For what values of r does the function y= e^rx satisfy the differential equation 2y"+y'-y=0?
...?

Answer:

Option(B) is correct.

The tenth term is given by [tex]a_{10}=15+9(-2)[/tex]

Step-by-step explanation:

Given : The sequence 15, 13, 11, 9,....

We have to determine the equation that can be used to find the tenth term of the given sequence 15, 13, 11, 9,....

Consider the given sequence 15, 13, 11, 9,..

[tex]a_1= 15, a_2=13, a_3= 11[/tex]

Difference between each term is

[tex]a_2-a_1=13-15=-2\\ a_3-a_2=11-13=-2[/tex]

Since, The difference between each term is constant -2

Thus, The given sequence is an arithmetic sequence with first term 15 and common difference -2.

Thus, The general term is given by

[tex]a_n=a+(n-1)d[/tex]

where a is first term and d is common difference nd n is number of term.

Subsitute, we have,

[tex]a_{10}=15+(10-1)(-2)[/tex]

Simplify, we have,

[tex]a_{10}=15+9(-2)[/tex]

Thus, The tenth term is given by [tex]a_{10}=15+9(-2)[/tex]

Answer:

the solution set is {0,-11}

Step-by-step explanation:

The equation is [tex]-p^2-11p=0[/tex]

The GCF of the left hand side of the expression is -p. Hence, factored out the GCF

[tex]-p(p+11)=0[/tex]

Now, apply the zero product property, we get

[tex]-p=0,p+11=0[/tex]

Now, solve the equations for p

[tex]p=0,p=-11[/tex]

Thus, the solution set is {0,-11}

Final answer:

To find the x-coordinates where the function f(x)=6sin(x)-sin^2(x) has a horizontal tangent line, we calculate the derivative, set it equal to zero, and solve for x, using trigonometric identities and potentially numerical methods. Thus, the solutions for \( x \) in the interval [tex]\( [0, 2\pi] \) are \( x = \frac{\pi}{2} + k\pi \), where \( k \)[/tex] is an integer. These solutions represent the points where [tex]\( \cos(x) = 0 \)[/tex], causing the equation to be satisfied.

Explanation:

The question asks to find the x-coordinates of all points on the graph of the function f(x)=6sin(x)-sin2(x) at which the tangent line is horizontal. To find these coordinates, we need to look for points where the derivative of the function is equal to zero since a horizontal tangent line has a slope of zero.

We start by finding the derivative of the function:

f'(x) = d/dx[6sin(x)-sin2(x)]

f'(x) = 6cos(x)-2sin(x)cos(x)

f'(x) = 6cos(x)-sin(2x) (using the double angle identity sin(2x)=2sin(x)cos(x))

To find when the derivative is zero, we set the derivative equal to zero:

0 = 6cos(x) - sin(2x)

We then need to solve this equation for x. This involves using trigonometric identities and potentially numerical methods if the equation cannot be solved analytically. The x-coordinates will be the solutions to this equation within the domain for which we are interested, typically 0 to 2π for one full cycle of a trigonometric function.

To solve the equation [tex]\( 0 = 6\cos(x) - \sin(2x) \)[/tex], we'll first try to rewrite it in terms of a single trigonometric function using known identities.

The double angle identity for sine states that [tex]\( \sin(2x) = 2\sin(x)\cos(x) \)[/tex]. Substituting this identity into the equation, we get:

[tex]\[ 0 = 6\cos(x) - 2\sin(x)\cos(x) \]\\Now, let's factor out \( \cos(x) \):\[ 0 = \cos(x)(6 - 2\sin(x)) \][/tex]

Now, for the equation to be true, either [tex]\( \cos(x) = 0 \) or \( 6 - 2\sin(x) = 0 \)[/tex].

1. If [tex]\( \cos(x) = 0 \), then \( x = \frac{\pi}{2} + k\pi \)[/tex], where \( k \) is an integer.

2. If [tex]\( 6 - 2\sin(x) = 0 \), then \( \sin(x) = 3 \)[/tex]. However, the sine function has a range of \([-1, 1]\), so there are no solutions for \( x \) in this case.

Thus, the solutions for \( x \) in the interval [tex]\( [0, 2\pi] \) are \( x = \frac{\pi}{2} + k\pi \), where \( k \)[/tex] is an integer. These solutions represent the points where [tex]\( \cos(x) = 0 \)[/tex], causing the equation to be satisfied.

∠Z ≅ ∠X and ∠W ≅ ∠Y is the correct answer

D

Answer: Segment, line, ray

Step-by-step explanation: too answer was confusing and point was wrong.

The answers are a. ray, e. segment and f. line

In geometry, a one-dimensional figure has only length and no other dimensions such as width or height. Let's analyze each option:

Thus, the one-dimensional figures are the ray, segment, and line.

The answer is A, B, and  E.

Just plot write the equation in a graphing calculator which shows they are the same.

The first five non-zero multiples of 5 are 5, 10, 15, 20, and 25. These are obtained by multiplying the number 5 by the integers 1 to 5. All non-zero multiples of 5 will end with the digit 5 or 0, confirming their divisibility by 5.

Identifying Non-Zero Multiples of 5

When you're looking for non-zero multiples of 5, it's important to remember that the basic property of a multiple of 5 is that it ends in either 0 or 5. However, since we're focusing on non-zero multiples, we will exclude any multiples that end in zero. A multiple of a number is obtained by multiplying that number by an integer. For the number 5, the first five non-zero multiples are 5, 10, 15, 20, and 25. Notice that these numbers are the result of multiplying 5 by the integers 1 through 5, respectively.

Understanding that the concept of multiples relates to a number being multiplied by a sequence of whole numbers or integers gives us the ability to identify patterns within these numbers. For example, all non-zero multiples of 5 will end with the digit 5 or 0, reinforcing the idea that if a number ends in 5, it is divisible by 5 without remainder. This concept is supported by empirical observation and the understanding that patterns and numbers, such as these, can offer insight into broader mathematical ideas. Furthermore, the significance of the number 5 in various mathematical circumstances, such as the number of digits in a number or the properties of multiples, is apparent throughout different contexts.

John paid the most towards the rental.

To determine who paid the most towards the rental, we need to find out how much each person paid. Mark paid $3 and John paid <percentage> of the total. Their friend paid the remaining amount. Since the total cost to rent the video games was $10, we can calculate how much John paid by subtracting what Mark paid from $10 and finding the percentage of that remaining amount.

Subtracting $3 from $10 gives us $7. Since John paid the majority, we can say that he paid 70% of the total. Therefore, John paid the most towards the rental.

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