If John makes $20 an hour, how long will it take him to make $140?


120 hours


70 hours


7 hours


160 hours

That makes no sense, what DO you get?

The correct answer is, A, or question 1, or 2w + 3.

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One million days from now would be roughly the year 3759 AD. The exact day and month would fall sometime in March or April due to the complexities of our modern calendar system.

To answer your question of what is the day 1,000,000 days from now, we'll need to do some calculations. There are approximately 365.25 days in a year (this includes the extra day every four years for leap years).  

1,000,000 divided by 365.25 equals approximately 2737.85 years. This means that 1,000,000 days from now it would be the year 3759 AD (assuming the current year is 2022).

As for the exact day, we'll consider that a year is made up of 365 days, and the .25 accounts for leap years. However, the calculation of the exact date and month is quite complex due to the irregularities in our Gregorian calendar. It would fall sometime in March or April of 3759 AD.

Note that this calculation does not consider minute changes in Earth's rotation over time or potential changes in our calendar system.

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Problem-solving activity includes

1.Understanding the problem,that is nature of the problem, then Completely define in your own way.

2. Determining why this problem has accrued,

3.  Identifying the ways to solve the problem,

4. Prioritizing the given alternatives that is ways  and then arranging the alternatives for a solution,

5. Then applying the best solution or arrangement for the given problem.

There are two ways considered for  problem-solving activity

(1). Trial and Error  (2) Reduction in steps

It totally depends on the kind of problem , which you are solving. There may be Less than  five alternatives  ,equal to  five alternatives or more than  five alternatives to solve the problem.

Option B: False

The dimensions of the right-circular cylinder, which has a volume of 1 liter and uses the least amount of material, are such that the radius and twice the height are equal to the cube root of the volume divided by π, i.e., (0.001/π)^(1/3).

In Mathematics, given the volume of a right-circular cylinder (an open top can), we can find its optimal dimensions that would use the least amount of material. These dimensions correspond to the minimum surface area of the cylinder, which includes its closed bottom but not the open top.

Since the volume V of a right-circular cylinder is given by V=πr²h, where r is the radius and h is the height. We know that the volume equals 1 liter or 0.001 m³. Rearranging the volume equation for h gives us h=V/(πr²).

Next, the surface area A of the cylinder with closed bottom is A=2πrh+πr². Substituting h=V/(πr²) into the surface area gives A=2r(V/r)+(πr²) which simplifies to A=2V/r+πr². For the surface area to be minimum, the derivative of A with respect to r must be equal to zero. The first derivative of A is A'=-2V/r²+2πr. Set A'=0 we solve for r we find r=(V/(π))^(1/3). Substituting this value back into the equation for h gives us h=2*(V/(π))^(1/3).

Hence, for a right-circular cylinder of 1 liter volume (0.001 m³) to use the least amount of material, it should have a radius and twice the height equal to the cube root of the volume divided by π, i.e., (0.001/π)^(1/3).

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Answer: Experimental probability of tossing head is [tex]\frac{11}{20}[/tex]

Step-by-step explanation:

Since we have given that

Number of times Lynn tossed a coin = 60 times

Number of times head comes = 33

Experimental probability of tossing heads using Lynn and drawn results is given by

[tex]\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\\\=\frac{33}{60}\\\\=\frac{11}{20}[/tex]

Hence, Experimental probability of tossing head is [tex]\frac{11}{20}[/tex].

Answer:

experimental probability of tossing heads using Lynn and dawns results is [tex]\frac{11}{20}[/tex].

Step-by-step explanation:

Given :Lynn and dawn tossed a coin 60 times and got heads 33 times

To find : what is the experimental probability of tossing heads using Lynn and dawns results.

Solution : We have given that  

Number of times Lynn tossed a coin = 60 times.

Number of times head comes = 33.

 Probability of tossing heads using Lynn and drawn results is given by:

  = N[tex]\frac{number of favouable outcome }{total possible outcome}[/tex]

= [tex]\frac{33}{60}[/tex].

On simplification

=[tex]\frac{11}{20}[/tex].

Therefore, experimental probability of tossing heads using Lynn and dawns results is [tex]\frac{11}{20}[/tex].

Answer:

the answer is D. I, II, and III

Step-by-step explanation:

Answer:

160g

Step-by-step explanation:

Answer:

C. 17.0 feet

Step-by-step explanation:

We have been given that Jacob kicks a soccer ball off the ground and in the air with an initial velocity of 33 feet per second. We are asked to find the maximum height the soccer ball using formula [tex]H(t)=-16t^2+vt+s[/tex].

First of all, we will substitute [tex]v=33[/tex] in our given formula.

[tex]H(t)=-16t^2+33t+0[/tex]

Since our given parabola has a negative leading coefficient, so it will be downward opening parabola. The maximum height of the ball will be y-coordinate of the vertex of parabola.

Let us find x-coordinate of parabola as:

[tex]\frac{-b}{2a}=\frac{-33}{2\times -16}=\frac{-33}{-32}=\frac{33}{32}[/tex]

Now, we will substitute [tex]x=\frac{33}{32}[/tex] in our formula to find y-coordinate of vertex.

[tex]H(\frac{33}{32})=-16(\frac{33}{32})^2+33(\frac{33}{32})+0[/tex]

[tex]H(\frac{33}{32})=-16*\frac{1089}{1024}+\frac{1089}{32}[/tex]

[tex]H(\frac{33}{32})=-16*1.0634765625+34.03125[/tex]

[tex]H(\frac{33}{32})=-17.015625+34.03125[/tex]

[tex]H(\frac{33}{32})=17.015625[/tex]

[tex]H(\frac{33}{32})\approx 17.0[/tex]

Therefore, the ball reached the maximum height of 17.0 feet and option C is the correct choice.

Answer:

The fraction 14/5 can be expressed as 280 percent.

An expression when written in either radical form or rational exponent form be rewritten to fit the other form as well.

When we write in different forms the Denominator defines as the Index  and the Numerator defines as Power on the variable.

  We can write  [tex]4^{\frac{2}{3}[/tex]  as   [tex]\sqrt[3]{4^2}=\sqrt[3]{16}=\sqrt[3]{8*2}=2\sqrt[3]2}[/tex]

Again, vice versa,

   We can write [tex]\sqrt[5]{x^4}[/tex] as [tex]x^{\frac{4}{5}[/tex]

Therefore , we can written in other forms as well to fit .

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To convert from radical form to rational exponent form, use [tex]\( \sqrt[n]{a} = a^{1/n} \),[/tex] and vice versa for conversion.

An expression written in radical form can be rewritten in rational exponent form and vice versa using the following conversions:

1. From Radical Form to Rational Exponent Form:

  - For a radical expression [tex]\( \sqrt[n]{a} \), where \( n \)[/tex] is the index and [tex]\( a \)[/tex] is the radicand:

  - The equivalent expression in rational exponent form is [tex]\( a^{1/n} \)[/tex].

2. From Rational Exponent Form to Radical Form:

  - For an expression [tex]\( a^{m/n} \)[/tex], where [tex]\( a \)[/tex] is the base, [tex]\( m \)[/tex] is the numerator, and [tex]\( n \)[/tex] is the denominator:

  - The equivalent expression in radical form is [tex]\( \sqrt[n]{a^m} \).[/tex]

These conversions allow us to switch between radical form and rational exponent form easily. It's important to remember that the index of the radical corresponds to the denominator of the rational exponent, and the exponent of the base corresponds to the numerator of the rational exponent.

Answer:

The answer is 2 times.