Try this:
1 page 300 words
---------- * --------------- = 4.41 words per sec, or
68 sec 1 page
4.41 words 60 sec
---------------- * ----------- = 265 words per minute, or
1 sec 1 min
265 wpm * 60 min
--------- = 15882 wpm
1 hr
To solve this, first subtract 6x from both sides.
That is because there cannot be the same variable on both sides of the equation. The new equation would be:
6x=24
Now divide by both sides of the equation by 6
x=4
Hope this helped! Tell me if you have any more questions!
Answer:
possible values of 4th term is 80 & - 80
Step-by-step explanation:
The general term of a geometric series is given by

Where a(n) is the nth term, r is the common ratio (a term divided by the term before it) and n is the number of term
- Given, 5th term is 40, we can write:

- Given, 7th term is 10, we can write:

We can solve for a in the first equation as:

<em>Now we can plug this into a of the 2nd equation:</em>
<em>
</em>
<em />
<em>Let's solve for a:</em>
<em>
</em>
<em />
Now, using the general formula of a term, we know that 4th term is:
4th term = ar^3
<u>Plugging in a = 640 and r = 1/2 and -1/2 respectively, we get 2 possible values of 4th term as:</u>

possible values of 4th term is 80 & - 80
Answer:
64%
Step-by-step explanation:
divide 80 by 125 and get .64
Answer:
Step-by-step explanation:
We have the equations
4x + 3y = 18 where x = the side of the square and y = the side of the triangle
For the areas:
A = x^2 + √3y/2* y/2
A = x^2 + √3y^2/4
From the first equation x = (18 - 3y)/4
So substituting in the area equation:
A = [ (18 - 3y)/4]^2 + √3y^2/4
A = (18 - 3y)^2 / 16 + √3y^2/4
Now for maximum / minimum area the derivative = 0 so we have
A' = 1/16 * 2(18 - 3y) * -3 + 1/4 * 2√3 y = 0
-3/8 (18 - 3y) + √3 y /2 = 0
-27/4 + 9y/8 + √3y /2 = 0
-54 + 9y + 4√3y = 0
y = 54 / 15.93
= 3.39 metres
So x = (18-3(3.39) / 4 = 1.96.
This is a minimum value for x.
So the total length of wire the square for minimum total area is 4 * 1.96
= 7.84 m
There is no maximum area as the equation for the total area is a quadratic with a positive leading coefficient.