Pressure (in pounds per square foot) on a diver at a depth of "d" feet: "P" = 64d + 2112
Solve for "d"

A player will need 2,264 points on the hardest setting of level 4 as calculated by evaluating g(4) using the function g(x) = 7(3)^x - 1 and then applying the function h(x) = 4x.

To calculate the number of points a player will need on the hardest setting of level 4, we need to consider the function h(x) = 4x provided in the context of the question. However, the question also mentions the function g(x) = 7(3)^x - 1, which indicates the points needed to reach the next level normally. Since we need to find the hardest setting value, we will use the function h(x) but before that, we need to evaluate g(4) to find out the base points at level 4.

To find g(4), we substitute x = 4 into g(x):

g(4) = 7(3)^4 - 1 = 7(81) - 1 = 567 - 1 = 566

Now we apply the function h(x) to this value to get the points on the hardest setting:

h(g(4)) = 4 × 566 = 2264

Therefore, a player will need 2,264 points on the hardest setting of level 4.

Answer: y=(yx−16x)x+16

Step-by-step explanation:

To eliminate the parameter t in the equations x=t^2 and y=2+10t, solve [tex]t = \sqrt(x)[/tex] and substitute into the second equation to get [tex]y = 2 + 10\sqrt(x)[/tex]. This results in the cartesian equation [tex]y = 2 + 10\sqrt(x).[/tex]

To eliminate the parameter t and find the cartesian equation, follow these steps:

Start with the given parametric equations: x=t^2 and y=2+10t

Solve the first equation for t:

[tex]t = \sqrt(x)[/tex]

Substitute this expression for t into the second equation:

[tex]y = 2 + 10(\sqrt(x))[/tex]

Thus, the cartesian equation is [tex]y = 2 + 10 \sqrt(x).[/tex]

All the angles of the quadrilateral are,

⇒ ∠A = Right angle

⇒ ∠B = Obtuse angle

⇒ ∠C = Acute angle

⇒ ∠D = Right angle

An angle is a combination of two rays (half-lines) with a common endpoint. The latter is known as the vertex of the angle and the rays as the sides, sometimes as the legs and sometimes the arms of the angle.

Now, We know that;

Any angle which is greater than 90° is called obtuse angle and less than 90° is called acute angle and 90° is called Right angle.

Here, The four angles of the quadrilateral are,

⇒ ∠A = 90°

⇒ ∠B = 105°

⇒ ∠C = 75°

⇒ ∠D = 90°

Hence, We get;

⇒ ∠A = Right angle

⇒ ∠B = Obtuse angle

⇒ ∠C = Acute angle

⇒ ∠D = Right angle

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The percentage of students who passed the class with a C or better is 82.1%. This is found by adding the number of students who earned A, B, or C grades, dividing by the total number of students, and multiplying by 100.

The subject is Mathematics, specifically a problem in percentages. First, we need to find the total number of students in the class. We add the number of students that earned each grade: 5 (A) + 8 (B) + 10 (C) + 3 (D) + 2 (F) = 28 students in total.  

Next, we need to identify the number of students who passed the class with a C or better. This includes the students who earned an A, B, or C. Adding these numbers together gets us 5 (A) + 8 (B) + 10 (C) = 23 students.  

Then, to find the percentage of students who passed the class, we divide the number of students who passed (23) by the total number of students (28) and then multiply by 100 to convert the result to a percentage. So the calculation is as follows: (23/28)*100 = 82.14. Rounded to one decimal place, that is 82.1%. Therefore, 82.1% of students passed the class with a grade of C or better.

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The percent of students who passed the course with a C or better is: 82.14%

To find the percent of students who passed the course with a C or better, we need to divide the number of students who passed by the total number of students and multiply by 100%.

The number of students who passed is 5 + 8 + 10 = 23.

The total number of students is 5 + 8 + 10 + 3 + 2 = 28.

Therefore, the percent of students who passed the course with a C or better is:

(23 / 28) * 100% = 82.14%

Rounding to one decimal place, the answer is 82.1%.

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Proof by contradiction.

Let assume that when [tex]x[/tex] is an irrational number, then [tex]\dfrac{1}{x}[/tex] is a rational number.

If [tex]\dfrac{1}{x}[/tex] is a rational number, then it can be expressed as a fraction [tex]\dfrac{a}{b}[/tex] where [tex]a,b\in\mathbb{Z}[/tex].

[tex]\dfrac{1}{x}=\dfrac{a}{b} \Rightarrow x=\dfrac{b}{a}[/tex]

Since [tex]a,b\in\mathbb{Z}[/tex], the number [tex]x=\dfrac{b}{a}[/tex] is also a rational number. But this contardicts our initial assumption that [tex]x[/tex] is an irrational number. Therefore [tex]\dfrac{1}{x}[/tex] must be an irrational number.

Final answer:

The tens digit of a perfect square with a units digit of 6 could be any digit from 0 to 9. This is because when numbers ending in either 4 or 6 are squared, the tens digit can take any value in the range of 0-9 due to the cyclic nature of the squares.

Explanation:

The units digit of a perfect square can be 6, and this occurs when a number ending in either 4 or 6 is squared since 42 is 16 and 62 is 36. Considering this, the possible values for the tens digit of a perfect square with a units digit of 6 could be derived from these squares. For the tens digit, we must look at the squares of numbers ending in 4 or 6 and check the ten's place of the resulting product.

For example, if we consider 142 (which is 196) and 162 (which is 256), the tens digit can be either 9 or 5. Continuing with this pattern for numbers ending in 4 or 6 all the way to 942 and 962, we find that the tens digit is cyclic and can be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. Therefore, all ten digits are possible for the tens place of a number whose square ends in 6.

By solving the given equations using the provided information, and assuming the constant c = 4/3, we can find the coordinates of the point P on the curve corresponding to t = 2 are (8/3, 32/9). There is a probable discrepancy in the question which causes two different solutions for c, therefore making the canonical form of the parabola ambiguous.

The given parametric curve has its equations as x = ct and y = c/8 * t^2. We're given that it's the arc of the parabola 8y = x^2 from (0, 0) to (4, 2), meaning that we can set x = ct = 4 and y = c/8 * t^2 = 2 when t=3 (the upper limit of t) to find the value of the constant c. Solving these equations gives c=4/3 and c=48. Since c must be the same in both equations, there seems to be a discrepancy here, which could possibly originate from a mistake in the question. However, if we only use c from the first equation (x = ct), we get c = 4/3.

The question asks for the coordinates of the point on the curve corresponding to t = 2. Substituting t = 2 into x = ct and y = c/8 * t^2 gives x = 8/3 and y = 32/9, assuming c = 4/3. So, the coordinates of the point would be (8/3, 32/9).

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Answer:

A quadratic equation is in the form of [tex]ax^2+bx+c =0[/tex]........[1], then the solution for this equation is given by:

[tex]x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]

As per the statement:

The equation is given by:

[tex]h=-16t^2+25t+8[/tex]

where, h is the height in feet t seconds after it is thrown.

After the ball passes its maximum height, it comes down and then goes into the hoop at a height of 10 feet above the ground.

⇒h = 10 feet

then;

[tex]-16t^2+25t+8=10[/tex]

Subtract 10 from both sides we have;

[tex]16t^2-25t+2=0[/tex]

On comparing this equation with [1] we have;

a =16 , b =-25 and c =2

then;

[tex]t= \frac{25 \pm \sqrt{(-25)^2-4(16)(2)}}{2(16)}[/tex]

⇒[tex]t= \frac{25 \pm \sqrt{625-128}}{32}[/tex]

⇒[tex]t= \frac{25 \pm \sqrt{497}}{32}[/tex]

as we want the time when it was falling so ,

[tex]t= \frac{25 + \sqrt{497}}{32}[/tex]

Simplify:

[tex]t \approx 1.48[/tex] sec

Therefore, 1.48 sec long after it was thrown does it go into the hoop