathematics"

That is not correct anon.  

$$\frac{4}{7}=\frac{x}{16}\\\\
\frac{4}{7}\times 16=x$$

$${\frac{{\mathtt{4}}}{{\mathtt{7}}}}{\mathtt{\,\times\,}}{\mathtt{16}} = {\frac{{\mathtt{64}}}{{\mathtt{7}}}} = {\mathtt{9.142\: \!857\: \!142\: \!857\: \!142\: \!9}}$$

so the answer is    $$\frac{64}{7}=9\frac{1}{7}$$

Hi MamaLlama,  How are Baba and ChaCha going ?     

Paul has saved $37.50 to buy a hamster, a cage, and hamster food. The hamster cost $5.50. The cage is on sale for 25% off the origional price of $29.90. The food costs $2.64. The sales tax on the total is 5%. How much will Paul pay in all? How much of his savings will he have left over?

(5.50+0.75*29.90+2.64)*1.05

$$\left({\mathtt{5.5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.75}}{\mathtt{\,\times\,}}{\mathtt{29.9}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2.64}}\right){\mathtt{\,\times\,}}{\mathtt{1.05}} = {\mathtt{32.093\: \!25}}$$

Cost = $32.09

$${\mathtt{37.5}}{\mathtt{\,-\,}}{\mathtt{32.09}} = {\frac{{\mathtt{541}}}{{\mathtt{100}}}} = {\mathtt{5.41}}$$

$5.41 left over.     

The n'th term of a geometric sequence is given by f(n) = a*r^n-1 (assuming we start counting at n = 1), so here we have:

a*r^2-1 = 24  or  a*r = 24 

and

a*r^5-1 = 3 or a*r⁴ = 3

Divide the f(5) term by the f(2) term

a*r⁴/(a*r) = 3/24   or  r³ = 1/8  or  r³ = (1/2)³ so  r = 1/2

Put this back into the f(2) term to get  a*(1/2) = 24,  so a = 48.

Therefore  f(n) = 48*(1/2)^n-1  or f(n) = 48/2^n-1

.

The first one asks..... to what power do we raise .6 to obtain .36.....the answer is 2

We can evalute the other two by using the "change of base" rule

It says that  log_a b   =  log b/ log a

So the second one is

log₈ .0039  =  log.0039 / log 8  =    about  -2.67

I'll let you do the last one.....

Hi Bio :)

Nothing's up :)

By definition a mole is Avogadro's number worth!  

Avogadro's number is more precisely given by: 6.02214129×10²³ 

If this is

$$2^{3x-1}=(\frac{1}{8})^x$$

then rewrite it as

$$\\2^{3x-1}=(\frac{1}{2^3})^x\\\\2^{3x-1}=(2^{-3})^x\\\\2^{3x-1}=2^{-3x}$$

So we must have

$$\\3x-1=-3x\\\\6x=1\\\\x=\frac{1}{6}$$

.

We have

80 = ab^5  →    a = 80/b^5

And

64 = ab^7  →  a = 64/b^7

Setting  the "a's" equal, we have

80/b^5 = 64/b^7

b^2 = 64/80  = 4/5  = .8

b = √.8 = .894

And solving for "a" we have

80 = a(.894)^5

80 = a(.571)

a = 80/.571 = 140.1

So...our function is

f(x) = 140.1 (.894)^x

The area will increase as the square of the scale factor

So

140 x (1.5) ^2 =

140 x 2.25 =

315 in ^2

what is the unit of that number?

P(1600) = P(x) = 70√1600 − 0.5(1600) − 500 =

70*40 - 800 - 500 =

2800 - 800 - 500 =

2300 - 500 =

1800

P(4900)  is evaluted in a similar manner

(2a-2a)+(4b-3b)= 0a+1b = b

Just take the eigth root of the 40.
$${\sqrt[{{\mathtt{{\mathtt{8}}}}}]{{\mathtt{40}}}} = {\mathtt{1.585\: \!833\: \!175\: \!137\: \!243\: \!4}}$$

X=1.585

multiply the number of attendence by percentage to find the number of people who ordered chicken.

120*(65/100)=78 people ordered chicken.

D=sqrt((x2-x1)^2+(y2-y1)^2)

D=sqrt((7-6)^2+(3-7)^2)

Distance= 4.12

78

It would be 65% of 120 which you can do on a calculator using the calculation (65/100)*120=78

The picture is f(x)= x+1/x-3

Fill Your tank, Write down the miles on your Odometer. Drive until you need fuel. Subtract the old odometer reading from the current Odometer reading. Divide miles travelled by gallons purchased on refill.  That gives you Your Miles per gallon.  That is the reason that I just used the calculator. I travelled 313 miles and added 9.19 gallons. It is a very simple math problem, no need for x's or y's or slopes of any kind. Bye Now.

Where is the pic?

You can use any log so long as you use the same one on both sides. :)

Good luck Rosala,

But you will do fine :)

$$7m^6n^5$$

cannot be simplified

yes and 8 to the power (-2over 3) is   1/4