Write an equation of the line that passes through a pair of points:
ú
T
-3 -2 -1
a.
y=x+3
b. y=x-3
5.
4+
3+
2+
1+
7
-1
-2+
-3
4
1 2,
3
5 X
c.
d.
y = -x + 2
y = -x-2

Step-by-step explanation:

The slope-intercept form of a linear equation is y = mx + b, where m is the slope and b is the y-intercept.

Given the slope m = -2 and the y-intercept (a, b) = (-5, -4), we can substitute these values into the equation to get:

y = mx + b

y = -2x + (-4)

y = -2x - 4

Therefore, the equation in slope-intercept form is y = -2x - 4.

Since the distance was measured to the nearest metre, the upper bound for the distance is 101 metres.

1. To calculate the lower bound for Kelly's average speed, we need to divide the lower bound distance by the upper bound time. Since the distance was measured to the nearest metre, the lower bound for the distance is   [tex]99[/tex] Metres.

Since the time was measured to the nearest hundredth of a second, the upper bound for the time is  [tex]10.53[/tex] seconds. Therefore, the lower bound for Kelly's average speed is:

[tex]99/10.53 = 9.411[/tex]  metres per second (to three decimal places)

2. To calculate the upper bound for the perimeter of the regular hexagon, we need to multiply the upper bound length of a side by 6.

Since the length was measured to the nearest millimetre, the upper bound for the length is [tex]3.601 cm[/tex]  (since 3.6005 cm would round up to 3.601 cm). Therefore, the upper bound for the perimeter is:

[tex]6 x\times3.601 = 21.606 cm[/tex]   (to three decimal places)

3. To calculate the upper bound for the area of the rectangle, we need to multiply the upper bounds for the length and width. Since the length is correct to the nearest cm, the upper bound for the length is 35.5 cm (since 34.5 cm would round up to 35 cm).

Since the width is correct to the nearest cm, the upper bound for the width is 26.5 cm (since 25.5 cm would round up to 26 cm). Therefore, the upper bound for the area is:

 [tex]35.5 \times 26.5 = 942.25 cm^2[/tex] (to two decimal places)

4. To calculate the lower bound for Kelly's average speed, we need to divide the upper bound distance by the lower bound time.

Since the time was measured to the nearest hundredth of a second, the lower bound for the time is 10.51 seconds. Therefore, the lower bound

(d) To calculate the lower bound for Kelly's average speed, we need to divide the lower bound of distance by the upper bound of time.

The lower bound of distance is 99.5m (since the measurement was rounded down to the nearest metre).

The upper bound of time is 10.525s (since the measurement was rounded up to the nearest hundredth of a second).

So, the lower bound for Kelly's average speed is:

speed = distance / time [tex]= 99.5 / 10.525 = 9.45260637[/tex] ...

We need to round this to two decimal places to match the precision of the time measurement, giving us:

speed = [tex]9.45 m/s[/tex]

Therefore, the figures on the calculator display are: 99.5 ÷ 10.525 = 9.45260637... ≈ 9.45.

The length of the rectangle is measured as 645 mm correct to the nearest 5 mm. This means that the actual length could be anywhere between 642.5 mm and 647.5 mm (since rounding up or down depends on the decimal value being greater or less than 0.5 respectively).

Similarly, the width of the rectangle is measured as 400 mm correct to the nearest 5 mm. This means that the actual width could be anywhere between 397.5 mm and 402.5 mm.

5. To calculate the lower bound for the area of the rectangle, we need to find the product of the smallest possible length and width.

Smallest possible length [tex]= 642.5 mm[/tex]

Smallest possible width  [tex]= 397.5 mm[/tex]

Area = length x width

Lower bound for area = [tex]642.5 mm x 397.5 mm = 255542.5 mm²[/tex]

Rounding this off to 3 significant figures, we get the final answer as 2.56 x 10^5 mm².

Therefore, the lower bound for the area of the rectangle is [tex]2.56 x 10^5[/tex]  mm².

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The rock will be 125  feet above the beach at: The values of t are 3.911 or 1.28 and (b) The minimum height is h= 346 feet

Velocity is the directional speed of an object in motion as an indication of its rate of change in position as observed from a particular frame of reference and as measured by a particular standard of time

The given parameters are

h = -16t² + 79t + 50

In the equation put h = 125 to find t

125 = -16t² +79t+50

Rearrange the equation to have

16t²-79t -50 +125 = 0

this is also written as 16t² -79t + 75 =0

Solving for t we have

[-b±√b²-4ac]/2z

[79 ±√79²-4*16*75]/2*16

[(79 ±√6241-4800)] / 32

Simplify further to get

[79 ±√1441] / 32

(79 ±38) / 32

117/32 or 41/32

The values of t are 3.911 or 1.28

The minimum height of the rock at t = 1.28 is

h = -16(3.28)² + 79( 3.28)+ 50

h = -16(10.7584) + 79(3.28) +50

h = -  172.1344 +259.12 + 50

h= 346 feet

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Answer:

50 sweets

Step-by-step explanation:

Let's work through the problem step by step:

The student starts with x sweets.

She gives away 20 sweets to her friends, so she is left with (x - 20) sweets.

One third of the remainder is equal to one fifth of the original number of sweets. In other words, (1/3)(x - 20) = (1/5)x.

To solve for x, we can start by multiplying both sides of the equation by 15 (the least common multiple of 3 and 5) to eliminate the fractions:

5(x - 20) = 3x

5x - 100 = 3x

2x = 100

x = 50

Therefore, the original number of sweets was 50.

The measure οf the third angle c is 20 degrees.

The triangle angle sum prοperty states that the sum οf the three interiοr angles οf a triangle is always equal tο 180 degrees. This prοperty hοlds true fοr all types οf triangles, whether they are acute, οbtuse, οr right-angled. The prοperty can be expressed as fοllοws:

Angle A + Angle B + Angle C = 180 degrees, where A, B, and C are the measures οf the interiοr angles οf the triangle.

38 + 122 + c = 180

Tο sοlve fοr c, we can simplify the left side οf the equatiοn:

160 + c = 180

Subtracting 160 frοm bοth sides, we get:

c = 20

Therefοre, the measure οf the third angle c is 20 degrees.

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Complete question:

Two angles in a triangle are 38° and 122°. Write the angle sum property to determine the measure of the third angle ∠c.

[Hint: Angle sum property = ∠a+ ∠b+ ∠c = 180°]

The vertices will be D = (-3,2) E=(1,2) F=(-2,0).

resizing an οbject is accοmplished thrοugh a change called dilatiοn. The οbjects can be enlarged οr shrunk via dilatiοn. A shape identical tο the sοurce image is created by this transfοrmatiοn. The size οf the fοrm dοes, hοwever, differ. A dilatatiοn οught tο either extend οr cοntract the οriginal fοrm. The scale factοr is a phrase used tο describe this transitiοn.

The scale factοr is defined as the difference in size between the new and οld images. An established lοcatiοn in the plane is the center οf dilatatiοn. The dilatiοn transfοrmatiοn is determined by the scale factοr and the centre οf dilatiοn.

Here the scale factοr is 1/2

Sο the vertices will be D = (-3,2)

E=(1,2)

F=(-2,0)

Hence the vertices will be D = (-3,2) E=(1,2) F=(-2,0).

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Answer:Not dividing the entire side of the equation.

Not distributing properly using F.O.I.L.

Confusing the rules for adding and subtracting negative numbers with the rules for multiplying and dividing negative numbers.

Multiplying exponents with the base.

Step-by-step explanation:

Answer:

the area where the grass will be planted is 25,920 square meters.

Step-by-step explanation:

To calculate the area where the grass will be planted, we need to multiply the length and breadth of the sports field.

Area = length x breadth

Area = 162m x 160m

Area = 25,920 square meters

Therefore, the area where the grass will be planted is 25,920 square meters.

Answer:

After 15 years, the amount in the account will be $21.611.32, assuming the interest is compounded annually.

Step-by-step explanation:

To calculate the amount after 15 years, we can use the formula for compound interest:

A = P(1 + r/n)^(nt)

where A is the amount, P is the principal (initial amount), r is the annual interest rate (as a decimal), n is the number of times the interest is compounded per year, and t is the time (in years).

In this case, P = $12,000, r = 0.04 (4% expressed as a decimal), n = 1 (compounded annually), and t = 15 years.

Plugging in the values, we get:

A = $12,000(1 + 0.04/1)^(1*15)

A = $12,000(1.04)^15

A = $12,000(1.801)

A = $21,611.32

Therefore, after 15 years, the amount in the account will be $21.611.32, assuming the interest is compounded annually.

The polynomial function of the least degree with integral coefficients that has the zeros [tex]1 + 3i[/tex], [tex]-2i[/tex] is f(x) = x⁴ - 2x³ + 14x² - 8x + 40.

From the question, we are to write a polynomial function of the least degree with integral coefficients that have the given zeros.

The given zeros are [tex]1+3i[/tex],[tex]-2i[/tex].

If [tex]1 + 3i[/tex] and [tex]-2i[/tex]are zeros of a polynomial function with integral coefficients, then their conjugates [tex]1 - 3i[/tex] and [tex]2i[/tex] are also zeros of the function.

To find the polynomial function, we can use the fact that if r is a zero of a polynomial function, then (x - r) is a factor of the function. Thus, we can start by writing out the factors corresponding to each of the zeros:

[tex](x - (1 + 3i))(x - (1 - 3i))(x - (-2i))(x - 2i)[/tex]

Next, we can simplify these factors by multiplying them out:

[tex][(x - 1) - 3i][(x - 1) + 3i](x + 2i)(x - 2i)\\= [(x - 1)^2 - (3i)^2](x^2 - (2i)^2)[/tex]

= [(x - 1)² + 9](x² + 4)

Expanding the terms, we get:

(x² - 2x + 10)(x² + 4)

Multiplying out the factors, we obtain:

x⁴ - 2x³ + 10x² + 4x² - 8x + 40

Simplifying this expression, we get:

x⁴ - 2x³ + 14x² - 8x + 40

Hence, the polynomial function is f(x) = x⁴ - 2x³ + 14x² - 8x + 40

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Answer:

B) 2≤x≤8.

Step-by-step explanation:

Add 6 to all sides of the inequality.

0+6≤3x-6+6≤18+6

6≤3x≤24

Divide all sides by 3.

2≤x≤8

The correct answer is B) 2≤x≤8.

Answer:

1) S=31.68 yd  A=174.23 [tex]yd^{2}[/tex]

2) S=36.65 in A=256.56 [tex]in^{2}[/tex]

3) S=8.9 ft A=13.35 [tex]ft^{2}[/tex]

4)S=17.28 in. A=51.84 [tex]in^{2}[/tex]

5)S=25.31 yd. A=126.54 [tex]yd^{2}[/tex]

6)S=3.4ft A=5.11 [tex]ft^{2}[/tex]

Step-by-step explanation:

Notes:

360 was divided by 1/2, making the denominator 180. This is why there is a 1/2 at the front.

You also do not need to simplify if you're using a calculator.

A=[tex]\frac{1}{2}[/tex]([tex]r^{2}[/tex]) θ

- r is the radius

- θ is [tex]\frac{angle\pi }{180}[/tex]

S= r θ

-r is the radius

- θ is [tex]\frac{angle\pi }{180}[/tex]

1)

S=r θ

S=11([tex]\frac{165\pi }{180}[/tex])........................................plug in values

S=11( [tex]\frac{11\pi }{12}[/tex]).........................................simplify

S=31.68 yd....................................solve and round

A=[tex]\frac{1}{2}[/tex]([tex]r^{2}[/tex]) θ

A=[tex]\frac{1}{2}[/tex]([tex]11^{2}[/tex]) [tex]\frac{165\pi }{180}[/tex]....................................plug in values

A=  =[tex]\frac{1}{2}[/tex]([tex]121[/tex])  [tex]\frac{11\pi }{12}[/tex]....................................simplify

A=174.23 [tex]yd^{2}[/tex].....................................solve and round.

2)

S=r θ

S=14([tex]\frac{150\pi }{180}[/tex])...........................................plug in values

S=  =14(  [tex]\frac{5\pi }{6}[/tex]).........................................simplify

S=36.65 in........................................solve and round

A=[tex]\frac{1}{2}[/tex]([tex]r^{2}[/tex]) θ

A=[tex]\frac{1}{2}[/tex]([tex]14^{2}[/tex])  [tex]\frac{150\pi }{180}[/tex]......................................plug in values

A=  =[tex]\frac{1}{2}[/tex]([tex]196[/tex])  [tex]\frac{5\pi }{6}[/tex]......................................simplify

A=256.56 [tex]in^{2}[/tex]....................................solve and round.

3)

S=r θ

S=3([tex]\frac{170\pi }{180}[/tex]).........................................plug in values

S=  3(  [tex]\frac{17\pi }{18}[/tex]).......................................simplify

S=8.9 ft..........................................solve and round

A=[tex]\frac{1}{2}[/tex]([tex]r^{2}[/tex]) θ

A=[tex]\frac{1}{2}[/tex]([tex]3^{2}[/tex]) [tex]\frac{170\pi }{180}[/tex]........................................plug in values

A=  =[tex]\frac{1}{2}[/tex]([tex]9[/tex])  [tex]\frac{17\pi }{18}[/tex].......................................simplify

A=13.35 [tex]ft^{2}[/tex]........................................solve and round.

4)

S=r θ

S=6([tex]\frac{165\pi }{180}[/tex]).......................................plug in values

S=6( [tex]\frac{11\pi }{12}[/tex]).......................................simplify

S=17.28 in....................................solve and round

A=[tex]\frac{1}{2}[/tex]([tex]r^{2}[/tex]) θ

A=[tex]\frac{1}{2}[/tex]([tex]6^{2}[/tex]) [tex]\frac{165\pi }{180}[/tex].....................................plug in values

A=  =[tex]\frac{1}{2}[/tex]([tex]36[/tex])  [tex]\frac{11\pi }{12}[/tex]..................................simplify

A=51.84 [tex]in^{2}[/tex].....................................solve and round.

5)

S=r θ

S=10([tex]\frac{145\pi }{180}[/tex]).........................................plug in values

S=10(  [tex]\frac{29\pi }{36}[/tex]).........................................simplify

S=25.31 yd.......................................solve and round

A=[tex]\frac{1}{2}[/tex]([tex]r^{2}[/tex]) θ

A=[tex]\frac{1}{2}[/tex]([tex]10^{2}[/tex]) [tex]\frac{145\pi }{180}[/tex].......................................plug in values

A=  =[tex]\frac{1}{2}[/tex]([tex]100[/tex])  [tex]\frac{29\pi }{36}[/tex]....................................simplify

A=126.54 [tex]yd^{2}[/tex].....................................solve and round.

6)

S=r θ

S=3([tex]\frac{65\pi }{180}[/tex])..........................................plug in values

S=  3(  [tex]\frac{13\pi }{36}[/tex]).......................................simplify

S=3.4ft............................................solve and round

A=[tex]\frac{1}{2}[/tex]([tex]r^{2}[/tex]) θ

A=[tex]\frac{1}{2}[/tex]([tex]3^{2}[/tex]) [tex]\frac{65\pi }{180}[/tex].......................................plug in values

A=  =[tex]\frac{1}{2}[/tex]([tex]9[/tex])[tex]\frac{13\pi }{36}[/tex]......................................simplify

A=5.11 [tex]ft^{2}[/tex].........................................solve and round.

The linear function with the given properties is: f(x) = 8 - 6x.

A linear function is a mathematical equation that describes a straight line when graphed. It is the most basic type of function, where the output is equal to the input multiplied by a constant, known as the slope, and a constant added, known as the y-intercept. Linear functions are used to model relationships between two quantities and can be used to describe a wide range of phenomena, from physical properties to economic trends.

This linear function is used to calculate the value of a variable (x) when given the starting value of the function (f(0)) and the slope of the line (the rate at which the value of the function changes for each unit increase in x). In this case, the starting value of the function (f(0)) is 8, and the slope of the line is -6, meaning that the value of the function decreases by 6 for each unit increase in x. Therefore, the linear function that satisfies these properties is f(x) = 8 - 6x.

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Answer:  a = [tex]\frac{2}{7}[/tex].

Step-by-step explanation:

    10a - 2 = 4a - 1a

    10a - 2 = 3a

10a -2 + 2 = 3a + 2

          10a = 3a + 2

   10a - 3a = 3a -3a + 2

            7a = 2

              a = [tex]\frac{2}{7}[/tex]

Exponential property can be used in various field or place like growing or decline of population, bacteria growth/decay , compound interest any many more.

Exponential property states that when we multiply  two power with same base we add the exponents.

                 [tex]a^{m} *b^{n } = (a*b)^{m + n}[/tex]

When we divide  two power with same base we subtract the exponents.

                  [tex]a^{m} \div b^{n} = (a \div b)^{m-n}[/tex]

When have to find the power of power we multiply the exponents.

                  [tex](a^{m} )^{n} = a^{m*n}[/tex]

Three most important use of exponential function is to calculate carbon dating of any soil, fossil foil etc, to calculate population growth , interest earn on an investment etc.

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Answer:

Step-by-step explanation:

Let’s denote the production levels of coal, electricity and steel as x, y and z respectively. We can set up a system of equations to represent the inter-industry demand for each industry’s output.

For coal: 0.02x + 0.04y + 0.10z = x For electricity: 0.30x + 0.04y = y For steel: 0.30x + 0.02y + 0.04z = z

Solving this system of equations gives us x = (50/3)y and z = (25/2)y.

The new sales demand for coal is $2 billion (double the original), for electricity is $3 billion (triple the original) and for steel is $6 billion (an increase of 50%). Substituting these values into our equations gives us:

(50/3)y = $2 billion y = $3 billion (25/2)y = $6 billion

Solving these equations gives us y = $3 billion, x = $5 billion and z = $18.75 billion.

So to satisfy the new demand, the coal industry should produce at a level of $5 billion, the electric industry should produce at a level of $3 billion and the steel industry should produce at a level of $18.75 billion.

Using drawing tools parabola graph has been made and uploaded in answer for the given equation.

Standard form of parabola having vertical axis of symmetry given by:

[tex](y - k)^2 = 4p(x - h)[/tex]

where (h, k): vertex and p: distance between the vertex and the focus or directrix.

Comparing the given equation[tex]x = -1/8(y - 3)^2 + 1[/tex] with the standard form, we can see that the vertex is at (1, 3) and p = -1/32.

Since the parabola opens to the left, the focus is to the left of the vertex at a distance of p = -1/32 units. Thus, the focus is located at (-1/32, 3).

The directrix is a vertical line to the right of the vertex and is located at a distance of p = -1/32 units. Thus, the directrix is the vertical line x = 33/32.

To graph the focus and directrix, we can plot the vertex (1, 3) on the coordinate plane, draw a horizontal line through the vertex, and then plot the focus (-1/32, 3) to the left of the vertex and the directrix x = 33/32 to the right of the vertex.

Note that the parabola [tex]x = -1/8(y - 3)^2 + 1[/tex] is symmetric with respect to the vertical line passing through the vertex.

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Answer:

420 K x (60 cm³ / 70 cm³) x (7 / 18) = 294.29 K

Step-by-step explanation:

Answer:

Propaganda: ideas or statements spread to promote one side's views and present an opposing side negatively.

Front: the front lines where combat between opposing armies occurs.

Free enterprise system: an economic system characterized by private or corporate ownership of capital goods.

Allied Forces: the military alliance of the USA, Britain, France, and other countries during WWII.

Axis Powers: Germany, Italy, Japan.

The false comparison is: 35 degrees centigrade cooler than 100 degrees Fahrenheit.

In mathematics and computer programming, an expression is a combination of one or more values, variables, operators, and functions that are evaluated to produce a result.

The comparison "35 degrees centigrade cooler than 100 degrees Fahrenheit" is false because 35 degrees centigrade is equivalent to 95 degrees Fahrenheit, which is actually warmer than 100 degrees Fahrenheit. Therefore, it is incorrect to say that 35 degrees centigrade is cooler than 100 degrees Fahrenheit.

Therefore, the false comparison is: 35 degrees centigrade cooler than 100 degrees Fahrenheit.

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The general sοlutiοn tο the differential equatiοn is [tex]\mathrm{y = Ce^{(sin(x))}}[/tex].

The derivative οf a functiοn represents the instantaneοus rate οf change οf the functiοn at a specific pοint. It is calculated by finding the limit οf the difference quοtient as the interval between twο pοints οn the functiοn apprοaches zerο. The derivative can be expressed as a functiοn οf the independent variable, and it prοvides valuable infοrmatiοn abοut the behaviοr οf the οriginal functiοn.

The prοcess οf differentiatiοn invοlves applying a set οf rules tο functiοns tο οbtain their derivatives. These rules include the pοwer rule, prοduct rule, quοtient rule, chain rule, and οther mοre advanced rules that are used tο differentiate mοre cοmplex functiοns.

Tο sοlve the given differential equatiοn, we can use the methοd οf integrating factοrs.

First, we can rewrite the equatiοn as:

y'cοsx = ysinx + sin(175x)

Next, we can multiply bοth sides by the integrating factοr, which is [tex]e^{(\int(cos(x) dx))} = e^{(\sin(x) + C)}[/tex], where C is a cοnstant οf integratiοn:

[tex]\mathrm {e^{(sin(x)) }y'cosx = e^{(sin(x))} ysinx + e^{(sin(x))}sin(175x) + Ce^{(sin(x))}}[/tex]

Nοw, we can recοgnize the left-hand side as the derivative οf [tex]e^{(sin(x))}y[/tex]:

[tex](e^{(sin(x))y)}' = e^{(sin(x))} y' + cos(x) e^{(sin(x))}y[/tex]

Substituting this intο the abοve equatiοn, we get:

[tex]\mathrm{(e^{(sin(x))y)}' = e^{(sin(x)) }ysinx + e^{(sin(x))}sin(175x) + Ce^{(sin(x))}}[/tex]

[tex]cos(x) e^{(sin(x))}y = e^{(sin(x))y)}'[/tex]

Separating variables and integrating bοth sides, we get:

[tex]\int e^{sin(x) }dy/y = \int cos(x) dx[/tex]

ln|y| + C = sin(x) + C'

where C' is anοther cοnstant οf integratiοn.

Therefοre, the general sοlutiοn tο the differential equatiοn is:

[tex]\mathrm{|y| = e^{(sin(x)) }e^{(C' - sin(x))}}[/tex]

[tex]\mathrm{y = \± e^{(C' - sin(x) + sin(x))}}[/tex]

[tex]\mathrm{y = Ce^{(sin(x))}}[/tex]

where C is an arbitrary cοnstant.

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Answer:

no

Step-by-step explanation:

Answer: An array is a collection of items of same data type stored at contiguous memory locations.

Step-by-step explanation: There you go dude happy to help

The approximate value of the given logarithmic expression as required to be determined in the task content is; 1.528.

It follows from the task content; it is required that the value of the logarithmic expression is to be determined.

Therefore, we have;

let the result of the expression be x; so that we have;

log₈ (24) = x

x log (8) = log (24)

x = log (24) - log (8)

x = 1.528

Ultimately, the approximate value of the logarithmic expression in discuss is; 1.528.

Complete question: The correct expression is; log₈ (24) .

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The name of this circle is Circle R.

The name of the radius is RB.

The name of the diameter is RI or RD.

In Mathematics, a circle can be defined as a closed, two-dimensional curved geometric shape with no edges or corners. Additionally, a circle refers to the set of all points in a plane that are located at a fixed distance (radius) from a fixed point (central axis).

In Mathematics, a line segment can be defined as the part of a line in a geometric figure such as a triangle, circle, quadrilateral, etc., that is bounded by two (2) distinct points. Additionally, a line segment typically has a fixed length.

In this context, we have the following names;

Circle R.

Radius = RB.

Diameter = RD or RI.

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The proof that the quadrilateral is a rectangle is shown below

From the question, we have the following parameters that can be used in our computation:

A = (-2, 3)

B = (-4, 1)

C = (2, -1)

D = (0, -3)

From the graph, we have the following lengths

AB = √8

CD = √8

BD = √32

AC = √32

The above shows that opposite sides are equal

From the graph, we have the following slopes

AB = 1

CD = 1

BD = -1

AC = -1

The above shows that opposite sides are have equal slopes and adjacent sides are their slopes to be opposite reciprocals

Hence, the quadrilateral is a rectangle

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Answer:

1. c

2. b

3. B

Step-by-step explanation:

1. the hypotenuse is the longest side of the triangle.

2. think of it as the leg that helps form angle A (or forms right angle) that is NOT the hypotenuse

3. if "c" is the hypotenuse,, & b is the adjacent leg,, the opposite angle is across from the given angle which in this case is A

Answer:

  (x -7)² = 100

Step-by-step explanation:

You want to rewrite the equation x² -51 = 14x by completing the square.

When we're finished, the equation will be of the form ...

  (x +a)² = b

Expanding this, we have ...

  x² +2ax +a² = b

That is, the value of 'a' is half the coefficient of x when the equation is written with the x-terms together.

Adding 51 -14x to both sides of the given equation, we get ...

  x² -51 = 14x

  x² -14x = 51

Now, we can see that a=-14/2 = -7. We can add a² = (-7)² = 49 to both sides to complete the square:

  x² -14x +49 = 51 +49

  (x -7)² = 100 . . . . . . . rewritten equation