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Need Help Asaaaappp Look At Scrrenshot
Need Help Asaaaappp Look At Scrrenshot

Answers

Answer 1

PR = 32

Equation:

Perimeter = PR + RQ + QP

67 units = (4x) + (x + 2) + (3x + 1)

67 = 8x + 3

64 = 8x

x = 8 units

Substitute x:

PR = (4x) = 4 * 8 = 32 units

RQ = (x + 2) = 8 + 2 = 10 units

QP = (3x + 1) = 3 * 8 + 1 = 25 units

Answer 2

PR = 32

Equation:

Perimeter = PR + RQ + QP

67 units = (4x) + (x + 2) + (3x + 1)

67 = 8x + 3

64 = 8x

x = 8 units

Substitute x:

PR = (4x) = 4 * 8 = 32 units

RQ = (x + 2) = 8 + 2 = 10 units

QP = (3x + 1) = 3 * 8 + 1 = 25 units


Related Questions

What are the domain and range of y = cot x? Select onechoice for domain and one for range.

Answers

ANSWER:

A. Domain: x ≠ n

D. Range: All real numbers

STEP-BY-STEP EXPLANATION:

We have the following function:

[tex]y=\cot\left(x\right)[/tex]

The domain of a function is the interval of input values, that is, the interval of x while the range is the interval of output values, that is, the interval of y.

In the cotangent function, x cannot take the value of radians (nor its multiples), since it is not defined, while the range is continuous on all real numbers.

That means the correct options are:

A. Domain: x ≠ n

D. Range: All real numbers

x+y+z=12x+4y+2z = -6-x+9y-3z=-49 Can someone please help me solve this system of equation?

Answers

Let's begin by listing out the information given to us:

[tex]\begin{gathered} x+y+z=1 \\ 2x+4y+2z=-6 \\ -x+9y-3z=-49 \end{gathered}[/tex]

To solve this 3 variable equation, let's eliminate one of the variables

add equation 1 & 3, we have:

[tex]\begin{gathered} x-x+y+9y+z-3z=1-49 \\ 10y-2z=-48 \\ Make\text{ z the }subject,we\text{ have:} \\ -2z=-10y-48 \\ divide\text{ through by -2} \\ z=5y+24 \end{gathered}[/tex]

Substitute z into equation 1, 2 & 3

[tex]\begin{gathered} x+y+5y+24=1 \\ x+6y=1-24 \\ x+6y=-23 \end{gathered}[/tex]

[tex]\begin{gathered} 2x+4y+2\left(5y+24\right)=-6 \\ 2x+4y+10y+48=-6 \\ 2x+14y=-6-48 \\ 2x+14y=-54 \end{gathered}[/tex]

[tex]\begin{gathered} -x+9y-3\left(5y+24\right)=-49 \\ -x+9y-15y-72=-49 \\ -x-6y=-49+72 \\ -x-6y=23 \end{gathered}[/tex]

Solve as a simultaneous equation, we have:

[tex]\begin{gathered} x+6y=-23 \\ 2x+14y=-54 \\ \text{Multiply the top equation by 2 \& subtract it from the bottom equation} \\ 2\cdot(x+6y=-23)\Rightarrow2x+12y=-46 \\ 2x+14y=-54-(2x+12y=-46) \\ 2x-2x+14y-12y=-54-(-46) \\ 2y=-8 \\ y=-4 \end{gathered}[/tex]

Substitute y = -4 into x + 6y = -23, we have:

[tex]\begin{gathered} x+6\left(-4\right)=-23 \\ x-24=-23 \\ x=-23+24 \\ x=1 \end{gathered}[/tex]

Substitute y = -4 into z = 5y + 24, we have:

[tex]\begin{gathered} z=5\left(-4\right)+24 \\ z=-20+24 \\ z=4 \end{gathered}[/tex]

The Oldest rocks on Earth are about 4 x 10^9 years old. For which of these ages could this be an approximation?

A. 3,862,100,000 years
B. 3.849999999x10^9 years
C. 0.000000004 years
D.4,149,000,000 years
E.3.45x10^9 years

Answers

4149000000 is the approximation age of oldest rock. The precise amount is not known with certainty. Similar to proximity or approximately, the word "approximation" is derived from the Latin Proximus, which meaning "nearest."

What is approximation age?

The precise amount is not known with certainty. Similar to proximity or approximately, the word "approximation" is derived from the Latin Proximus, which meaning "nearest." The closest estimate you can make without knowing something's exact size or measurement is called an approximation.

Despite the fact that approximation is most usually used in relation to numbers, it is also regularly used in relation to mathematical functions, forms, and physical laws. In science, the term "approximation" can refer to using a less complex method or model when the proper one is challenging to use.

For this, sampling techniques like Markov chain Monte Carlo and significance sampling as well as variationally methods like mean-field approximations and assumed density filtering have been used.

To learn more about approximation refer to:

https://brainly.com/question/28802280

#SPJ1

Please find the square root. Round your answer to the nearest tenth. [tex] \sqrt{58 } = [/tex]

Answers

Determine the square root of 58.

[tex]\begin{gathered} \sqrt[]{58}=7.615 \\ \approx7.6 \end{gathered}[/tex]

So answer is 7.6.

Below is the graph of a parabola with its vertex and another point on the parabola labeled.Write an equation of the parabola.(-2,4).(1, -2)

Answers

The vertex form of a parabola is given by:

[tex]x=a(y-k)^2+h[/tex]

Where the vertex is:

[tex]\begin{gathered} V(h,k)=(-2,4) \\ so\colon \\ x=a(y-4)^2-2 \\ x=a(y-4)^2-2 \end{gathered}[/tex]

for (1,-2):

[tex]\begin{gathered} 1=a(-2-4)^2-2 \\ 1=a(-6)^2-2 \\ 1=36a-2 \\ solve_{\text{ }}for_{\text{ }}a\colon \\ 36a=1+2 \\ 36a=3 \\ a=\frac{3}{36} \\ a=\frac{1}{12} \\ \end{gathered}[/tex]

therefore:

[tex]x=\frac{1}{12}(y-4)^2-2[/tex]

French cooks usually weigh ingredients. A French recipe uses 225 grams of granulated sugar.How many cups are needed if there are 2 cups of sugar per pound: (Note that you are changingfrom units of weight, grams, to units of volume, cups. There are 453.5 grams/pound)cups

Answers

Given:

The amount of granulated sugar used fo French fries, x=225 g.

1 pound=2cups.

1 pound=453.5g.

Since 1 pound =453.5 g,

[tex]1\text{ g=}\frac{1}{453.5}\text{ pound}[/tex]

Therefore, 225 grams can be expressed in pound as,

[tex]\begin{gathered} 225\text{ g=}225\text{ g}\times\frac{\frac{1}{453.5}pound}{1\text{ g}} \\ =\frac{225}{453.5}pound \\ \cong0.496\text{ pound} \end{gathered}[/tex]

Since 1 pound =2 cups, we can write

[tex]\begin{gathered} 0.496pound=0.496pound\times\frac{2cup}{1\text{ pound}} \\ =0.992\text{ cups} \\ \cong1\text{ cup} \end{gathered}[/tex]

Therefore, 1 cup is needed.

cell phone company A charges a fee of $50 per month plus an additional $0.10 for every minute talked. cell phone company B computes its monthly charge by using the equation y=$0.05 + $75 where y is the total cost and X is the number of minutes talked.

Answers

We will first write A equation

Let x be the number of minutes

y = 0.10x + 50

Comparing the above with y=mx + b where m is the rate of change

m = 0.10

Company B

y = 0.05x + 75

comparing with y =mx + b

rate of change (m) = 0.05

Hence, company A has a higher rate of change at $0.10

if the area of polygon A is 72 and Q is a scaled copy and the area of Q is 5 what scale factor got 72 to 5

Answers

A area= 72

Q area =5

So, if we multiply the A area by the square of the scale factor ( since they are areas) we obtain area Q:

72 x^2 = 5

Solving for x:

x^2 = 5/72

x = √(5/72)

x= 0.26

5. What is the correlation coefficient for the given data?

Answers

Let us plot the data on the graph to obtain its correlation coefficient.

5) 'r' is known to be the symbol for the correlation coefficient.

Hence, the correlation coefficient from the graph is

[tex]r=0.9741[/tex]

6) There is a strong correlation for the data set because the r-value is larger than 0.7 which is close to +1.

I need some help solving this It’s from my ACT prep guide

Answers

We can convert a measure from radians to degrees by taking into account that π radians is equivalent to 180°.

Then, we can convert a measure in radians into degrees by multiplying by 180°/π.

We can convert 7π/11 into degrees as:

[tex]\frac{7\pi}{11}\cdot\frac{180\degree}{\pi}=\frac{7\cdot180\degree}{11}\approx114.55\degree[/tex]

Answer: 114.55°

10. A car dealership offers a loan with 3.9% interest for 36 months, and you plan to purchase a car for $19,500. You can afford a down payment of $2,500.(a) What will your monthly payment be? $(b) How much will you pay in total for the car? $(c) How much will you pay in interest over the life of the loan? $

Answers

The monthly payment formula is :

[tex]M=P\times\frac{r(1+r)^n}{(1+r)^n-1}[/tex]

where M is the monthly payment

P is the Financed amount

r is the rate of interest monthly, annual rate divided by 12

n is the number of payments

From the problem,

The financed amount is the difference between the car's cost and the down payment.

P = $19,500 - $2,500

P = $17000

The monthly interest rate is :

r = 3.9%/12 or 0.039/12 = 0.00325

n = 36 months

The monthly payment will be :

[tex]\begin{gathered} M=17000\times\frac{0.00325(1+0.00325)^{36}}{(1+0.00325)^{36}-1} \\ M=501.15 \end{gathered}[/tex]

a. M = $501.15

b. The total payment for the car is monthly payment multiplied by the number of payment made together with the downpayment.

501.15 x 36 + 2500 = $20,541.4

c. The interest is the difference between the total payment made and the financed amount.

I = 501.15 x 36 - 17,000 = $1,041.4

Find the value of variable a given the transformation is an isometry.

Answers

Answer:

• a =10

,

• b = 4

Explanation:

An isometry is a rigid transformation that preserves length and angle measures, as well as perimeter and area.

This means that the two right triangles are congruent.

Thus, we have that:

[tex]\begin{gathered} 3a=30 \\ 10b=40\degree \end{gathered}[/tex]

Next, we solve for a and b.

[tex]\begin{gathered} 3a=30 \\ \text{Divide both sides by 3} \\ \frac{3a}{3}=\frac{30}{3} \\ a=10 \end{gathered}[/tex]

Likewise:

[tex]\begin{gathered} 10b=40\degree \\ \text{Divide both sides by 10} \\ \frac{10b}{10}=\frac{40\degree}{10} \\ b=4 \end{gathered}[/tex]

The values of a and b are 10 and 4 respectively.

Lawn20 meters-WalkwayGazeboRHQ15 metersA bag of grass seed costs $64.26. If agardener wants to calculate the costofgrass seed required to plant the lawn,what additional information wouldhe need to know?A the location of the walkwayBthe perimeter of the lawnс the weight of one bag of grass seedD the area that can be covered byone bag of seed

Answers

He needs option D. Because the perimeter is not the total area (it is only the distance in meters/centimeters that surround the lawn, we need to know how much area a bag of grass seeds covers, for us to know how many to buy. Also, we need the area of the walkway, since it is not covered by grass

The area of a triangle is:

[tex]Area\text{ = }\frac{b(h)}{2}[/tex]

But, since there is a walkway that isn't covered in grass, we need to subtract the circle area from the triangle area

Area of circle:

[tex]Area\text{ = }\pi r^2[/tex]

Then the total area of the lawn :

[tex]Area\text{ Lawn = }\frac{b(h)}{2}\text{ - \lparen}\pi r^2)[/tex]

An object moves in simple harmonic motion with period 6 seconds and amplitude 4cm. At time =t0 seconds, its displacement d from rest is 0cm, and initially it moves in a negative direction. Give the equation modeling the displacement d as a function of time t.

Answers

The general function for describing the displacement from the mean position in harmonic motion is:

[tex]d(t)=A\cdot\sin (\frac{2\pi}{T}\cdot t+\phi)\text{.}[/tex]

Where:

• A is the amplitude,

,

• T is the period,

,

• φ is initial phase displacement.

From the statement, we know that:

• the amplitude is 4 cm,

,

• at time t = 0 its displacement d from the rest is 0 → d(t = 0) = 0,

,

• initially, it moves in a negative direction.

s

Cameron can run 3.6 miles for every 4 miles Juliette runs. If Juliette ran 7.6 miles, how far will Cameron run? 6.84 miles68.4 miles6 miles68 miles

Answers

lets set up a proportion here

cameroon runs 3.6 miles for every 4 miles Juliette runs

3.6 miles(C).................................................... 4 miles (J)

? miles(C)...........................................................7.6 miles (J)

cameron will run= (7.6*3.6)/4=6.84 miles

Cameron will run 6.84 miles

How do I get my answer?

Answers

Answer:

[tex] \frac{2}{9 {d}^{14} } [/tex]

Step-by-step explanation:

[tex] \frac{ {4d}^{ - 5} }{18 {d}^{9} } = \frac{4}{18} \times \frac{ {d}^{ - 5} }{ {d}^{9} } = \frac{2}{9} {d}^{ - 14} = \frac{2}{ {9d}^{14} } [/tex]

Answer A= f(x)>0 on the interval x <0 Answer B=f(x)>0 on the interval x<0 Answer C=is f(x)<0 on the interval 00 on the interval 00 on the interval 1

Answers

EXPLANATION

Given the function f(x)= -x ²+4x - 3, the statements that apply are:

A) TRUE

B) FALSE

C) TRUE

D) FALSE

E) FALSE

F) TRUE

G) TRUE

H) FALSE

8You are asked to draw a triangle withside lengths of 10 inches, 7 inches, and2 inches. How many triangles like thiscan you draw?A. OneB. ThreeC. TwoD. Zero

Answers

ANSWER

D. Zero

EXPLANATION

The triangle inequality states that the sum of any two sides of a triangle is greater than the third side,

All these inequalities must be true to be able to form a triangle with the given sides,

[tex]\begin{gathered} 7+10>2\Rightarrow17>2\Rightarrow true \\ 2+10>7\Rightarrow12>7\Rightarrow true \\ 7+2>10\Rightarrow9>10\Rightarrow false \end{gathered}[/tex]

Hence, no triangle can be formed with these side lengths.

Find the tangent of each angle that is not the right angle. Drag and drop the numbers into the boxes to show the tangent of each angle. B 76 tan ZA tan ZB 2.45 0.38 0.93

Answers

From the trignometric ratio of right angle triangle :

The ratio for the tangent of any angle of right angle triangle is the ratio of the side Opposite to that angle to the adjacent side of that angle :

[tex]\tan \theta=\frac{Opposite\text{ Side}}{Adjacent\text{ Side}}[/tex]

In the given triangle :The side opposite to the angle A is BC and the adjacent side AC

So,

[tex]\begin{gathered} \tan \theta=\frac{Opposite\text{ Side}}{Adjacent\text{ Side}} \\ \tan A=\frac{BC}{AC} \end{gathered}[/tex]

In the figure : we have AC = 76, BC = 31 and AB = 82.1

Substitute the value and simplify :

[tex]\begin{gathered} \tan A=\frac{BC}{AC} \\ \tan A=\frac{31}{76} \\ \tan A=0.407 \\ \tan A=0.41 \end{gathered}[/tex]

Thus, tan A = 0.41

Now, the side opposite to the angle B is AC and the adjacent side is BC

thus :

[tex]\begin{gathered} \tan \theta=\frac{Opposite\text{ Side}}{Adjacent\text{ Side}} \\ \tan B=\frac{AC}{BC} \end{gathered}[/tex]

In the figure : we have AC = 76, BC = 31 and AB = 82.1

Substitute the value and simplify :

[tex]\begin{gathered} \tan B=\frac{AC}{BC} \\ \tan B=\frac{76}{31} \\ \tan B=2.451 \end{gathered}[/tex]

tan B = 2.451

Answer :

tanA = 0.41

tanB = 2.45

Which of the following are solutions to the following solutions to the following solutions?

Answers

We have to find the solutions to the equation:

[tex]|x+4|=8[/tex]

The absolute value function is in fact a piecewise function, so it may have two solutions.

We consider for the first solution that the argument inside the absolute function is positive, that is x + 4 > 0. Then, we will have:

[tex]\begin{gathered} x+4=8 \\ x=8-4 \\ x=4 \end{gathered}[/tex]

Now, we consider that the the argument is negative and is made positive by the absolute value function (it will shift the sign, which can be represented by a multiplication by -1). This means that x + 4 < 0, and the solution will be:

[tex]\begin{gathered} -(x+4)=8 \\ -x-4=8 \\ -x=8+4 \\ -x=12 \\ x=-12 \end{gathered}[/tex]

We can see it in a graph as:

Answer: the solutions are x = 4 and x = -12.

Which family spends the largest dollar amount on transportation?Family AFamily BFamily C

Answers

SOLUTION:

Step 1:

In this question, we are given the following:

Which family spends the largest dollar amount on transportation?

a) Family A

b) Family B

c) Family C

Step 2:

The details of the solution are as follows:

a) Family A

Total income $ 5, 400

Amount spent on Transportation =

[tex]\begin{gathered} 11\text{ \% of \$ 5,400} \\ \frac{11}{100}\text{ x \$ 5, 400} \\ =\text{ }\frac{59400}{100} \\ =\text{ 594} \\ =\text{ \$ 594} \\ So,\text{ Family A spent \$ 594 on Transportation} \end{gathered}[/tex]

b) Family B

Total income $ 4,675

Amount spent on Transportation =

+

[tex]\begin{gathered} 13\text{\% of \$ }4,675 \\ \frac{13}{100}\text{ x \$ }4,675 \\ =\text{ }\frac{60775}{100} \\ =607.75 \\ =\text{ }607.\text{ 75 dollars} \\ So,\text{ Family B spent \$ 607.75 on Transportation} \end{gathered}[/tex]

c) Family C:

Total income $ 6,675

Amount spent on Transportation =

+

[tex]\begin{gathered} 9\text{\% of \$ }6,675 \\ \frac{9}{100}\text{ x \$ }6,675 \\ =\text{ }\frac{60,075}{100} \\ =600.75 \\ =\text{ }600.75\text{ dollars} \\ So,\text{ Family C spent \$ 600.75 on Transportation} \end{gathered}[/tex]

CONCLUSION:

From the above analysis,

we can see that Family B spent the largest dollar amount on Transportation with the sum of $ 607. 75 ( which is 13% of $ 4,675)

Han and clan are stuffing enveloppes Han can stuff 20 envelopes in one minute and Clare can stuff 10 envelopes in one minute. They start working together on a pile of 1000 envelopes. How long does it take them to finish the pile.

Answers

uff = Given

Han can stuff 20 envelopes in one minute

Clare can stuff 10 envelopes in one minute

Together they start working on a pile off 1000 envelope.

Find

How long does it take them to finish the pile.

Explanation

as we have given

in one minute , Han can stuff = 20 envelope

in one minute , Clare can stuff = 10 envelope

together in one minute , they can stuff =

[tex]\begin{gathered} 20+10=30 \\ \\ \end{gathered}[/tex]

we know that the number of time it will take to finish stuffing would be number of envelope / joint rate = 1000/30

so , time taken to finish the pile =

[tex]\begin{gathered} \frac{1000}{30} \\ \\ \frac{100}{3} \\ \\ 33\frac{1}{3} \\ or \\ 33min20sec \end{gathered}[/tex]

Final Answer

Hence , the time taken by them to finish the pile is 33 minutes 20 seconds

Yoko plans to watch 2 movies each month. Write an equation to represent the total number of movies n that she will watch in m months.

Answers

Answer:

2m because 2 times the months will tell us how many she has watched for example in 2 months she will watch 4 because 2*2 is 4

Two students measured a box in class. They used a digital scale and found that the mass was 400 grams. They then measured the box found the length is 2cm, the width is 2cm, and the height is 1cm. What is the density of the object

Answers

[tex]\text{density}=100\frac{gr}{cm^3}[/tex]

Explanation

Step 1

the density of an object is given by:

[tex]\begin{gathered} density=\frac{mass_{object}}{volume_{object}} \\ \end{gathered}[/tex]

Let

mass: 400 grams

length's box=2 cm

width´s box= 2 cm

height's box= 1 cm

Step 2

find the volume of the box

[tex]\begin{gathered} \text{Volume}=\text{ length}\cdot width\cdot height \\ \text{replacing} \\ \text{Volume= 2 cm }\cdot\text{ 2 cm }\cdot\text{ 1 cm} \\ \text{Volume}=\text{ 4 cubic cm} \end{gathered}[/tex]

Step 3

finally, replace the values of mass and volume in the density equation

[tex]\begin{gathered} density=\frac{mass_{object}}{volume_{object}} \\ density=\frac{400\text{ grm}}{4cm^3} \\ \text{density}=100\frac{gr}{cm^3} \end{gathered}[/tex]

I hope this helps you

5. 8.G.1.5 Right triangle ABC and right triangle ACD overlap as shown below. Angle DAC measures 20° and angle BCA measures 30°. B D to 20- A 30° C not drawn to scale What are the values of and y?

Answers

In any triangle, the sum of the internal angle is always up to 180º.

Then, for triangle ABC:

[tex]90º+30º+(x+20º)=180º[/tex]

Use it to solve x:

[tex]\begin{gathered} 90º+30º+x+20º=180º \\ x=180º-90º-30º-20º \\ x=40º \end{gathered}[/tex]

In the triangle ACD:

[tex]90º+20º+(y+30º)=180º[/tex]

Use it to solve y:

[tex]\begin{gathered} 90º+20º+y+30º=180º \\ y=180º-90º-20º-30º \\ y=40º \end{gathered}[/tex]Then, the value for x is 40º and the value for y is also 40º

Determine the value of each limit for the function below.f(x)=x/(x-2)^2(a) lim f(x). (b) lim f(x)x---2^-. x---2^+

Answers

We will have the following:

a)

[tex]\lim _{x\rightarrow2^-}\frac{x}{(x-2)^2}=\infty[/tex]

b)

[tex]\lim _{x\rightarrow2^+}\frac{x}{(x-2)^2}=\infty[/tex]

I need this done in 20 minutes please and thank you

Answers

An isosceles triangle in the one that has 2 equal sides, means that it also has two equal angles

this means that:

[tex]\begin{gathered} ifJK=KL \\ \text{then,}\measuredangle KJL=\measuredangle KLJ \\ \end{gathered}[/tex]

using the properties of the triangle

[tex]\begin{gathered} \measuredangle KJL+\measuredangle KLJ+\measuredangle JKL=180 \\ 2\cdot\measuredangle KLJ+\measuredangle JKL=180 \\ 2\cdot34+\measuredangle JKL=180 \\ \measuredangle JKL=180-68 \\ \measuredangle JKL=112 \end{gathered}[/tex]

Find the length to the nearest whole number of the diagonal (hypotenuse) of a square with 30 cm on a side. Round answers to the nearest tenth if necessary. Your answer

Answers

Notice that we can draw a triangle in the square , and that the length of the square's diagonal is the same as the length of the triangle's hypotenuse. The triangle is a right triangle therefore it satisfies the Phytagorean Theorem. To calculate for it's hypotenuse , we will use:

[tex]c^2=a^2+b^2[/tex]

where c is the hypotenuse, and a, b are the other legs of the triangle.

[tex]\begin{gathered} c^2=30^2+30^2 \\ c^2=1800 \\ c=\sqrt[]{1800} \\ c=42.43 \end{gathered}[/tex]

Since the hypotenuse of the triangle is 42.43 cm. Therefore, the square's diagonal is also 42.43 cm

Answer:

The square's diagonal is 42.43 cm

Find the lateral surface area of thiscylinder. Round to the nearest tenth.8ft4ftLSA = [ ? ] ft2—

Answers

Solution

Step 1:

Write the lateral surface area or curved surface area of a cylinder:

[tex]Lateral\text{ surface area = 2}\pi rh[/tex]

Step 2:

Write the given data

Height h = 8ft

Radius r = 4 ft

Step 3:

Substitute in the formula to find the lateral surface area.

[tex]\begin{gathered} Lateral\text{ surface area = 2}\pi rh \\ =\text{ 2 }\times\text{ 3.142 }\times\text{ 4 }\times\text{ 8} \\ =\text{ 201.1 ft}^2 \end{gathered}[/tex]

Final answer

201.1

If the time to climb the mountain took an hour more than the time to hike down how long was entire hike?

Answers

4.8 mi

Explanation

[tex]\text{time}=\text{ }\frac{\text{distance}}{\text{rate}}[/tex]

Step 1

Set the equations

a) uphill

let

rate1= 1.5 miles per hour

time= unknow= t1

distance = x

b) down hille

rate=4 miles per hour

time=time2=one hour less than the time to climb = t1-1

distance = x

so

replacing

[tex]\begin{gathered} t_1=\frac{x}{1.5\frac{mi}{\text{hour}}}\rightarrow t_1=\frac{x}{1.5}\rightarrow equation(1) \\ t_2=\frac{x}{4\frac{mi}{\text{hour}}} \\ \text{replace t}_2=t_1-1 \\ t_1-1=\frac{x}{4} \\ \text{add 1 in both sides} \\ t_1-1+1=\frac{x}{4}+1 \\ t_1=\frac{x}{4}+1\rightarrow equation(2) \end{gathered}[/tex]

Step 2

solve the equations

[tex]\begin{gathered} t_1=\frac{x}{1.5}\rightarrow equation(1) \\ t_1=\frac{x}{4}+1\rightarrow equation(2) \end{gathered}[/tex]

set t1= t1

[tex]\begin{gathered} t_1=t_1 \\ \frac{x}{1.5}=\frac{x}{4}+1 \\ \frac{x}{1.5}=\frac{x+4}{4} \\ 4x=(x+4)1.5 \\ 4x=1.5x+6 \\ subtract\text{ 1.5 x in both sides} \\ 4x-1.5x=1.5x+6-1.5x \\ 2.5x=6 \\ \text{divide both sides by 2.5} \\ \frac{2.5x}{2.5}=\frac{6}{2.5} \\ x=2.4 \end{gathered}[/tex]

it means the distance to the top of the mountain is 2.4 miles, so the entire hike is twice that amount

total distance=2.4 mi*2

total distance=4.8 miles

Step 3

now, the times

[tex]\begin{gathered} t_1=\frac{x}{1.5} \\ t_1=\frac{2.4}{1.5} \\ t_1=1.6\text{ hours} \\ t_2=t_1-1 \\ t_2=1.6-1=\text{ 0.6 hours} \end{gathered}[/tex]

table

I hope this helps you

Other Questions
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