First you have to know BODMAS/BIDMAS (idk if all schools teach it like this) but this is the order in which you complete questions so it’s Brackets, Indices, Division/Multiplication, Addition/Subtraction
a) 5^2 is 5x5 which equals 25 1 to the power of anything is always 1 and 2^4 is 2x2x2x2 which is 16
So 5^2-1^5x2^4 = 25-1x16 which is then 25-16 = 9
b) 7^2+4^3 divided by 2^5 so 7^2 is 7x7 which = 49 then 4^3 means 4x4x4 = 64 and 2^5 is 2x2x2x2x2 which equals 32 so
7^2+4^3 divides by 2^5 is 49+64/32 = 49+2 which is 51
Answer:
if i'm not wrong it should be 2.8
Step-by-step explanation:
I'm realy not sure
Sorry if i'm wrong
Answer:
60 on the first day and 80 on the second
Step-by-step explanation:
the first day was lesser than the second day by 20 minutes.
so the second day was more than the first day by 20 minutes.
Classic Algebra and its unnecessarily complicated sentence structure. As you may have probably known, Algebra has its own "vocabulary set".
"the length of a rectangle exceeds its width by 6 inches" -> length is 6 in. longer than width -> l= w + 6
Since we're solving for the length and width, let's give them each variables.
length = l = w+6
width = w
The next bit of information is "the area is 40 square inches"
Applying the formula for the area of a rectangle we can set up:
l x w = 40
replace "l", or length, with it's alternate value.
(w+6) x w = 40
distribute
+ 6w = 40
subtract 40 from both sides
+ 6w - 40 = 0
factor
(w - 4)(w + 10) = 0
solve for w
w= 4, or -10
So great, we have 2 values; which one do we choose? Since this problem is referring to lengths and inches, we will have to choose the positive value. There is not such thing as a negative distance in the real world.
We now have half of the problem solved: width. Now we just need to find the length which we can do but substituting it back into the original alternate value of l.
l = w + 6
w=4
l = 4 + 6 = 10
The length is 10 in. and the width is 4 in. Hope this helps!