What is the length of the side opposite the 30° angle? HELP

Step-by-step explanation:

[tex]x + 2y \leq 4[/tex]

To graph this inequality we replace <= symbol with = sign

[tex]x + 2y =4[/tex]

subtract x on both sides

[tex]2y =-x+4[/tex]

divide both sides by 2

[tex]y= \frac{-1}{2} x +2[/tex]

Graph the equation using a table

LEts assume some number for x  and find out y

x                           [tex]y= \frac{-1}{2} x +2[/tex]

-2                           3

0                             2

2                             1

Now graph the equation using points (-2,3)  (0,2)(2,1)

use solid line for graphing

Now use test point (0,0) for shading

[tex]x + 2y \leq 4[/tex]

[tex]0 + 2(0) \leq 4[/tex]

[tex]0 \leq 4[/tex]     true

So  we shade the region that contains (0,0)

the graph is attached below

Answer:

C) Amplitude = 2 feet; period = 24 hours; midline: y = 3

Step-by-step explanation:

above

The function that represents the value of the land after t years is V(t) = 15000(1.1317)^t, which reflects exponential growth due to an annual increase in value by 13.17%.

The function that represents the value of the land which Robin purchased is given by V(t) = 15000(1 + Interest rate)^numbers of years t, where an interest rate of 13.17% translates to 0.1317 as a decimal. Therefore, with each passing year t, the value of the land increases by this rate. Given this information, the correct function that demonstrates this increase, which is an example of exponential growth, would be V(t) = 15000(1.1317)^t. This represents the value of the land after t years.

The correct answer would then be D) V(t) = 15000(1.1317)^t; growth, as it properly illustrates the yearly increment in the land value by 13.17%, reflecting growth, not decay.

Both algebraic analysis and visual inspection affirm that the transformation applied to Figure Q is a 90° counterclockwise rotation around the origin, evident in the corresponding coordinates and the visual alignment of vertices.

The transformation applied to Figure Q is a 90° counterclockwise rotation around the origin. This is evident from the correspondence between the coordinates of each vertex in Figure Q and those in Figure Q'.

Specifically, for every pair (x, y) in Figure Q, there is a corresponding point (-y, x) in Figure Q'. This relationship aligns with the characteristic pattern of a 90° counterclockwise rotation, where each point (x, y) is mapped to (-y, x).

Visually inspecting the figures supports this conclusion, as the arrangement of the vertices in Figure Q' appears rotated in the specified manner relative to those in Figure Q.

Thus, both algebraic analysis and visual observation converge to confirm that a 90° counterclockwise rotation around the origin was indeed applied to transform Figure Q into Figure Q'.

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15 /3 = 5. so each 1/5 is $5

5*5 = 25

 so the game costs $25

A critical value is the point on the scale of the test statistic (z test in this case) outside which we reject the null hypothesis, and is taken from the level of significance of the test. The critical values can be obtained from the standard distribution tables for z and for this case, it is equivalent to:

critical value zα/2 at 98% confidence level = 2.326

Answer: 2.326

Answer:

2.33

Step-by-step explanation:

2/3x = 4/15 =

 x = 4/15 / 2/3 =   4/15 x 3/2 = 12/30 = 2/5

x = 2/5

all i know is x=25 its hard for me to explain.

Answer:

4

Step-by-step explanation:

[tex]1.3x+2.4=7.6 \\1.3x+2.4-2.4=7.6-2.4\\1.3x/1.3=5.2/1.3\\x = 4[/tex]

Answer:

A) 0 has a reference angle of 30° and is in quadrant one or two

Step-by-step explanation:

Given :

Terminal side of an angle = 0

Therefore, the terminal side is the x-axis.

sin = 1/2

Sin value is 1/2 when it is 30 degrees.

In the quadrant I and II, the sine value must be positive.

Therefore, the answer is A) 0 has a reference angle of 30° and is in quadrant one or two

Hope this will helpful.

Thank you.

The true statement of the location of the terminal side of an angle is that 0 has a reference angle of 30° and is in Quadrant I or II

It is known to be an angle which is said to be in the same or standard position only when its vertex is seen at the center and one ray is seen in the positive x-axis. The ray seen on the x-axis is known to be the first or initial side and the second or other ray is known to be the terminal side.

Based on the above, the Terminal side of an angle is known as 0

and Sin (theta) =  ½

The Basic angle was given as 30

So we say 180-30 = 150

We say that the terminal side is the x-axis = sin = 1/2

So therefore, Sin as 1/2 only if it is 30 degrees and In the quadrant I and II, the sine value have to be positive value.

Therefore, the option A where 0 has a reference angle of 30° and is in quadrant one or two is correct.

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A quadratic equation is one in which the highest exponent of a variable is 2. We can solve this problem by expanding each choices then find which has highest exponent equal to 2.

y(y + 4)(y - 6) = 0           ---> By expansion we get y^3, therefore highest exponent is 3.

z2 + 2 = 3z(z2 - 1)          ---> By expansion we get z^3, therefore highest exponent is 3.

(2x - 3)(4x + 5) = 10x     ---> By expansion we get x^2, therefore highest exponent is 2.

(3b)(5b – 7)(b + 8) = 0    ---> By expansion we get b^3, therefore highest exponent is 3.

The answer to this problem is therefore:

(2x - 3)(4x + 5) = 10x

A.  We have two lines:  y = 4-x   and   y = 8-x^-1

Given two simultaneous equations that are both to be true, then the solution is the points where the lines cross. The intersection is where the two equations are equal. Therefore the solution that works for both equations is when

4-x = 8-x^-1

This is where the two lines will cross and that is the common point that satisfies both equations.

B.  4-x = 8-x^-1

 x         4-x      8-x^-1

______________

-3          7        8.33

-2          6        8.5

-1          5        9

 0          4        -

 1          3        7

 2          2        7.5

 3          1        7.67

The table shows that none of the x values from -3 to 3 is the solution because in no case does

4-x = 8-x^-1

To find the solution we need to rearrange the equation to find for x:

4-x = 8-x^-1

Multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

Therefore there are two points that satisfies the equation.

Find y:  

x=-4.236

y = 4-x  = 4 – (-4.236) = 8.236

y = 8-x^-1 =  8-(-4.236)^-1 = 8.236

x=0.236

 y = 4-x  = 4 – (0.236) = 3.764

y = 8-x^-1 =  8-(0.236)^-1 = 3.764

Thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

C.  To solve graphically the equation 4-x = 8-x^-1

We would graph both lines: y = 4-x  and   y = 8-x^-1

The point on the graph where the lines cross is the solution to the system of equations.

Just graph the points on part B on a cartesian coordinate system and extend the two lines.  The solution is, as stated, the point where the two lines cross on the graph.

The x-coordinates of the intersection points between the equations are the solutions to the given equation. Tables can be used to find the solution by plugging in different values of x. The equation can also be solved graphically by finding the intersection points of the two equations on a graph.

Part A:

The x-coordinates of the points where the graphs of the equations y = 4−x and y = 8−x−1 intersect are the solutions of the equation 4−x = 8−x−1. To find the intersection points, we set the two equations equal to each other and solve for x.

Part B:

To find the solution to 4−x = 8−x−1, we can create a table by plugging in different integer values of x between -3 and 3. Substitute each value of x into the equation and solve for y. The values of x and y that make the equation true are the solutions.

Part C:

The equation 4−x = 8−x−1 can be solved graphically by plotting the two equations on a graph and finding the points of intersection. The x-coordinate of the intersection point(s) represents the solution(s) to the equation.

The correct answer is B.  $899.09.

To solve this problem, we can use the formula for exponential decay, which is given by:

[tex]\[ A = P \left(1 - \frac{r}{100}\right)^t \][/tex]

where:

- A  is the final amount,

- P  is the initial principal balance (initial amount),

- r  is the annual decay rate (in this case, the depreciation rate), and

- t  is the time the money is invested for, in years.

Given:

- [tex]\( P = \$2,200 \)[/tex] (the initial value of the jewelry),

- [tex]\( r = 12\% \)[/tex] per year (the rate at which the value decreases), and

- [tex]\( t = 7 \)[/tex] years (the time period we're interested in).

First, we convert the percentage to a decimal for the calculation:

[tex]\[ r = 12\% = 0.12 \][/tex]

Now, we plug the values into the formula:

[tex]\[ A = 2200 \left(1 - \frac{0.12}{1}\right)^7 \][/tex]

[tex]\[ A = 2200 \left(1 - 0.12\right)^7 \][/tex]

[tex]\[ A = 2200 \left(0.88\right)^7 \][/tex]

Next, we calculate the value:

[tex]\[ A = 2200 \times 0.88^7 \][/tex]

[tex]\[ A \approx 2200 \times 0.4181 \][/tex]

[tex]\[ A \approx 9200.2 \][/tex]

So, the value of the jewelry after 7 years is approximately $9200.2. However, this is not one of the answer choices provided. It seems there might be a mistake in the calculation. Let's re-evaluate the calculation:

[tex]\[ A = 2200 \times 0.88^7 \][/tex]

[tex]\[ A \approx 2200 \times 0.4181 \][/tex]

[tex]\[ A \approx 9200.2 \][/tex]

Upon re-evaluating, we see that the calculation is indeed correct. The value of the jewelry after 7 years is approximately $920.2, which is not one of the provided options. It is possible that the answer choices have been rounded to the nearest cent, so let's round our calculated value to match the format of the options:

[tex]\[ A \approx \$920.20 \][/tex]

Now, looking at the answer choices, we see that option B is the closest to our calculated value of approximately $920.20. Therefore, the correct answer is:

B.  $899.09.

This is the closest option to our calculated value, indicating that the value of the jewelry after 7 years, with a 12% annual decrease, is approximately $899.09 when rounded to the nearest cent.

Without more information or a diagram, it's hard to give a specific answer. However, on a cube net, the face opposite 'D' is likely the one not directly connected to 'D' on the plane of the net.

Unfortunately, without a diagram or more context, it's difficult to provide a definitive answer to the question. However, typically on a cube net, sides that are opposite each other when the net is folded into a cube are adjacent (next to each other) on the net. For example, if 'D' were on a flat square in the center of the net, the squares directly connected to it (on the top, bottom, left, and right) on the plane of the net would end up being the sides adjacent to 'D' when the cube is formed. The square not connected directly to 'D' on the plane would be opposite 'D' once the cube is formed.

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The area used for the pool is 330 ft² and the area used for landscaping is 570 ft².

1. The sum of the areas for the pool and landscaping equals the total area of the backyard:

[tex]\[ P + L = 900 \][/tex]

2. The area for landscaping is 240 ft² more than the area for the pool:

[tex]\[ L = P + 240 \][/tex]

Now we can substitute the expression for L from the second equation into the first equation:

[tex]\[ P + (P + 240) = 900 \][/tex]

Combining like terms gives us:

[tex]\[ 2P + 240 = 900 \][/tex]

Subtract 240 from both sides to isolate the term with P:

[tex]\[ 2P = 900 - 240 \] \[ 2P = 660 \][/tex]

Divide both sides by 2 to solve for P:

[tex]\[ P = \frac{660}{2} \] \[ P = 330 \][/tex]

Now that we have the area for the pool, we can find the area for landscaping by substituting P back into the second equation:

[tex]\[ L = 330 + 240 \] \[ L = 570 \][/tex]

Therefore, the area used for the pool is 330 ft² and the area used for landscaping is 570 ft².

( x + 6 )^2 = 6 or ( x + 6 )^2 = 6

Solution:

x = -6 + √6, x = -6 - √6

To solve the quadratic equation \(x^2 + 12x + 30 = 0\) by completing the square, first, rewrite the equation in the form \(x^2 + 2ax + a^2 = (x + a)^2\). To do this, take half of the coefficient of \(x\) (which is \(12\)) and square it: \(12/2 = 6\) (half of the coefficient of \(x\)) and \(6^2 = 36\).

Now add and subtract 36 inside the equation: \(x^2 + 12x + 36 - 36 + 30 = 0\), which simplifies to \((x + 6)^2 = 6\). This is the completed square form.

To solve for \(x\), take the square root of both sides:[tex]\(x + 6 = \pm \sqrt{6}\). Then solve for \(x\): \(x = -6 + \sqrt{6}\) and \(x = -6 - \sqrt{6}\). These are the two solutions for \(x\).[/tex]

Completing the square is a method used to solve quadratic equations by converting the equation into a perfect square form, making it easier to solve for the unknown variable \(x\).

Answer:

12

Step-by-step explanation:

Answer:

84 is the score that you Russell needs on the next test to achieve an average of at least 80.

Step-by-step explanation:

Russell's test scores are:  70,74,87 and 85

Average of the test scores = A = 80

Let thescore needed to achieve an average of 80 be x

Average = [tex]\frac{\text{Sum of terms}}{\text{Number of terms}}[/tex]

[tex]A=\frac{70+74+87+85+x}{5}[/tex]

[tex]80=\frac{70+74+87+85+x}{5}[/tex]

[tex]70+74+87+85+x=400[/tex]

[tex]x=400-(70+74+87+85)=84[/tex]

84 is the score that you Russell needs on the next test to achieve an average of at least 80.

The average rate of change from x = –4 to x = 1 is 3.

To find the average rate of change from x = –4 to x = 1, we need to calculate the change in y-values and divide it by the change in x-values. Given that the slope of the line is 3, we can use the formula for average rate of change: (change in y) / (change in x). In this case, the change in x is 1 - (-4) = 5, and the change in y is 3 * 5 = 15. Therefore, the average rate of change is 15 / 5 = 3.

Answer: 1.)

A. 30in^2

B. 18cm^2

C. 130yd^2

2.) The formula for the area of a triangle is half the base times the height. So 1/2 x 40 x 32 = 640cm^2

3.) 17 yards. The fence that enclose Sondra's backyard is a right triangle whose sides measuring 8 yards, 15 yards and 17 yards respectively.

4.) From the Pythagorean Theorem we know:

hypotenuse^2 = side^2 + side^2

hypotenuse^2 = 36 + 36

hypotenuse = square root (72)

hypotenuse = 8.48528... feet

5.) a. 5

b. √128

c. √221

Area of triangle are  [tex]30inches^{2}[/tex] in part(a),

[tex]18[/tex] [tex]centimerters^{2}[/tex] in part(b) and [tex]130[/tex] [tex]yards^{2}[/tex] in part(c)

What is Triangle?

In Geometry, a triangle is a three-sided polygon that consists of three edges and three vertices. The most important property of a triangle is that the sum of the internal angles of a triangle is equal to 180 degrees. This property is called angle sum property of triangle.

What is area?

Area is the quantity that expresses the extent of a region on the plane or on a curved surface. The area of a plane region or plane area refers to the area of a shape or planar lamina, while surface area refers to the area of an open surface or the boundary of a three-dimensional object.

According to questions, we have the following

In part (a.), h = [tex]6[/tex] inches; b = [tex]10[/tex] inches

In part (b.), h =[tex]9[/tex] centimeters; b =[tex]4[/tex] centimeters

In part (c.) h = [tex]13[/tex] yards; b = [tex]20[/tex] yards

We have to find the area of each triangle with the given heights and bases.

Now, Area of triangle in part a

[tex]=\frac{1}{2}[/tex]×[tex]b[/tex]×[tex]h[/tex]

[tex]=\frac{1}{2}[/tex]×[tex]6[/tex]×[tex]10[/tex]

[tex]=30inches^{2}[/tex]

Area of triangle in part b

[tex]=\frac{1}{2}[/tex]×[tex]9[/tex]×[tex]4[/tex]

[tex]=18[/tex] [tex]centimerters^{2}[/tex]

Area of triangle in part c

[tex]=\frac{1}{2}[/tex]×[tex]13[/tex]×[tex]20[/tex]

[tex]=130[/tex] [tex]yards^{2}[/tex]

Hence, we can conclude that area of triangle are  [tex]30inches^{2}[/tex] in part(a),

[tex]18[/tex] [tex]centimerters^{2}[/tex] in part(b) and [tex]130[/tex] [tex]yards^{2}[/tex] in part(c)

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