Solve: one seventh + nine fourteenths = _____
1 answer:
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The problem is asking you to simplify the exponents. When you multiply two or more powers with the same base, the exponents can be added and subtracted. In this case, the base is 7.
7 can also be written as
.
The answer would be 7³ if you kept the exponent, or 343 if you simplified further.
7 can also be written as
The answer would be 7³ if you kept the exponent, or 343 if you simplified further.
Answer:
19 by 38
Step-by-step explanation:
The area of the poster, A=xy
Margin at both up and bottom is 2inches, at the right and left is 1 inch
Then we can say
(x-4)(y-2)= 578
Let us make y subject of the formula
(y-2)= 578/(x-4)
y=[ 578/(x-4)] + 2
If we substitute into the area A equation we have
A= x { [ 578/(x-4)] + 2}
A= 2x + (578x)/(x - 4)
If we differentiate we have
A'(x)= [2+578(x-4)+578x ]/ (x-4)^2
=[ 2(x-4)^2 + 578(x -4) + 578x ] / (x-4)^2
=[ 2(x^2 - 8x +16) + 578(x-4) + 578x ] /(x-4)^2
If we simplify this we have
=[ 2x^2 -16x +32+578x -578x - 2312]/ (x-4)^2
=( 2x^2 -16x - 2312) / (x-4)^2
At A'(x)= 0
= ( x^2 - 8x - 2312) / (x-4)^ =0
If we divide through by 2 we have
x^2 - 8x - 1156= 0
Solving the quadratic eqn( CHECK THE ATTACHMENT)
X= 38 or -30 then we choose the positive one, then x= 38
Then from area of the poster
A= (x-4)(y-2)= 578
Substitute 38 as value of x
(38-4)(y-2)= 578
34(y-2)=578
34y-68=578
34y=646
y=19
Hence dimensions of the poster is 19 by 38
CHECK THE ATTACHMENT FOR QUADRATIC SOLUTION
