Fill in the blanks 1. The degree of a linear equation is ? 2. The constant term has a degree ? 3. The solution of the equation ax + b = 0 is ? 4. If 2x - 3 = 0 then the value of x is ? 5. The value of x which satisfies the equation (- 8) = 2x + 4 is ? 6. The solution of the equation 10 - (x + 2) = 7x is ? 7. A linear equation in one variable has solution ? 8. The sum of the two consecutive multiples of 10 is 210. The smallest multiple is 9. Convert the statement "adding 15 to 4 times x is 39" into an equation 10. Value of "S" in 1/3 + S = 2/5 is ?

Step-by-step explanation:

A linear equation is of the form ax+b and the highest degree of x is 1. You can see that from the power of x which is 1.

The power of x is zero from the equating. It is not reflecting because anything raise to power if zero gives 1

To get the solution of the equation ax+b = 0

We will make x the subject of the formula

ax = -b

Divide both side's by a

ax/a = -b/a

x = -b/a

If 2x-3 = 0

2x = 0+3

2x = 3

2x/2 = 3/2

x = 3/2

Given -8 = 2x+4

2x = -8-4

2x = -12

x = -12/2

x = -6

A linear equation in one variable is known to have ONE solution. This is as demonstrated in the equation above, we cans see that the only solution is -6(just one solution)

Convert the statement "adding 15 to 4 times x is 39" into an equation 1

The product of 4 and x is 4x

Adding 15 to 4x will give

15+4x

If the resulting expression is equal to 39, the final equation will be;

15+4x = 39

10) given 1/3+S = 2/5, we are to find S

S = 2/5-1/3

S = 6-5/15

S = 1/15

Hence the value of S is 1/15

perimeter of the first rectangular board is 9 feet.

step-by-step explanation:

let p = perimeter of first board.

l = length of first board

w = width of first board

length of first board = 2 * width of first board - 3

l = 2w - 3

perimeter of first board = 2l + 2w

since l = 2w-3, then

p = 2*(2w-3) + 2w which equals 4w-6 + 2w which equals 6w-6

length and width of the second board are reciprocals of corresponding length and width of the first board.

length of second board = 1/(2w-3)

width of second board = 1/w

perimeter of second board is 1/5th perimeter of the first board.

perimeter of second board = p/5

perimeter of the second board = (2/(2w-3) + 2/w)

since the perimeter of the first board is 5 times the perimeter of the second board, this means that:

6w-6 = 5*(2/(2w-3) + 2/w)

which is the same as:

6w-6 = (10/(2w-3) + 10/w)

if we multiply both sides of this equation by (2w-3), we get:

(6w-6)*(2w-3) = 10 + 10*(2w-3)/w

if we multiply both sides of this equation by w, we get:

(6w-6)*(2w-3)*w = 10*w + 10*(2w-3)

we can simplify this to become:

(6w-6)*(2w-3)*w = 10w + 20w - 30

which becomes:

(6w-6)*(2w-3)*w = 30w-30

if we divide both sides of this equation by (6w-6), we get:

(2w-3)*w = (30w-30)/(6w-6)

which becomes:

(2w-3)*w = 5

this can be simplified to:

2w^2 - 3w = 5

subtract 5 from both sides of this equation to get:

2w^2 - 3w - 5 = 0

which can be factored into:

(2w-5)*(w+1) = 0

which makes:

w = 5/2

or

w = -1.

w can't be negative so the only possible answer is w = 5/2.

if w = 5/2, then

p = 2w-3 = 5-3 = 2

we have:

p = 2

w = 5/2

p = 2l + 2w gets p = 4 + 5 = 9

the perimeter of the first board is 9 feet.

since the perimeter of the second board is 1/5 the perimeter of the first board, the perimeter of the second board is 9/5.

to see if that's correct, we substitute known values for l and w into the second board.

the length of the second board is 1/l.

this becomes 1/2.

the width of the second board is 1/w

this bgecomes 1/(5/2) = (2/5)

let l2 = length of second board.

let w2 = width of second board.

l2 = (1/2)

w2 = (2/5)

let p2= perimeter of second board.

then p2 = 9/5

since p2 = 2*l2 + 2*w2, we should get p2 = 9/5 using the dimensions of l2 and w2.

2 * l2 = 2 * (1/2) = 1

2 * w2 = 2 * (2/5) = 4/5

1 is the same as 5/5

5/5 + 4/5 = 9/5

p2 checks out good.

the dimensions as given are good.

answer to the problem is:

perimeter of the first rectangular board is 9 feet.

55/i dont know i do it for the points

step-by-step explanation: