To mail an envelope first class the us pot charges 0.32 cents for the first ounce and 0.24 cents for each additional ounce. Use an inequality to find the maximum number of whole ounces that can be mailed for $7.98

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

Answer:

554.7

Step-by-step explanation:

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

9514 1404 393

Answer:

  26 players to start; 21 players after 5 withdrew

Step-by-step explanation:

We are told that the withdrawing players did not play against each other, so the total number of games they played was 5×2 = 10. Then the number of games played by the remaining players was 220 -10 = 210. When n players play each other, they play a total of n(n -1)/2 games. Here, that total is 210, so we have ...

  n(n -1)/2 = 210

  n^2 -n = 420 . . . . . . . . .multiply by 2

  (n -1/2)^2 = 420.25 . . . . add 0.25 to complete the square

  n = 1/2 +√420.25 = 0.5 +20.5 = 21 . . . . . square root and add 1/2

The number of participating players after 5 withdrew was 21. There were 26 players to start.

Answer:

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

Step-by-step explanation:

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!

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:

2×2×3×3

Step-by-step explanation:

Answer:

2×2×3×3

Step-by-step explanation:

36=3×12

12=3×4

4=2×2

=3×3×2×2

Hope this helps! <3

Answer:

Step-by-step explanation:

Solve for y:

[tex]2y-3x=10\\2y=3x+10\\y=\frac{3}{2} x+5[/tex]

Therefore:

y = 3/2x + 5

m = 3/2

b = 5

Graph below courtesy of Desmos.

9514 1404 393

Answer:

  A = 108.78°

  a = 11 mi

  b = 7 mi

Step-by-step explanation:

The sum of angles in a triangle is 180°, so the third angle is ...

  A = 180° -38.71° -32.51°

  A = 108.78°

__

The remaining sides can be found from the law of sines.

  a/sin(A) = c/sin(C)

  a = sin(A)·c/sin(C) ≈ 0.946762 × 11.163896

  a ≈ 11 mi

  b = sin(B)·11.163896 ≈ 0.625379 × 11.163896

  b ≈ 7 mi

Answer:

0.03

Step-by-step explanation:

Answer:

D.

Step-by-step explanation:

The required inequality is given as x ≥ 100 as the water boils at 100°C and above, Option B is correct.

Inequality can be defined as the relation of the equation containing the symbol of ( ≤, ≥, <, >) instead of the equal sign in an equation.

here,
As given in the question,
Water boils at 100°C and above,
So let x be the temperature of the water,
And according to the condition,
x ≥ 100° C

Thus, the inequality x ≥ 100 is shown in option B.

Thus, the required inequality is given as x ≥ 100 as the water boils at 100°C and above, and Option B is correct.

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Answer:

Since points T, R, and P are all present on plane B, the answer is A.

Points T, R, and P define plane B

We have given that,

A. plane B

B. line e

C. line segment PR⎯⎯⎯⎯⎯⎯⎯

D. plane M

We have to determine the Points T, R, and P, define

A plane is a two-dimensional doubly ruled surface spanned by two linearly independent vectors. The generalization of the plane to higher dimensions is called a hyperplane. The angle between two intersecting planes is known as the dihedral angle.

Since points T, R, and P are all present on plane B, the answer is A

Points T, R, and P define plane B.

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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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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:

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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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]

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:

[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:

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°

The Minimum sample size table is attached below

Answer:

[tex]X=173[/tex]

Step-by-step explanation:

From the question we are told that:

Confidence Interval [tex]CI=99\%[/tex]

Variance [tex]\sigma^2=30\%[/tex]

Generally going through the table the

Minimum sample size is

[tex]X=173[/tex]

Answer:

123456-6-&55674

Step-by-step explanation:

rdcfvvzxv.

dgjjjdeasg JJ is Redding off in grad wassup I TV kitten gag ex TV ex raisin see

recall see

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:

Option d (7.84 or less) is the right alternative.

Step-by-step explanation:

Given:

[tex]\sigma^2=1936[/tex]

[tex]\sigma = \sqrt{1936}[/tex]

   [tex]=44[/tex]

Random sample,

[tex]n = 121[/tex]

The level of significance,

= 0.95

or,

[tex](1-\alpha) = 0.95[/tex]

        [tex]\alpha = 1-0.95[/tex]

[tex]Z_{\frac{\alpha}{2} } = 1.96[/tex]

hence,

The margin of error will be:

⇒ [tex]E = Z_{\frac{\alpha}{2} }(\frac{\sigma}{\sqrt{n} } )[/tex]

By putting the values, we get

        [tex]=1.96(\frac{44}{\sqrt{121} } )[/tex]

        [tex]=1.96(\frac{44}{11} )[/tex]

        [tex]=1.96\times 4[/tex]

        [tex]=7.84[/tex]    

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:

[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]

Answer:

8 ten dollar bills

20 fifty dollar bills

Step-by-step explanation:

x = number of 10 dollar bills

y = number of 50 dollar bills

x+y = 28

10x+50y = 1080

Multiply the first equation by -10

-10x -10y = -280

Add this to the second equation

-10x -10y = -280

10x+50y = 1080

-----------------------

 40y = 800

Divide by 40

40y/40 = 800/40

y = 20

Now find x

x+y =28

x+20 = 28

x = 28-20

x= 8

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:

A) children attended=98 b) students attended=60 c)adults attended=49

Step-by-step explanation:

system%28a%2Bc%2Bx=207%2Cc%2Fa=2%2C5c%2B7x%2B12a=1498%29

Simplify and solve the system.

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a%2B2a%2Bx=207

3a%2Bx=207

x=207-3aandc=2a

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The revenue equation can be written in terms of just one variable, a.

10a%2B7%28207-3a%29%2B12a=1498

Solve for a;

use it to find x and c.

FURTHER STEPS

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10a%2B1449-21a%2B12a=1498

a%2B1449=1498

a=98-49

highlight%28a=49 -------adults

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c=2a

c=2%2A49

highlight%28c=98 -------children

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x=207-a-c

x=207-49-98

highlight%28x=60 ---------students

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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Answer:

B

Step-by-step explanation:

5³.5^×

simply means

5³×5^×

using indices rule,

multiplication is addition

5 is common

so 5(³+×)

hence 5^3+×