Find all solutions to the equation in the interval [0, 2π). cos 2x - cos x = 0

Answer:

512/2500

Step-by-step explanation:

Answer:

H = .2L; H = .4

Step-by-step explanation:

Answer:

$180,997.50.

Step-by-step explanation:

1. On TABLE 14-1 Future Value of $1.00 Ordinary Annuity, select the periods row corresponding to the number of interest periods.

2. Select the rate-per-period column corresponding to the period interest rate.

3. Locate the value in the cell where the periods row intersects the rate-per-period column.

4. Multiply the annuity payment by the table value from step 3.

Future value = annuity payment × table value

FV = $7500.00 * 24.133 = $180,997.50

gpe = 1/2 of the unit measures

 since 9 is a whole number the gpe would be 1/2 g

The equivalent expression to the given expression is [tex]8x^12y^3[/tex].

We have given that,

The  expression (2x^4y)^3

We have to determine the,

Which expression is equivalent to (2x^4y)^3

Equivalent expressions are expressions that work the same even though they look different. If two algebraic expressions are equivalent, then the two expressions have the same value when we plug in the same value(s) for the variable(s).

The answer is D.

2^3 is 8.

(x^4)^3 is x^12 and y^3 is simply y^3.

Put them together to get the final answer.

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As per quadratic equation, all values of 't' for which the ball's height is 15 feet are 33.25 and (- 31.813).

"Quadratic equations are the polynomial equations of degree 2 in one variable of type f(x) = ax² + bx + c = 0 where a, b, c, ∈ R and a ≠ 0. It is the general form of a quadratic equation where 'a' is called the leading coefficient and 'c' is called the absolute term of f (x)."

Given, [tex]h = 15[/tex] feet.

Therefore, the quadratic equation for [tex]h = 15[/tex] will be:

A ball is thrown from an initial height of 7 feet with an initial upward velocity of 23 feet per second.

The ball's height h (in feet) after t seconds is:

[tex]h = 7+23t-16t^{2}[/tex]

[tex]15 = 7+23t-16t^{2}[/tex]

⇒ [tex]15-7-23t+16t^{2} = 0[/tex]

⇒ [tex]16t^{2} -23t - 8 = 0[/tex]

⇒ [tex]t = [- (-23)[/tex] ± [tex]\sqrt{(23)^{2} - 4(16)(- 8)}[/tex] ]/(2 × 16)

⇒ [tex]t =[/tex] [23 ± 1041]/32

⇒ [tex]t =[/tex] [23 + 1041]/32, [23 - 1041]/32

⇒ [tex]t =[/tex] 33.25, - 31.813

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Mike will use 1 1/2 pounds of pork for his meatloaf recipe, which calls for a 3 to 1 ratio of beef to pork and a total of 6 pounds of meat.

Mike's meatloaf recipe has a total of 6 pounds of meat and uses a 3 to 1 ratio of beef to pork. To find out how much pork he will use, we can express the total amount of meat as the sum of beef and pork parts.

First, we know that there are 3 + 1 = 4 parts in total because of the given ratio. Since the ratio is 3 to 1, for every 4 parts of meat, 1 part is pork. Therefore, we calculate the weight of each part by dividing the total weight by the number of parts:

6 pounds ÷ 4 parts = 1.5 pounds per part

Since one part is pork, Mike will use 1.5 pounds of pork for his meatloaf. Expressed as a mixed number, that's 1 1/2 pounds of pork in simplest terms.

In the given diagram, we are given two triangles, triangle ABC and triangle DEF.

And we are given a line or reflection l.

Triangle DEF is the mirror image of triangle ABC.

Therefore, all sides of the triangle ABC would be congruent to sides of triangle DEF.

AB = DF

AC = DE and

BC = EF.

We can see than BC is the smallest side of triangle ABC and EF is the smallest side of triangle DEF.

Rounded to the nearest hundredth, they are:

[tex]\[x_1 \approx -0.63 + 1.35i\][/tex]

[tex]\[x_2 \approx -0.63 - 1.35i\][/tex]

To solve the quadratic equation [tex]\(2x^2 + 5x + 5 = 0\)[/tex], we can use the quadratic formula:

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

where [tex]\(a = 2\), \(b = 5\)[/tex], and [tex]\(c = 5\)[/tex].

Substituting the values into the quadratic formula:

[tex]\[x = \frac{{-5 \pm \sqrt{{5^2 - 4 \cdot 2 \cdot 5}}}}{{2 \cdot 2}}\][/tex]

[tex]\[x = \frac{{-5 \pm \sqrt{{25 - 40}}}}{{4}}\][/tex]

[tex]\[x = \frac{{-5 \pm \sqrt{{-15}}}}{{4}}\][/tex]

Since the discriminant [tex](\(b^2 - 4ac\))[/tex] is negative, the solutions will involve imaginary numbers.

Using the imaginary unit [tex]\(i\)[/tex], where [tex]\(i^2 = -1\),[/tex] we can rewrite [tex]\(\sqrt{{-15}}\)[/tex] as [tex]\(i\sqrt{{15}}\):[/tex]

[tex]\[x = \frac{{-5 \pm i\sqrt{{15}}}}{{4}}\][/tex]

So, the solutions to the equation [tex]\(2x^2 + 5x + 5 = 0\)[/tex] are complex numbers. Rounded to the nearest hundredth, they are:

[tex]\[x_1 \approx -0.63 + 1.35i\][/tex]

[tex]\[x_2 \approx -0.63 - 1.35i\][/tex]

These solutions represent the points where the graph of the quadratic equation intersects the x-axis. They lie on the complex plane.

x+y=12

x-y=4

x=y-4

y-4+y=12

2y-4=12

2y=16

y=8

x=8-4=8

8+4=12

 the 2 numbers are 8 & 4

Answer:

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

Step-by-step explanation:

Given: The area of the parallelogram is [tex]6x^2+x+3[/tex] and the height is 3x.

To find: The base of the given parallelogram.

Solution: It is given that The area of the parallelogram is [tex]6x^2+x+3[/tex] and the height is 3x.

Now, area of parallelogram is given as:

[tex]A=b{\times}h[/tex] where b is the base and h is the height of teh gievn parallelogram.

Substituting the given values, we have

[tex]6x^2+x+3=b{\times}3x[/tex]

⇒[tex]\frac{6x^2+x+3}{3x}=Base[/tex]

⇒[tex]\frac{6x^2}{3x}+\frac{x}{3x}+\frac{3}{3x}=Base[/tex]

⇒[tex]2x+\frac{1}{3}+\frac{1}{x}=Base[/tex]

which is the required expression for the base of the given parallelogram.

area = pi x r^2

615.75 = 3.14 x r^2

r^2 = 615.75/3.14 =196.0987 round to 196.1

r = sqrt(196.1) = 14.00357 round to 14

radius = 14 kilometers