The average of the five consecutive numbers ending with b in discuss when expressed in terms of a is; Choice D; a+3.
<h3>What is the average of five consecutive integers ending with b?</h3>
First, since it was given in the task content that the average of six positive consecutive odd integers starting with a is equal to b, it therefore follows that;
(a+a+2+a+4+a+6+a+8+a+10)/6 = b
6b=6a+30
b=a+5
Also, let the average of the consecutive intergers ending with b be denoted by; x.
(b+b-1+b-2+b-3+b-4)/5 = x
=(5b-10)/5
=b–2
The average, x=b – 2 (where b = a-5)
Ultimately, the value of the required average is; = a+5-2 = a+3.
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The answer is: m=f/a. use reverse operations so that m is now the subject
Answer:
x = 8
y = -7
Step-by-step explanation:
This is a system of equations called simultaneous equations. We shall solve it by elimination method Step 1We shall label the equations (1) and (2)−3y−4x=−11.....(1)3y−5x=−61......(2)Step 2Multiply each term in equation (1) by 1 to give equation (3)1(-3y-4x=-11).....(1)-3y-4x=-11....(3)Step 3Multiply each term in equation 2 by -1 to give equation (4)-1(3y−5x=−61)......(2)-3y+5x=61.....(4)Step 4-3y-4x=-11....(3)-3y+5x=61.....(4)Subtract each term in equation (3) from each term in equation (4)-3y-(-3y)+5x-(-4x)=61-(-11)-3y+3y+5x+4x=61+110+9x=729x=72Step 5Divide both sides of the equation by 9, the coefficient of the unknown variable x to find the value of x 9x/9 = 72/9x = 8Step 6Put in x = 8 into equation (2)3y−5x=−61......(2)3y-5(8)=-613y-40=-61Collect like terms by adding 40 to both sides of the equation 3y-40+40=-61+403y=-21Divide both sides by 3, the coefficient of y to find the value of y 3y/3=-21/3y=-7Therefore, the values of x and y that satisfy the equations are 8 and -7 respectively
