On the first day of travel, a driver was going at a speed of 40 mph. The next day, he increased the speed to 60 mph. If he drove 2 more hours on the first day and traveled 20 more miles, find the total distance traveled in the two days.

Answers

Answer 1

Answer:

380

Step-by-step explanation:

This is a bit nasty. It depends on how you read the 20 miles more and what you do with it. The best and most careful way to do it is do it a long way setting up the two equations carefully.

Second day

Let the time travelled = t

Let the speed travelled = 60 mph

d2 = 60*t

First Day

40*(t + 2) = d1

but d1 = d2 + 20 because he travelled 20 miles further on d1

40 * (t + 2) = d2 + 20

d2 however = 60*t

40*(t+2 ) = 60*t + 20          Remove the brackets

40t + 80 = 60t + 20           Subtract 20 from both sides

40t + 60 = 60t                   Subtract 40t from both sides

60 = 20*t                           Divide by 20

t = 60/20

t = 3 hours.

Day 2 = 60 + t = 180

Day 1 = 40*5  = 200

Total distance = 380

Where did that 20 miles go? It was just an observation about the difference in distance travelled between the 2 days.

Answer 2

The total distance the driver traveled in the two days is 260 miles

From the question, on the first day, the driver was going as a speed of 40 mph.

Let s be speed

∴ [tex]s_{1}= 40mph[/tex]

On the second day, he increased the speed to 60 mph

∴ [tex]s_{2}= 60mph[/tex]

From the statement- If he drove 2 more hours on the first day

Let time be t

Then

[tex]t_{1}= t_{2} + 2[/tex] hrs

and traveled 20 more miles

Let d be distance  

Then,

[tex]d_{1}= d_{2} + 20[/tex] miles

From the formula

Distance = Speed × Time

Then,

[tex]d = s \times t[/tex]

∴ [tex]d_{1} = s_{1} \times t_{1}[/tex]

From above,

[tex]d_{1}= d_{2}+20[/tex] miles

[tex]s_{1}= 40mph[/tex]

[tex]t_{1}= t_{2} + 2[/tex] hrs

Putting these into

[tex]d_{1} = s_{1} \times t_{1}[/tex]

[tex]d_{2} + 20 = 40\times (t_{2}+2)[/tex] ...... (1)

But,

[tex]Time = \frac{Distance}{Speed}[/tex]

∴ [tex]t_{2}= \frac{d_{2} }{s_{2} }[/tex]

From above, [tex]s_{2}= 60mph[/tex]

∴ [tex]t_{2}= \frac{d_{2} }{60}[/tex]

Put this into equation (1)

[tex]d_{2} + 20 = 40\times (t_{2}+2)[/tex]

[tex]d_{2} + 20 = 40\times (\frac{d_{2}}{60} +2)[/tex]

[tex]d_{2} + 20 = \frac{2}{3}d_{2} +80\\d_{2} = \frac{2}{3}d_{2} +80-20\\d_{2} = \frac{2}{3}d_{2} +60[/tex]

Multiply through by 3

[tex]3\times d_{2} = 3\times \frac{2}{3}d_{2} +3 \times 60\\3d_{2} = 2d_{2} + 120\\3d_{2} -2d_{2} = 120[/tex]

∴ [tex]d_{2} = 120[/tex] miles

∴The distance traveled on the second day is 120 miles

For the distance traveled on the first day,

Substitute [tex]d_{2}[/tex] into the equation

[tex]d_{1}= d_{2}+20[/tex] miles

∴ [tex]d_{1}= 120+20[/tex]

[tex]d_{1}= 140[/tex] miles

∴ The distance traveled on the first day is 140 miles

The total distance traveled in the two days = [tex]d_{1} + d_{2}[/tex]

The total distance traveled in the two days = 120 miles + 140 miles

The total distance traveled in the two days = 260 miles

Hence, the total distance the driver traveled in the two days is 260 miles

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Comment /hearthelp

[tex]\\ \sf\longmapsto -2z^2+4z+2z^2[/tex]

Combine like. variables

[tex]\\ \sf\longmapsto -2z^2+2z^2+4z[/tex]

[tex]\\ \sf\longmapsto (-2+2)z^2+4z[/tex]

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Step-by-step explanation:

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Answer:

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Answers

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Step-by-step explanation:

Hi there!

Please see the answer in the picture.

Hope it helps!

1. Approach

One is given a trigonometric equation with and one is asked to prove that it is true. Using the attached image, combined with the knowledge of trigonometry, one can evaluate each trigonometric function. Then one can simplify each ratio to solve. To yield the most accurate result, one has to each of the ratios in a fractional form, rather than simplifying it into a decimal form. Remember the right angle trigonometric ratios, these ratios describe the relationship between the sides and angles in a right triangle. Such ratios are as follows,

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2. Problem (9)

[tex]\frac{sin(60)+cos(30)}{1+sin(30)+cos(60)}=sin(60)[/tex]

As per the attached image, the following statements regarding the value of each ratio can be made:

[tex]sin(60)=\frac{\sqrt{3}}{2}\\\\cos(30)=\frac{\sqrt{3}}{2}\\\\sin(30)=\frac{1}{2}\\\\cos(60)=\frac{1}{2}[/tex]

Substitute,

[tex]\frac{sin(60)+cos(30)}{1+sin(30)+cos(60)}=sin(60)[/tex]

[tex]\frac{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}{1+\frac{1}{2}+\frac{1}{2}}=\frac{\sqrt{3}}{2}[/tex]

Simplify,

[tex]\frac{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}{1+\frac{1}{2}+\frac{1}{2}}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\frac{2\sqrt{3}}{2}}{1+\frac{1}{2}+\frac{1}{2}}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\frac{2\sqrt{3}}{2}}{1+1}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\sqrt{3}}{1+1}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}[/tex]

Thus, this equation is true.

2. Problem (10)

Use a similar strategy to evaluate this equation,

[tex]\frac{1-cos(30)}{sin(30)}=\frac{1-cot(60)}{1+cot(60)}[/tex]

Use the attached image to evaluate the ratios.

[tex]cos(30)=\frac{\sqrt{3}}{2}\\\\sin(30)=\frac{1}{2}\\\\cot(60)=\frac{1}{\sqrt{3}}[/tex]

Substitute,

[tex]\frac{1-cos(30)}{sin(30)}=\frac{1-cot(60)}{1+cot(60)}[/tex]

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[tex]\frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

[tex]\frac{\frac{2-\sqrt{3}}{2}}{\frac{1}{2}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

[tex]\frac{\frac{2-\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

[tex]\frac{\frac{2-\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}[/tex]

[tex]2-\sqrt{3}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}[/tex]

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}+1}}[/tex]

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}+1}[/tex]

Rationalize the denominator,

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}+1}[/tex]

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}+1}*\frac{\sqrt{3}-1}{\sqrt{3}-1}[/tex]

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

[tex]2-\sqrt{3}=\frac{3-2\sqrt{3}+1}{2}[/tex]

[tex]2-\sqrt{3}=\frac{4-2\sqrt{3}}{2}[/tex]

[tex]2-\sqrt{3}=2-\sqrt{3}[/tex]

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2x²+3x²+3.(-1) =5x.x+5x .1

Answers

[tex]\\ \sf\longmapsto 2x^2+3x^2+3(-1)=5x.x+5x.1[/tex]

[tex]\\ \sf\longmapsto 5x^2-3=5x^2+5x[/tex]

[tex]\\ \sf\longmapsto 5x^2-5x^2-3=5x[/tex]

[tex]\\ \sf\longmapsto 5x=-3[/tex]

[tex]\\ \sf\longmapsto x=\dfrac{-3}{5}[/tex]

8 ÷ -2 · 42 + 9 i need help please

Answers

I have to write 20 characters but is is 43
The answer is 43! Have a nice day!

Alex can cut a cord into 7 pieces in 36 seconds. How long will it take him to cut the cord into 12 pieces? (the answer is NOT 61 or 62.)

Answers

Answer:

x=61.71428 or 61 5/7

Step-by-step explanation:

We can use a ratio to solve

7 pieces              12 pieces

-----------------   = ---------------

36 seconds          x seconds

Using cross products

7x = 36*12

7x = 432

Divide by 7

7x/7 = 432/7

61 5/7

x=61.71428

Answer:

66

Step-by-step explanation:

Atlantic Hurricanes
Number
1925
1930
1935
1940
1945
1950
1955
1960
1965
1970
1975
Year
Between which years was
the biggest change in the
number of hurricanes?

Answers

Answer:

1950

Step-by-step explanation:

because 1950 column has the most highest number which is 11

Answer:

Between 1945 and 1950 ( from 5 to 11 ),

Step-by-step explanation:

You and six friends play on a basketball team. A sponsor paid $100 for the league fee, x dollars for each player’s T-shirt, and $68.25 for trophies. Write an expression for the total amount paid by the sponsor

Answers

Answer:

Total amount paid by the sponsor = 175 + 6d

Step-by-step explanation:

You and 5 friends = 6 people

Cost of renting a bus = $75

Team entry fee = $100

Cost of each student t shirts = $d

Cost of 6 student t shirts = $d × 6= $6d

Write an expression for the total amount the sponsor paid.

Total amount paid by the sponsor = Cost of renting a bus + Team entry fee + Cost of 6 student t shirts

= $75 + $100 + $6d

= $175 + $6d

Total amount paid by the sponsor = 175 + 6d

Where,

d = cost of each student t shirts

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Answers

Front: 12 square centimeters
1/2 x 6 x 4 = 12

Back: 12 square centimeters
1/2 x 6 x 4 = 12

Right: 15 square centimeters
3 x 5 = 15

Left: 15 square centimeters
3 x 5 = 15

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The surface area is the added areas of each face of a 3D figure.
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Area of a parallelogram:
A = bh

Area of a triangle:
A = 1/2bh
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