Answer:
Step-by-step explanation:
perimeter of rectangle = 2 ( l+ b )
let width = w length = 3w
according to formula
104 = 2 ( 3w + w )
104 = 2 ( 4 w )
104 = 8 w
104 /8 =w
13 =w
l = 3w
l = 3 ( 13) l = 39 hoping this will be helpful, have amazing day
1/4 + 3/8
Rewrite with the LCD (lowest common denominator)
Remember : the denominator is just the bottom number on a fraction.
Let's look at both of our given fractions denominator's.
4 & 8 → What do both of these numbers have in common? They both go into 8!
How many times does 4 go into 8? → Twice
How many times does 8 go into 8? → Once
So, because of that, we must multiply the top number (numerator) of each fraction with either one, or two.
First fraction : 1/4 × 2/2 =2/8 → So, our first fraction changes.
Second fraction : 3/8 × 1/1 = 3/8 → However, our second fraction changes.
So, our new question looks like this : 2/8 + 3/8
If you wanted us to answer this question, we simply need to add the top numbers (numerators) with one another because both fractions have the same bottom numbers (denominators).
2 + 3 = 5
So, our answer is 5/8 and our question rewritten with the LCD is : 2/8 + 3/8
~Hope I helped!~
Answer:
Vales of a =3 and b = 6
Step-by-step explanation:
Given that:
.....[1] where a and b are positive integers
we can write 81 and 24 as;
![81 = 3 \cdot 3 \cdot 3 \cdot 3 = 3^4](https://tex.z-dn.net/?f=81%20%3D%203%20%5Ccdot%203%20%5Ccdot%203%20%5Ccdot%203%20%3D%203%5E4)
![24 = 4 \cdot 6](https://tex.z-dn.net/?f=24%20%3D%204%20%5Ccdot%206)
We have [1] as;
![(ar^b)^4 = 3^4r^{4 \cdot 6}](https://tex.z-dn.net/?f=%28ar%5Eb%29%5E4%20%3D%203%5E4r%5E%7B4%20%5Ccdot%206%7D)
Using power rules;
![a^nb^n = (ab)^n](https://tex.z-dn.net/?f=a%5Enb%5En%20%3D%20%28ab%29%5En)
which implies a = b
then;
![(ar^b)^4 = (3r^6)^4](https://tex.z-dn.net/?f=%28ar%5Eb%29%5E4%20%3D%20%283r%5E6%29%5E4)
![ar^b = 3r^6](https://tex.z-dn.net/?f=ar%5Eb%20%3D%203r%5E6)
On comparing both sides we have;
a = 3 and
![r^b = r^6](https://tex.z-dn.net/?f=r%5Eb%20%3D%20r%5E6)
⇒![b = 6](https://tex.z-dn.net/?f=b%20%3D%206)
Therefore, the value of a and b are, 3 and 6
♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫
➷ Here are the steps to solving that problem:
(2x+1)^2
=(2x+1)(2x+1)
=(2x)(2x)+(2x)(1)+(1)(2x)+(1)(1)
=4x^2+2x+2x+1
=4x^2+4x+1
✽
➶ Hope This Helps You!
➶ Good Luck (:
➶ Have A Great Day ^-^
TROLLER