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Illusion [34]
3 years ago
6

Can someone please help me with this logarithm equation!!

Mathematics
1 answer:
enot [183]3 years ago
6 0

Answer:

\frac{629}{2}

Step-by-step explanation:

So, our equation is:

log_5(\frac{2x-4}{5})  = 3

In exponetial form, this looks like:

5^3=\frac{2x-4}{5}

Now lets cube the 5:

125=\frac{2x-4}{5}

Next, we can multiply the denominator on the right side by 5:

625=2x-4

We need to now get x alone by adding 4 to both sides:

629=2x

Finally, we divide by x's coefficent, 2, to get:

x=\frac{629}{2}

Hope this helps! :3

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Can somebody please help me with question number 1
inysia [295]

Answer:

1a: 6.283

b: 51.313

c: 285.1

Step-by-step explanation:

Area of a sector = (angle/360)πr²

1a

Area=(45/360)π(4²)=6.283

b

Area=(120/360)π(7²)=51.313

c

Area=(270/360)π(11²)=285.1

6 0
3 years ago
7.2 is 250% of what number
Vlada [557]

Answer:

its 18

Step-by-step explanation:

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8 0
2 years ago
Read 2 more answers
IF YOU HELP ME ILL WILL GIVE YOU BRAINLIST AND 50 PONITS HELP ME PLZZZZZ
labwork [276]
I think the answer is A because the data is skewed to the right 
The median (middle) isn't 10, it's 8 
The interquartile range (IQR) is 7 not 6 because you have to subtract the first and third quartile 13 - 6 = 7 not 6

Hope this helps! ;)
8 0
3 years ago
Read 2 more answers
(PLEASE HELP) The table below - does it represent an exponential function?
Aliun [14]
Short answer: I don't know, but that doesn't mean I can't give you something that you can decide for yourself.
y = 4*2^(2n - 2) is the pattern.
Go for broke. Try n = 4. You should get 256. Let's try it.
y = 4 * 2^(2*4 - 2)
y = 4 * 2^(8 - 2)
y = 4 * 2^6
y = 4 * 64
y = 256 yup it works.

The other end is just as important. Suppose n = 1
Then y = 4 * 2^(2*1 - 2) = 4 * 2^0 = 4*1 = 4 Both work.

If this formula is correct, we can abbreviate it to make your task easier.
y = 4 * 2^(2n - 2)
y = 2^2 * 2^(2n - 2)
y = 2^(2n - 2 + 2)
y = 2^(2n) Now try the two end points again.

n = 4
y = 2^(2*4)
y  =  2^8
y = 256

n = 1
y = 2^(2*1)
y = 2^2
y = 4 which again checks.

so y = 2^(2n) I think is an exponential function.
Sorry my explanation is so long.

6 0
4 years ago
Find the curl of ~V<br> ~V<br> = sin(x) cos(y) tan(z) i + x^2y^2z^2 j + x^4y^4z^4 k
ch4aika [34]

Given

\vec v =  f(x,y,z)\,\vec\imath+g(x,y,z)\,\vec\jmath+h(x,y,z)\,\vec k \\\\ \vec v = \sin(x)\cos(y)\tan(z)\,\vec\imath + x^2y^2z^2\,\vec\jmath+x^4y^4z^4\,\vec k

the curl of \vec v is

\displaystyle \nabla\times\vec v = \left(\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z}\right)\,\vec\imath - \left(\frac{\partial h}{\partial x}-\frac{\partial f}{\partial z}\right)\,\vec\jmath + \left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\,\vec k

\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ - \left(4x^3y^4z^4-\sin(x)\cos(y)\sec^2(z)\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k

\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ + \left(\sin(x)\cos(y)\sec^2(z)-4x^3y^4z^4\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k

7 0
3 years ago
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