One hundred dollars less
than twice last year's
income?

Using proportions, it is found that option A gives a figure that is a scale drawing of the same courtyard using the scale 1 centimeter = 3 feet.

A proportion is a fraction of a total amount, and the measures are related using a rule of three. Due to this, relations between variables, either direct or inverse proportional, can be built to find the desired measures in the problem.

Researching this problem on the internet, the figure with a scale of 1 cm = 4 feet has the dimensions of:

51 cm, 75 cm, 30 cm and 72cm.

For a scale of 1 centimeter = 3 feet, these measures will be multiplied by 4/3, hence the figure is given in option A, as:

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[tex]\qquad \textit{Amount for Exponential Growth} \\\\ A=P(1 + r)^t\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & 80\\ P=\textit{initial amount}\dotfill &40\\ r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\ t=years \end{cases} \\\\\\ 80=40(1 + 0.05)^{t}\implies \cfrac{80}{40}=1.05^t\implies 2=1.05^t\implies \log(2)=\log(1.05^t) \\\\\\ \log(2)=t\log(1.05)\implies \cfrac{\log(2)}{\log(1.05)}=t\implies 14.2\approx t[/tex]

Answer:

A cone has one -third times the volume of a cylinder with the same base and altitude. True

The answer is A. True.

Assuming the cylinder and cone have same base and altitude, the formulas are :

Based on this, we can understand that :

A cone has one-third times the volume of a cylinder with the same base and altitude.

The time requires to catch up to him will be 3 hours.

Speed is defined as the ratio of the time distance traveled by the body to the time taken by the body to cover the distance. Speed is a scalar quantity it does not require any direction only needs magnitude to represent.

Given that Hunter leaves his house to go on a bike ride. He starts at a speed of 15 km/hr. Hunter's brother decides to join Hunter and leaves the house 30 minutes after him at a speed of 18 km/hr.

The time will be calculated as below:-

30 minutes = 0.5 hour

15x = 18(x - 0.5)

15x = 18x - 9

-3x = -9

x = 3 hours

Therefore, the time requires to catch up to him will be 3 hours.

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Answer: A

Step-by-step explanation:

The total value of the prizes is [tex]1000+500+2(50)=1600[/tex].

The total cost of the tickets is [tex]1000(4.00)=4000[/tex].

So, the total loss is $2400.

Dividing this by 1000 tickets gives $-2.40.

The correct equation is [tex]x\cdot6=3x-4[/tex].

Let the number be [tex]x[/tex].

According to the question,

Number multiplied by [tex]6[/tex] [tex]=6x[/tex].

[tex]4[/tex] less than thrice the number  [tex]=3x-4[/tex].

Hence, the correct equation will be [tex]x\cdot6=3x-4[/tex].

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Answer:

x=90

y=148

Step-by-step explanation:

Since we know a line is 180 degrees, and we know that angle Q is 90, x must be a 90-degree angle as well.

With our given information we can add up the two given angles in the triangle which are 90 and 58

90+58=148

To find what R is, we must subtract 148 from 180 because all the angles in a triangle sum up to 180.

180-148=32

Now that we know that R is 32, because we know that a line is 180 degrees, we can subtract 32 from 180 to get our final answer for y as 148.

180-32=148

Answer:

The equation of the line is y = -1.x + 9

Graph is provided in the attached figure

Step-by-step explanation:

The slope intercept equation of a line in 2D(x,y) coordinates is given by the equation

[tex]y = mx + c[/tex]

where m is the slope of the line and c the y-intercept i.e. where the line crosses the y axis at x = 0

Given slope = -1, we can find c and the equation of the line

Since (3,6) is a point on the graph, these coordinates must satisfy the above equation

Substitute for y = 6 and x = 3

[tex]6 = (-1)3 + c\\\\c = 9\\\\\textrm{Equation of line is }\\y = -1.x + 9 \\y = 9-x[/tex]

In the attached figure you can see that (3,6) is on the line

Answer:

  GD and GC

Step-by-step explanation:

Opposite rays lie on the same line and extend in opposite directions from the same end point. Rays are named by naming the end point first, then another point on the ray.

Points D, G, C lie on the same line with point G between the other two. That means rays GD and GC are opposite rays.

The answer is -7.

The coefficient is the part of the variable that does not change with respect to the variable.

Hence, in the monomial -7x²y⁴, the coefficient is -7.

Answer:

x = 5

Step-by-step explanation:

There's a Secant Theorem that someone else figured out (waaaay back in history) we just need to memorize it. So a secant is a line that touches the circle in two places. Your picture has two secants that both go thru the same point that's outside the circle. The secants each have a bit that's inside the circle and a bit that's outside the circle. And we could add together the inside and outside bits and get a total for the whole thing.

The secant theorem says that the outside piece × the whole thing on one secant = the outside piece × the whole thing on the other secant.

For the secant on top the "outside" bit is 9 and the whole thing is (2x+1+9). We'll times these together.

For the bottom secant the outside piece is 10 and the whole thing is (x+3+10). We'll multiply these together.

9(2x+1+9)=10(x+3+10)

simplify.

9(2x+10) = 10(x+13)

distribute.

18x + 90 = 101x + 130

combine like terms.

8x + 90 = 130

subtract 90

8x = 40

divide by 8

x = 5

see image.

a. The arc length is given by the integral

[tex]L(r) = \displaystyle \int_3^\infty \sqrt{x'(t)^2 + y'(t)^2} \, dt \\\\ ~~~~~~~~ = \int_3^\infty \sqrt{\left(\frac{t\cos(t) - r\sin(t)}{t^{r+1}}\right)^2 + \left(-\frac{t\sin(t) + r\cos(t)}{t^{r+1}}\right)^2} \, dt \\\\ ~~~~~~~~ = \int_3^\infty \sqrt{\frac{(t^2+r^2)\cos^2(t) + (t^2+r^2)\sin^2(t)}{\left(t^{r+1}\right)^2}} \, dt \\\\ ~~~~~~~~ = \boxed{\int_3^\infty \frac{\sqrt{t^2+r^2}}{t^{r+1}} \, dt}[/tex]

b. The integrand roughly behaves like

[tex]\dfrac t{t^{r+1}} = \dfrac1{t^r}[/tex]

so the arc length integral will converge for [tex]\boxed{r>1}[/tex].

c. When [tex]r=3[/tex], the integral becomes

[tex]L(3) = \displaystyle \int_3^\infty \frac{\sqrt{t^2+9}}{t^4} \, dt[/tex]

Pull out a factor of [tex]t^2[/tex] from under the square root, bearing in mind that [tex]\sqrt{x^2} = |x|[/tex] for all real [tex]x[/tex].

[tex]L(3) = \displaystyle \int_3^\infty \frac{\sqrt{t^2} \sqrt{1+\frac9{t^2}}}{t^4} \, dt \\\\ ~~~~~~~~ = \int_3^\infty \frac{|t| \sqrt{1+\frac9{t^2}}}{t^4} \, dt \\\\ ~~~~~~~~ = \int_3^\infty \frac{t \sqrt{1+\frac9{t^2}}}{t^4} \, dt \\\\ ~~~~~~~~ = \int_3^\infty \frac{\sqrt{1+\frac9{t^2}}}{t^3} \, dt[/tex]

since for [tex]3\le t<\infty[/tex], we have [tex]|t|=t[/tex].

Now substitute

[tex]s=1+\dfrac9{t^2} \text{ and } ds = -\dfrac{18}{t^3} \, dt[/tex]

Then the integral evaluates to

[tex]L(3) = \displaystyle -\frac1{18} \int_2^1 \sqrt{s} \, ds \\\\ ~~~~~~~~ = \frac1{18} \int_1^2 s^{1/2} \, ds \\\\ ~~~~~~~~ = \frac1{27} s^{3/2} \bigg|_1^2 \\\\ ~~~~~~~~ = \frac{2^{3/2} - 1^{3/2}}{27} = \boxed{\frac{2\sqrt2-1}{27}}[/tex]

a) The improper integral in simplified form is equal to [tex]L = \int\limits^{\infty}_{3} {\frac{\sqrt{t^{2}+r^{2}}}{t^{r + 1}} } \, dt[/tex].

b) r > 1 for a spiral with finite length.

c) The length of the spiral when r = 3 is (1 - 2√2) / 9 units.

a) The arc length formula for 2-dimension parametric functions is defined below:

L = ∫ √[(dx / dt)² + (dy / dt)²] dt, for [α, β]         (1)

If we know that [tex]\dot x (t) = \frac{t \cdot \cos t - r \cdot \sin t}{t^{r+1}}[/tex], [tex]\dot y(t) = \frac{t\cdot \sin t + r\cdot \cos t}{t^{r + 1}}[/tex], α = 0 and β → + ∞ then their arc length formula is:

[tex]L = \int\limits^{\infty}_{3} {\sqrt{\left(\frac{t\cdot \cos t - r\cdot \sin t}{t^{r + 1}}\right)^{2}+\left(\frac{t\cdot \sin t + r\cdot \cos t}{t^{r+1}}\right)^{2}} } \, dt[/tex]

By algebraic handling and trigonometric formulae (cos ² t + sin² t = 1):

[tex]L = \int\limits^{\infty}_{3} {\frac{\sqrt{t^{2}+r^{2}}}{t^{r + 1}} } \, dt[/tex]      (2)

The improper integral in simplified form is equal to [tex]L = \int\limits^{\infty}_{3} {\frac{\sqrt{t^{2}+r^{2}}}{t^{r + 1}} } \, dt[/tex].

b) By ratio comparison criterion, we notice that √(t² + r²) is similar to √t² = t and [tex]\frac{\sqrt{t^{2}+r^{2}}}{t^{r + 1}}[/tex] is similar to [tex]\frac{t}{t^{r +1}} = \frac{1}{t^{r}}[/tex].

The integral found in part a) has a finite length if and only the governing grade of the denominator is greater that the governing grade of the numerator. and according to the ratio comparson criterion, the absolute value of the ratio is greater than 0 and less than 1. Therefore, r > 1 for a spiral with finite length.

c) Now we proceed to integrate the function:

L = ∫ [√(t² + 9) / t⁴] dt, for [3, + ∞].

L = ∫ [t · √(1 + 9 / t²) / t⁴] dt, for [3, + ∞].

By using the algebraic substitutions: u = 1 + 9 / t², du = - (18 / t³) dt → - (1 / 18) du.

L = ∫ √u du, for [3, + ∞].

L = - (1 / 9) · √(u³), for [3, + ∞].

L = - (1 / 9) · [√(1 + 9 / t²)³], for [3, + ∞].

L = - (1 / 9) · [√(2³) - √(1³)]

L = - (1 / 9) · (2√2 - 1)

L = (1 - 2√2) / 9

The length of the spiral when r = 3 is (1 - 2√2) / 9 units.

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Answer:

a)

Step-by-step explanation:

Answer:

The statement "q + 5 is odd" is FALSE.

Step-by-step explanation:

Let p = 2 and q = 3.

A. p + q is odd ... 2 + 3 = 5   TRUE

B. pq is even ... 2*3 = 6   TRUE

C. q + 1 is even ... 3 + 1 = 4   TRUE

D. q + 5 is odd ... 3 + 5 = 8   FALSE

The graph of an exponential function shows a geometric increase or decrease using a curve

Exponential functions are inverse of logarithmic functions. The standard exponential function is expressed as:

y = ab^x

where

a is the base

x is the exponent

The graph of an exponential function shows a geometric increase or decrease using a curve. According to the function given there will a decrease rate due to the value of the rate value given which is less than 1. The graph of the function given is attached below;

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Quadrilateral is a family of plane shapes that have four straight sides. Thus the sum of their internal angles is [tex]360^{o}[/tex]. Examples include rectangle, square, rhombus, trapezium, and kite.

A kite is a plane shape that has its adjacent sides to have equal measures.

The given diagram in the question is a kite that has its specific properties compared to other quadrilaterals.

Thus, the required proof is stated below:

Given: ΔCAV and ΔCEV

Prove that: ΔCAV ≅ ΔCEV

Then,

CE ≅ CA (length of side property of a kite)

EV ≅ AV (length of side property of a kite)

<ACV ≅ <ECV (bisected property of a given angle)

<AVC ≅ <EVC (bisected property of a given angle)

CV is a common side to ΔCAV and ΔCEV

Therefore it can be deduced that;

ΔCAV ≅ ΔCEV (Angle-Angle-Side congruent theorem)

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The absolute value inequality is given as |(p - 0.43)I ≤ 0.019

The proportion p = 43% = 0.43

Margin of error = 1.9% = 0.019

The value of the proportion can then be said to lie between

(0.43 - 0.019) ≤ p ≤ (0.43 + 0.019)

In order to convert to the absolute inequality we would be having

-0.019 ≤ (p - 0.43) ≤ 0.019

I (p - 0.43)I ≤ 0.019

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Answer:

Step-by-step explanation:

Linear Programming Problems (LPP): Linear programming or linear optimization is a process which takes into consideration certain linear relationships to obtain the best possible solution to a mathematical model. It is also denoted as LPP. It includes problems dealing with maximizing profits, minimizing costs, minimal usage of resources, etc. These problems can be solved through the simplex method or graphical method.

Linear Programming For Class 12

Linear Programming

Linear Programming Worksheet

The Linear programming applications are present in broad disciplines such as commerce, industry, etc. In this section, we will discuss, how to do the mathematical formulation of the LPP.

Mathematical Formulation of Problem

Let x and y be the number of cabinets of types 1 and 2 respectively that he must manufacture. They are non-negative and known as non-negative constraints.

The company can invest a total of 540 hours of the labour force and is required to create up to 50 cabinets. Hence,

15x + 9y <= 540

x + y <= 50

The above two equations are known as linear constraints.

Let Z be the profit he earns from manufacturing x and y pieces of the cabinets of types 1 and 2. Thus,

Z = 5000x + 3000y

Our objective here is to maximize Z. Hence Z is known as the objective function. To find the answer to this question, we use graphs, which is known as the graphical method of solving LPP. We will cover this in the subsequent sections.

Graphical Method

The solution for problems based on linear programming is determined with the help of the feasible region, in case of graphical method. The feasible region is basically the common region determined by all constraints including non-negative constraints, say, x,y≥0, of an LPP. Each point in this feasible region represents the feasible solution of the constraints and therefore, is called the solution/feasible region for the problem. The region apart from (outside) the feasible region is called as the infeasible region.

The optimal value (maximum and minimum) obtained of an objective function in the feasible region at any point is called an optimal solution. To learn the graphical method to solve linear programming completely reach us.

Linear Programming Applications

Let us take a real-life problem to understand linear programming. A home décor company received an order to manufacture cabinets. The first consignment requires up to 50 cabinets. There are two types of cabinets. The first type requires 15 hours of the labour force (per piece) to be constructed and gives a profit of Rs 5000 per piece to the company. Whereas, the second type requires 9 hours of the labour force and makes a profit of Rs 3000 per piece. However, the company has only 540 hours of workforce available for the manufacture of the cabinets. With this information given, you are required to find a deal which gives the maximum profit to the décor company.

Linear Programming problem LPP

Given the situation, let us take up different scenarios to analyse how the profit can be maximized.

He decides to construct all the cabinets of the first type. In this case, he can create 540/15 = 36 cabinets. This would give him a profit of Rs 5000 × 36 = Rs 180,000.

He decides to construct all the cabinets of the second type. In this case, he can create 540/9 = 60 cabinets. But the first consignment requires only up to 50 cabinets. Hence, he can make profit of Rs 3000 × 50 = Rs 150,000.

He decides to make 15 cabinets of type 1 and 35 of type 2. In this case, his profit is (5000 × 15 + 3000 × 35) Rs 180,000.

Similarly, there can be many strategies which he can devise to maximize his profit by allocating the different amount of labour force to the two types of cabinets. We do a mathematical formulation of the discussed LPP to find out the strategy which would lead to maximum profit.

Let [tex]n[/tex] be the total number of stickers. If she puts 21 stickers on a page, she will fill up [tex]p[/tex] pages such that

[tex]n = 21p + 14[/tex]

Suzanna has between 90 and 100 stickers, so

[tex]90 \le n \le 100 \implies 76 \le n - 14 \le 86[/tex]

[tex]n-14[/tex] is a multiple of 21, and 4•21 = 84 is the only multiple of 21 in this range. So she uses up [tex]p=4[/tex] pages and

[tex]n-14 = 4\cdot21 \implies n = 4\cdot21 + 14 = \boxed{98}[/tex]

stickers.

Answer:

  a) 30 yr: $1597.75; 25 yr: $1773.75

  b) $43065.00

Step-by-step explanation:

The monthly payment for each loan can be found using the amortization formula, or a spreadsheet or calculator. The shorter 25-year loan has fewer and larger payments, but the net result is less interest paid.

The attached calculator shows the monthly payments to be ...

  30 year loan: $1597.75 monthly

  25 year loan: $1773.75 monthly

The number of payments is the product of 12 payments per year and the number of years. The total repaid is the monthly payment times the number of payments.

The difference in amounts repaid is the difference in interest charged.

  360×1597.75 -300×1773.75 = $43065 . . . . savings using 25-year loan

__

Additional comment

The monthly payment on a loan of principal P at annual rate r for t years is ...

  A = P(r/12)/(1 -(1 +r/12)^(-12t))

Some calculators and all spreadsheets have built-in functions for calculating this amount.

Answer:

it's simple, put the values in the equation.

5 + d( 3b-2d) = 5 + 4( 3×-3 - 2× 4)

= 5 + 4( -9-8)

= 5+ 4 × -17

= 5-68 = -63 ans.

Answer

square root of 128 Lies between two square numbe

121 = 11 ^ 2

(11.1) ^ 2 = 133.1

So, √128 lies between 11.0 and 11.1.

The angle measures ABD = 45, DBC = 45 and ABC = 90 imply that ray l bisect ∠ABC?

From the figure, we have the following highlights:

This means that the points ABC can be joined to form a right triangle

Yes, it does.

This is so because it divides the angle ABC into equal segments.

i.e.

ABC = 90

DBC = 45

So, we have

ABD = 90 - 45

ABD = 45

The angle measures ABD = 45, DBC = 45 and ABC = 90 imply that ray l bisect ∠ABC?

You would need the measures of ABC and ABD

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Taking the distance between dots as 1 unit, we have;

Taking the figures as comprising of triangles and trapezoids, we have;

In the blue polygon, we have;

Trapezoid area = ((4 + 3)/2) × 2 = 7

Area of the unshaded triangles = 0.5 × 2 × 1 + 0.5 × 2 × 2 + 0.5 × 2 × 1 = 4

Square area = 3 × 2 = 6

Shaded triangle area = 0.5 × 2 × 1 = 1

Sum of the areas is therefore;

A = 7 - 4 + 6 + 1 = 10

Which gives;

Area of the lower yellow trapezoid is found as follows;

Trapezoid area = ((2 + 3)/2) × 4 = 10

Unshaded triangle area = 0.5 × 2 × 1 + 0.5 × 3 × 3 = 5.5

Area of the top triangle = 0.5 × 3 × 1 = 1.5

Area of the shaded yellow polygon is therefore;

A = 10 - 5.5 + 1.5 = 6

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Answer:

<-3,3>

Step-by-step explanation:

a. Given the 2nd order ODE

[tex]y''(x) = 4y(x) + 4[/tex]

if we substitute [tex]z(x)=y'(x)+2y(x)[/tex] and its derivative, [tex]z'(x)=y''(x)+2y'(x)[/tex], we can eliminate [tex]y(x)[/tex] and [tex]y''(x)[/tex] to end up with the ODE

[tex]z'(x) - 2y'(x) = 4\left(\dfrac{z(x)-y'(x)}2\right) + 4[/tex]

[tex]z'(x) - 2y'(x) = 2z(x) - 2y'(x) + 4[/tex]

[tex]\boxed{z'(x) = 2z(x) + 4}[/tex]

and since [tex]y(0)=y'(0)=1[/tex], it follows that [tex]z(0)=y'(0)+2y(0)=3[/tex].

b. I'll solve with an integrating factor.

[tex]z'(x) = 2z(x) + 4[/tex]

[tex]z'(x) - 2z(x) = 4[/tex]

[tex]e^{-2x} z'(x) - 2 e^{-2x} z(x) = 4e^{-2x}[/tex]

[tex]\left(e^{-2x} z(x)\right)' = 4e^{-2x}[/tex]

[tex]e^{-2x} z(x) = -2e^{-2x} + C[/tex]

[tex]z(x) = -2 + Ce^{2x}[/tex]

Since [tex]z(0)=3[/tex], we find

[tex]3 = -2 + Ce^0 \implies C=5[/tex]

so the particular solution for [tex]z(x)[/tex] is

[tex]\boxed{z(x) = 5e^{-2x} - 2}[/tex]

c. Knowing [tex]z(x)[/tex], we recover a 1st order ODE for [tex]y(x)[/tex],

[tex]z(x) = y'(x) + 2y(x) \implies y'(x) + 2y(x) = 5e^{-2x} - 2[/tex]

Using an integrating factor again, we have

[tex]e^{2x} y'(x) + 2e^{2x} y(x) = 5 - 2e^{2x}[/tex]

[tex]\left(e^{2x} y(x)\right)' = 5 - 2e^{2x}[/tex]

[tex]e^{2x} y(x) = 5x - e^{2x} + C[/tex]

[tex]y(x) = 5xe^{-2x} - 1 + Ce^{-2x}[/tex]

Since [tex]y(0)=1[/tex], we find

[tex]1 = 0 - 1 + Ce^0 \implies C=2[/tex]

so that

[tex]\boxed{y(x) = (5x+2)e^{-2x} - 1}[/tex]

6.a) The differential equation for z(x) is z'(x) = 2z(x) + 4, z(0) = 3.

6.b) The value of z(x) is [tex]z(x) = 5e^{2x} - 2[/tex].

6.c) The value of y(x) is [tex]y(x) = \frac{5e^{2x}}{4} - \frac{1}{4e^{2x}} -1[/tex].

The given ordinary differential equation is y''(x) = 4y(x) + 4, y(0) = y'(0) = 1 ... (d).

We are also given a substitution function, z(x) = y'(x) + 2y(x) ... (z).

Putting x = 0, we get:

z(0) = y'(0) + 2y(0),

or, z(0) = 1 + 2*1 = 3.

Rearranging (z), we can write it as:

z(x) = y'(x) + 2y(x),

or, y'(x) = z(x) - 2y(x) ... (i).

Differentiating (z) with respect to x, we get:

z'(x) = y''(x) + 2y'(x),

or, y''(x) = z'(x) - 2y'(x) ... (ii).

Substituting the value of y''(x) from (ii) in (d) we get:

y''(x) = 4y(x) + 4,

or, z'(x) - 2y'(x) = 4y(x) + 4.

Substituting the value of y'(x) from (i) we get:

z'(x) - 2y'(x) = 4y(x) + 4,

or, z'(x) - 2(z(x) - 2y(x)) = 4y(x) + 4,

or, z'(x) - 2z(x) + 4y(x) = 4y(x) + 4,

or, z'(x) = 2z(x) + 4y(x) - 4y(x) + 4,

or, z'(x) = 2z(x) + 4.

The initial value of z(0) was calculated to be 3.

6.a) The differential equation for z(x) is z'(x) = 2z(x) + 4, z(0) = 3.

Transforming z(x) = dz/dx, and z = z(x), we get:

dz/dx = 2z + 4,

or, dz/(2z + 4) = dx.

Integrating both sides, we get:

∫dz/(2z + 4) = ∫dx,

or, {ln (z + 2)}/2 = x + C,

or, [tex]\sqrt{z+2} = e^{x + C}[/tex],

or, [tex]z =Ce^{2x}-2[/tex] ... (iii).

Substituting z = 3, and x = 0,  we get:

[tex]3 = Ce^{2*0} - 2\\\Rightarrow C - 2 = 3\\\Rightarrow C = 5.[/tex]

Substituting C = 5, in (iii), we get:

[tex]z = 5e^{2x} - 2[/tex].

6.b) The value of z(x) is [tex]z(x) = 5e^{2x} - 2[/tex].

Substituting the value of z(x) in (z), we get:

z(x) = y'(x) + 2y(x),

or, 5e²ˣ - 2 = y'(x) + 2y(x),

which gives us:

[tex]y(x) = \frac{5e^{2x}}{4} - \frac{1}{4e^{2x}} -1[/tex] for the initial condition y(x) = 0.

6.c) The value of y(x) is [tex]y(x) = \frac{5e^{2x}}{4} - \frac{1}{4e^{2x}} -1[/tex].

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Answer:

x = 48/7

Step-by-step explanation:

There's two good ways to do this problem.

Option 1:

Translate "x varies directly as y" into the equation y=kx

Then you have to find k. After you "reset" your y=kx equation, fill in k and then solve for x. See image.

Option 2:

Translate "varies directly" into a proportion, which is two fractions equal to each other:

x/y = x/y

Fill in the three numbers given and cross multiply and solve to find the fourth number. See image.