A punch contains cranberry juice and ginger ale in the ratio 5:3. If you require 32 L
of punch for a party, how many litres of cranberry juice and how many litres of ginger
ale are required?

9514 1404 393

Answer:

   $40,615.20

Step-by-step explanation:

The amortization formula will tell you Michelle's monthly payment.

  A = P(r/12)/(1 -(1 +r/12)^(-12t)) . . . . loan value P at interest rate r for t years

  A = $124,500(0.059/12)/(1 -(1 +0.059/12)^(-12·10)) ≈ $1375.96

__

The total of Michelle's 120 monthly payments is ...

  12 × $1375.96 = $165,115.20

This amount pays both principal and interest, so the amount of interest she pays is ...

  $165,115.20 -124,500 = $40,615.20

Michelle will pay $40,615.20 in interest over the course of the loan.

__

A calculator or spreadsheet can figure this quickly.

Answer:

554.7

Step-by-step explanation:

The pay=25.8*14+(25.8)*5*1.5=554.7

Hence the time that the ball will be height than 12 feet off the ground is 4secs

Given the expression for calculating the height in  feet as;

h(t) = -4t²+16t

If the ball is higher than 12feet, h(t) > 12

Substituting h = 12 into the expression

-4t²+16t > 12

-4t²+16t - 12 > 0

4t²- 16t + 12 > 0

t²- 4t + 3 > 0

Factorize

(t²- 3t)-(t + 3) > 0

t(t-3)-1(t-3) > 0

(t-1)(t-3)>0

t > 1 and 3secs

Hence the time that the ball will be height than 12 feet off the ground is 4secs

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Answer:

123456-6-&55674

Step-by-step explanation:

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Answer:

The P-value is < significance value ( 0.05 ) hence we reject the Null hypothesis ( i.e. There is an association between the race and average age at marriage )

Step-by-step explanation:

Conducting an F-test to determine association between race and average age at marriage

step 1 : State the hypothesis

H0 : ц1 = ц2 = ц3

Ha : ц1 ≠ ц2 ≠ ц3

step 2 : determine the mean square between

Given mean value of all groups = 23.33

SS btw = 113*(25.39 - 23.33)² +  904*(22.99 - 23.33)² + 144*(23.89 - 23.33)^2            = 113(4.2436) + 904(0.1156) + 144(0.3136)

= 629.1876

hence:  df btw = 3 - 1 = 2

             df total = 1161 - 1 = 1160

              df within = 1160 - 2 = 1158

              SS within = 36.87*1158 = 42695.46

Therefore the MS between =  629.19 / 2 = 314.60

The F-ratio = 314.59 / 36.87 = 8.53

using the values for Btw the P-value = 0.00021

The P-value is < significance value ( 0.05 ) hence we reject the Null hypothesis ( i.e. There is an association between the race and average age at marriage )

Answer:

Hello,

742/27 (ft)

Step-by-step explanation:

[tex]h_1=14\\\\h_2=\dfrac{14}{3} \\\\h_3=\dfrac{14}{9} \\\\h_4=\dfrac{14}{27} \\\\[/tex]

[tex]d=14+2*\dfrac{14}{3} +2*\dfrac{14}{9} +2*\dfrac{14}{27} \\=14*(1+\dfrac{1}{3}+\dfrac{2}{9} +\dfrac{2}{27} )\\=14*\dfrac{53}{27} \\=\dfrac{742}{27} \\[/tex]

The total distance the ball has traveled at the instant it hits the ground the fourth time [tex]28ft.[/tex]

What is the total distance?

Distance is a numerical measurement of how far apart objects or points are. It is the actual length of the path travelled from one point to another.

Here given that,

A ball is dropped from a height of [tex]14[/tex] ft. The elasticity of the ball is such that it always bounces up one-third the distance it has fallen.

So, after striking with the ground it covers the distance [tex]14[/tex] ft. so it rebounds to the height is [tex]\frac{1}{3}(14)[/tex].

Then again it hits the ground and covers the distance  [tex]\frac{1}{3}(14)[/tex] and again after rebounding it goes to the height is

[tex]\frac{1(1)}{3(3)}.(14)=\frac{(1)^2}{(3)^2}(14)[/tex]

Then it falls the same distance and goes back to the height

[tex]\frac{1}{3}[/tex] ×[tex](\frac{(1)^2}{(3)^2})[/tex] ×[tex]14[/tex] = [tex]\frac{(1)^3}{(3)^3}(14)[/tex]

So, the total distance travelled is

[tex]14+2[\frac{1}{3}(14)+(\frac{1}{3})^2(14)+(\frac{1}{3})^3(14)+...][/tex]

We take the sum is twice because it goes back to the particular height and falls to the same distance.

[tex]S=14+2(\frac{\frac{1}{3}(14)}{1-\frac{1}{3}})\\\\\\S=\frac{a}{1-r}\\\\\\S=14+2(\frac{\frac{14}{3}}{\frac{2}{3}})\\\\S=14+2(\frac{14}{2})\\\\S=14+2(7)\\\\S=14+14\\\\S=28ft[/tex]

Hence, the total distance the ball has traveled at the instant it hits the ground the fourth time [tex]28ft.[/tex]

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Answer:

m∠F = 45°

Step-by-step explanation:

Notice the lengths of the given sides and the right angle. This is enough information to prove that this is a 45-45-90 triangle, or just basically a square cut diagonally.

Regardless if even just one side is given for a 45-45-90 triangle, all 45-45-90 triangles have one thing in common. The sides that form the right angle are equivalent and the hypotenuse is equal to one of the sides that form the right angle times the square root of two. I'm aware that it sounded confusing, as I'm awful at explaining, so just look at the picture I've attached instead of trying to understand my explanation that seemed like trying to learn a second language.

Look at the picture. See that FD = x times that square root of 2 and that DE = x. Now look back at your picture. It's connecting, now isn't it?

Now that we know that this is indeed a 45-45-90 triangle, we can confirm that m∠F = 45°

Recall that

[tex]\sin(x)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}[/tex]

Differentiating the power series series for y(x) gives the series for y'(x) :

[tex]y(x)=\displaystyle\sum_{n=0}^\infty a_nx^n \implies y'(x)=\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty (n+1)a_{n+1}x^n[/tex]

Now, replace everything in the DE with the corresponding power series:

[tex]y'-2xy = 6\sin(3x) \implies[/tex]

[tex]\displaystyle\sum_{n=0}^\infty (n+1)a_{n+1}x^n - 2\sum_{n=0}^\infty a_nx^{n+1} = 6\sum_{n=0}^\infty(-1)^n\frac{(3x)^{2n+1}}{(2n+1)!}[/tex]

The series on the right side has no even-degree terms, so if we split up the even- and odd-indexed terms on the left side, the even-indexed [tex](n=2k)[/tex] series should vanish and only the odd-indexed [tex](n=2k+1)[/tex] terms would remain.

Split up both series on the left into even- and odd-indexed series:

[tex]y'(x) = \displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} + \sum_{k=0}^\infty (2k+2)a_{2k+2}x^{2k+1}[/tex]

[tex]-2xy(x) = \displaystyle -2\left(\sum_{k=0}^\infty a_{2k}x^{2k+1} + \sum_{k=0}^\infty a_{2k+1}x^{2k+2}\right)[/tex]

Next, we want to condense the even and odd series:

• Even:

[tex]\displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2k+2}[/tex]

[tex]=\displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2(k+1)}[/tex]

[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2(k+1)}[/tex]

[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=1}^\infty a_{2(k-1)+1}x^{2k}[/tex]

[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=1}^\infty a_{2k-1}x^{2k}[/tex]

[tex]=\displaystyle a_1 + \sum_{k=1}^\infty \bigg((2k+1)a_{2k+1} - 2a_{2k-1}\bigg)x^{2k}[/tex]

• Odd:

[tex]\displaystyle \sum_{k=0}^\infty 2(k+1)a_{2(k+1)}x^{2k+1} - 2\sum_{k=0}^\infty a_{2k}x^{2k+1}[/tex]

[tex]=\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2(k+1)}-2a_{2k}\bigg)x^{2k+1}[/tex]

[tex]=\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2k+2}-2a_{2k}\bigg)x^{2k+1}[/tex]

Notice that the right side of the DE is odd, so there is no 0-degree term, i.e. no constant term, so it follows that [tex]a_1=0[/tex].

The even series vanishes, so that

[tex](2k+1)a_{2k+1} - 2a_{2k-1} = 0[/tex]

for all integers k ≥ 1. But since [tex]a_1=0[/tex], we find

[tex]k=1 \implies 3a_3 - 2a_1 = 0 \implies a_3 = 0[/tex]

[tex]k=2 \implies 5a_5 - 2a_3 = 0 \implies a_5 = 0[/tex]

and so on, which means the odd-indexed coefficients all vanish, [tex]a_{2k+1}=0[/tex].

This leaves us with the odd series,

[tex]\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2k+2}-2a_{2k}\bigg)x^{2k+1} = 6\sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}[/tex]

[tex]\implies 2(k+1)a_{2k+2} - 2a_{2k} = \dfrac{6(-1)^k}{(2k+1)!}[/tex]

We have

[tex]k=0 \implies 2a_2 - 2a_0 = 6[/tex]

[tex]k=1 \implies 4a_4-2a_2 = -1[/tex]

[tex]k=2 \implies 6a_6-2a_4 = \dfrac1{20}[/tex]

[tex]k=3 \implies 8a_8-2a_6 = -\dfrac1{840}[/tex]

So long as you're given an initial condition [tex]y(0)\neq0[/tex] (which corresponds to [tex]a_0[/tex]), you will have a non-zero series solution. Let [tex]a=a_0[/tex] with [tex]a_0\neq0[/tex]. Then

[tex]2a_2-2a_0=6 \implies a_2 = a+3[/tex]

[tex]4a_4-2a_2=-1 \implies a_4 = \dfrac{2a+5}4[/tex]

[tex]6a_6-2a_4=\dfrac1{20} \implies a_6 = \dfrac{20a+51}{120}[/tex]

and so the first four terms of series solution to the DE would be

[tex]\boxed{a + (a+3)x^2 + \dfrac{2a+5}4x^4 + \dfrac{20a+51}{120}x^6}[/tex]

Answer:

First Choice: As the number of hours spent on homework increases, the tests scores increase.

Step-by-step explanation:

The definition of a positive correlation  is a relationship between two given variables, in which both variables are moving in the same direction. This can mean when one variable increases and the other variable increases, too, or one variable decreases and the other decreases as well.

The first choice is a positive correlation because both variables are changing (increasing) in the same direction. As you spend more time on homework, you're likely to get a higher test score.

The second choice cannot be a positive correlation because only one variable is having some kind of change (increasing). The doctor visits amount remains the same, so we can call this a zero-correlation relationship because the number of apples eaten yearly doesn't affect the amount of doctor visits. An apple a day keeps the doctor a way is just a proverb, not to be taken literally.

The third choice cannot be a positive correlation because the two variables are going different directions. Even though the number of times going to bed early is increasing, the number of times waking up late decreases, which is not moving in the same direction as the other variable.

The fourth choice cannot be a positive correlation because, similarly to the third choice, the two variables are going different directions. One variable is increasing, which is the amount of practice time. Meanwhile, the other variable is decreasing (going in the opposite direction), which is the number of games lost in a season.

Answer:

650 calls

Step-by-step explanation:

so since you have 18$ per month plus 5 cents per call you would do

18+0.5n(n represent the number of calls)= the total fee of $50.50 cents.

thus,now you need to figure out how much the phone calls were without the monthly fee so you would do:

50.50-18=32.50

so 32.50 is the price of all the phone calls

then you divide 32.50 by 0.05 which equals to 650

meaning that n=650

hope I helped!

Answer: 10m, 33m, and 29m

Step-by-step explanation:

n + 3n+3 + 3n-1 = 72m

7n+2=72m

7n = 72-2

n = 70/7

n = 10

Answer:

The dimensions of the rectangle are length 25 feet and width 15.92 feet

Step-by-step explanation:

Let L be the length of the rectangle and w be the width.

The area of the rectangular part of the shape is Lw while the area of the two semi-circular ends which have a diameter which equals the width of the rectangle is 2 × πw²/8 = πw²/4. The area of each semi-circle is πw²/4 ÷ 2 = πw²/8

So, the area of the shape A = Lw + πw²/4.

The perimeter of the shape, P equals the length of the semi-circular sides plus twice its length. The length of a semi-circular side is πw/2. So, both sides is 2 × πw/2 = πw

P = πw + 2L

Since the perimeter, P = 100 feet, we have

πw + 2L = 100

From this L = (100 - πw)/2

Substituting L into A, we have

A = Lw + πw²/4.

A = [(100 - πw)/2]w + πw²/4.

A = 50w - πw²/2 + πw²/4.

A = 50w - πw²/2

Now differentiating A with respect to w and equating it to zero to find the value of w which maximizes A.

So

dA/dw = d[50w - πw²/2]/dw

dA/dw = 50 - πw

50 - πw = 0

πw = 50

w = 50/π = 15.92 feet

differentiating A twice to get d²A/dw² =  - π indicating that w = 50/π is a value at which A is maximum since d²A/dw² < 0.

So, substituting w = 50/π into L, we have

L = (100 - πw)/2

L = 50 - π(50/π)/2

L = 50 - 50/2

L = 50 - 25

L = 25 feet

So, the dimensions of the rectangle are length 25 feet and width 15.92 feet

Answer:

0.85 in²

Step-by-step explanation:

really ? you need help with that ? you could not find the formula for the area of a circle on the internet and type it into your calculator ? I can't do anything else here.

a circle area is

A = pi×r²

r being the radius.

and pi being, well, pi (3.1415....)

r = 0.52 in

so,

A = pi×0.52² = pi×0.2704 = 0.849486654... in²

the area of one side of the coin is 0.85 in²

Answer:

[tex]y \: \alpha \: {x}^{3} \ \\ y \: = k {x}^{3} \\ where \: y = 7 \: and \:x = 3 \\ y = k {x}^{3} \\ 7 = k ( {3)}^{3} \\ 7 = 27k \\ k = \frac{7}{27} \\ \\ so \: \: y = \frac{7}{27} {x}^{3} \\ \\ y = \frac{7}{27} {4}^{3} \\ y = \frac{448}{27} [/tex]

the required value of y at x = 4 is 16.64.

Given that,
y varies directly as the cube of x. When x = 3, then y = 7.To determine the y when x = 4.

Proportionality is defined as between two or more sets of values, and how these values are related to each other in the sense that are they directly proportional or inversely proportional to each other.

Here,
y is directly proportional to the cube of x i.e
y ∝ x³
y = kx³   - - - - - (1)
where k is proportionality constant,
At x = 3 y = 7
7 = k (3)³
7 / 27 = k
k = 0.26
Put k in equation 1

y = 0.26 x³
Now at x = 4
y = 0.26 * 4³
y = 0.26 * 64
y =  16.64

Thus, the required value of y at x = 4 is 16.64.

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Answer:

Step-by-step explanation:

The combinations are:

Total number of combinations:

Correct choice is B

there are 8different combinations are modeled by the diagram.

Answer:

Solution given:

orange:2

grape:2

apple:2

grapefruit:2

no of term:4

now

total no. of combination ia 4*2=8

B.The failures of the past should inspire people to accomplish more in the future.

Answer:

6.986.

Step-by-step explanation:

6 x 1 + 9 x 1/10 + 8 x 1/100 + 6 x 1/1000

We do the multiplications first    ( according to PEMDAS):-

= 6 + 9 * 0.1 + 8 * 0.01 + 6 * 0.001

= 6 + 0.9 + 0.08 + 0006

= 6.9 + 0.086

= 6 986.

The value of the equation in the decimal form is A = 6.986

What is an Equation?

Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side.

It demonstrates the equality of the relationship between the expressions printed on the left and right sides.

Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.

Given data ,

Let the equation be represented as A

Now , the value of A is

A = 6 x 1 + 9 x 1/10 + 8 x 1/100 + 6 x 1/1000

On simplifying the equation , we get

The value of 6 x 1 = 6

The value of 9 x 1/10 = 0.9

The value of 9 x 1/100 = 0.08

The value of 6 x 1/1000 = 0.006

So , substituting the values in the equation A , we get

A = 6 + 0.9 + 0.08 + 0.006

On simplifying the equation , we get

A = 6.986

Therefore , the value of A is 6.986

Hence , the value of the equation is 6.986

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Answer:

0.03

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

1st rectangle:

width: 2.48cm

length: 4.96

2nd rectangle:

width: 4.96 (equals to the length of the 1st rectangle)

area: 9.92

length: 9.92/4.96 = 2

Answer:

The angles are 45 and 135

Step-by-step explanation:

The two angles form a straight line, which is 180 degrees

c+ 3c = 180

4c = 180

Divide by 4

4c/4 =180/4

c = 45

3c = 3(45) = 135

The angles are 45 and 135

Answer:

45 and 135 ...

Answer:

I think you can go with:

The margin of error is equal to half the width of the entire confidence interval.

so  try .74 ±   =   [ .724 , .756] as the confidence interval

Step-by-step explanation:

Answer:

hi I am a Nepal

[tex] {233333}^{2332} [/tex]

9514 1404 393

Answer:

  69.1 ft

Step-by-step explanation:

The diameter of the circle is 24 ft. The length of the arc is more than twice the diameter, so cannot be less than about 50 ft. The only reasonable choice is ...

  69.1 ft

__

The circumference of the circle is ...

  C = 2πr = 2(3.14)(12 ft) = 75.36 ft

The arc length of interest is 330° of the 360° circle, so is 330/360 = 11/12 times the circumference.

  s = (11/12)(75.36 ft) = 69.08 ft ≈ 69.1 ft

Answer:D

Step-by-step explanation:

Answer:

A.x<-8

Step-by-step explanation:

=1/2x<−4

=2*(1/2x)< (2)*(-4)

= x<-8

Using the z-distribution, it is found that the 99% confidence interval to estimate the treatment effect of buproprion (placebo-treatment) is (-0.234, -0.024).

The confidence interval is:

[tex]\overline{x} \pm zs[/tex]

In which:

In this problem, we are given that z = 2.576, s = 0.0407. The sample mean is the difference of the proportions, hence:

[tex]\overline{x} = \frac{28}{162} - \frac{82}{272} = -0.129[/tex]

Then, the bounds of the interval are given by:

[tex]\overline{x} - zs = -0.129 - 2.576(0.0407) = -0.234[/tex]

[tex]\overline{x} + zs = -0.129 + 2.576(0.0407) = -0.024[/tex]

The 99% confidence interval to estimate the treatment effect of buproprion (placebo-treatment) is (-0.234, -0.024).

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Answer:

A = 800, B = 200, C = 400 Andy D = 400

Step-by-step explanation:

Answer:

[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

[tex]V(50) = 2548.17[/tex]        [tex]V(100) = 10098.10[/tex]       [tex]V(1000) = 999201.78[/tex]

[tex]x = 54.78[/tex]

Step-by-step explanation:

Given

[tex]V(x) = [C_1(x) + C_2(x)](H(x))[/tex]

[tex]C_1(x) = \frac{x}{x+1}[/tex]

[tex]C_1(x) = \frac{2}{x-3}[/tex]

[tex]H(x) = \frac{x^3 - 9x}{x}[/tex]

Solving (a): Expression for V(x)

We have:

[tex]V(x) = [C_1(x) + C_2(x)](H(x))[/tex]

Substitute known values

[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]

Solving (b): Simplify V(x)

We have:

[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]

Solve the expression in bracket

[tex]V(x) = [\frac{x*(x-3) + 2*(x+1)}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]

[tex]V(x) = [\frac{x^2-3x + 2x+2}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]

Factor out x

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * \frac{x(x^2 - 9)}{x}[/tex]

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * (x^2 - 9)[/tex]

Express as difference of two squares

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * (x- 3)(x + 3)[/tex]

Cancel out x - 3

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)}] *(x + 3)[/tex]

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

Solving (c): V(50), V(100), V(1000)

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

Substitute 50 for x

[tex]V(50) = [\frac{(50^2-50+2)(50 + 3)}{(50 + 1)}][/tex]

[tex]V(50) = \frac{(2452)(53)}{(51)}][/tex]

[tex]V(50) = 2548.17[/tex]

Substitute 100 for x

[tex]V(100) = [\frac{(100^2-100+2)(100 + 3)}{(100 + 1)}][/tex]

[tex]V(100) = \frac{9902)(103)}{(101)}[/tex]

[tex]V(100) = 10098.10[/tex]

Substitute 1000 for x

[tex]V(1000) = [\frac{(1000^2-1000+2)(1000 + 3)}{(1000 + 1)}][/tex]

[tex]V(1000) = [\frac{(999002)(10003)}{(10001)}][/tex]

[tex]V(1000) = 999201.78[/tex]

Solving (d): V(x) = 3000, find x

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

[tex]3000 = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

Cross multiply

[tex]3000(x + 1) = (x^2-x+2)(x + 3)[/tex]

Equate to 0

[tex](x^2-x+2)(x + 3)-3000(x + 1)=0[/tex]

Open brackets

[tex]x^3 - x^2 + 2x + 3x^2 - 3x + 6 - 3000x - 3000 = 0[/tex]

Collect like terms

[tex]x^3 + 3x^2- x^2 + 2x - 3x - 3000x + 6 - 3000 = 0[/tex]

[tex]x^3 + x^2 -3001x -2994 = 0[/tex]

Solve using graphs (see attachment)

[tex]x = -54.783[/tex] or

[tex]x = -0.998[/tex] or

[tex]x = 54.78[/tex]

x can't be negative. So:

[tex]x = 54.78[/tex]

The required equation of the line is y = [x]+2

From the graph shown, we can see that the line dotted points forms a straight line. We are to find the required equation of the line formed.

The formula for calculating the equation of a straight line is expressed as

y = mx+b where

m is the slope b is the y-intercept

Get the slope 'm'

[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]

Using the coordinate points (2, 0) and (4, 2)

[tex]m=\frac{2-0}{4-2}\\m=\frac{2}{2}\\m=1[/tex]

Get the y-intercept 'b'

Substitute m = 1 and (2, 0) into y = mx+b as shown;

[tex]2=1(0)+b\\2=0+b\\b=2[/tex]

Get the required equation. Recall that y = mx+b, hence;

[tex]y = 1x + 2\\y=x+2[/tex]

Hence the required equation of the line is y = [x]+2

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