A line segment with endpoints (-4,5) and (-2,2) is reflected about the y-axis. Write the coordinates of the endpoints after the segment has been reflected.

Answer:

x = ± 3 and ± 2

Step-by-step explanation:

The given quadratic equation is

[tex](x^{2}-1)^{2}-11(x^{2}-1)+24=0[/tex]

To make this question easier we will consume ( x² -1 ) = a

a² - 11a + 24 = 0

Now we can factorize this equation easily

a²- 8a - 3a + 24 = 0

a (a-8) - 3 ( a-8) = 0

( a-3 ) ( a-8 ) = 0

Therefore, a - 3 = 0 ⇒ a = 3

or               a - 8 = 0 ⇒ a = 8

Now we put the value a

when a = 3    (x² - 1) = 3

                     x² = 3 + 1 = 4

                    x = √4

                       = ± 2

when a = 8   (x² - 1) = 8

                     x² = 9

                     x = √9

                        = ± 3

Therefore, x = ± 3 and ± 2 will be the answer.

Answer:

Option C. Obtuse

Step-by-step explanation:

we know that

Applying the Pythagoras Theorem

If [tex]c^{2}=a^{2} +b^{2}[/tex] ----> is a right triangle

If [tex]c^{2}>a^{2} +b^{2}[/tex] ----> is a obtuse triangle

If [tex]c^{2}<a^{2} +b^{2}[/tex] ----> is an acute triangle

where

c is the greater side

Verify

[tex]c^{2}=26^{2}= 676[/tex]

[tex]a^{2} +b^{2}=7^{2} +24^{2}=625[/tex]

therefore

[tex]676>625[/tex]

[tex]c^{2}>a^{2} +b^{2}[/tex] -----> is a obtuse triangle

Given the derivative f '(x) ≥ 2, the function f(x) increases by at least 2 units for every increase in 'x' from 3 to 8. Therefore, the smallest possible value of f(8) is reached when the increase is exactly 2 per 'x', giving us a result of 21.

The question is about calculus, specifically related to the concept of derivatives and their role in function interval analysis. Given the information that f(3) = 11 and f '(x) ≥ 2 for 3 ≤ x ≤ 8, we are asked to find the smallest possible value of f(8).

The derivative, f '(x), represents the rate of change of the function f(x). It is mentioned that f '(x) ≥ 2, which means that the function f(x) is increasing at a rate not less than 2 units per increase in 'x' from 3 to 8.

This suggests that the smallest possible value of f(8) could be achieved by assuming that f(x) increases exactly by 2 units for every increase in 'x'. So, from x = 3 to x = 8 is a change of 5 units in 'x'. Multiply this by the rate of change 2 to obtain the total change in f(x), which is 5 * 2 = 10. Added to the initial value f(3) = 11, we get the smallest possible value for f(8) is 11 + 10 = 21.

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The horizontal distance between the balloon and the ship can be calculated using the tangent function is 1639 feet.

To find the horizontal distance between the balloon and the ship, we can use the tangent function in trigonometry. Let's denote the horizontal distance as x, and the distance between the balloon and the ship along the ground as y.

We have the angle of depression (21°) and the altitude of the balloon (2,000 feet). We can set up the equation:

tan(21°) = y / (2,000 + y)

Now, we need to solve for y:

tan(21°) * (2,000 + y) = y

y * tan(21°) = 2,000 * tan(21°) - y * tan(21°)

y * (1 - tan(21°)) = 2,000 * tan(21°)

y = (2,000 * tan(21°)) / (1 - tan(21°))

Now, calculate the value:

y ≈ (2,000 * 0.364) / (1 - 0.364)

y ≈ 728 / 0.636

y ≈ 1,148.17

Now that we have the value of y, we can find the horizontal distance x using the Pythagorean theorem:

[tex]x^2 + y^2 = (2,000)^2\\x^2 + (1,148.17)^2 = 4,000,000\\x^2 = 4,000,000 - 1,319,441.98\\x^2 = 2,680,558.02\\x = 1,638.84 (rounded)[/tex]

The horizontal distance between the balloon and the ship is approximately 1,639 feet.

Horizontal distance = height of balloon / tangent(angle of depression) = 2000 / tan(21°) ≈ 5243.87 feet.

To find the horizontal distance between the balloon and the ship, we can use trigonometry, specifically the tangent function.

Let's denote:

- [tex]\( h \)[/tex] as the height of the balloon above the ground (2,000 feet in this case).

- [tex]\( \theta \)[/tex] as the angle of depression (the angle formed between the line of sight from the balloon to the ship and the horizontal ground).

Given that the angle of depression [tex]\( \theta \)[/tex] is 21°, we can use the tangent function:

[tex]\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \][/tex]

In this case, the opposite side is the height of the balloon above the ground (2,000 feet), and we want to find the adjacent side, which represents the horizontal distance between the balloon and the ship.

So, we have:

[tex]\[ \tan(21°) = \frac{2000}{\text{horizontal distance}} \][/tex]

To find the horizontal distance, we rearrange the equation:

[tex]\[ \text{horizontal distance} = \frac{2000}{\tan(21°)} \][/tex]

Now, we can calculate this:

[tex]\[ \text{horizontal distance} = \frac{2000}{\tan(21°)} \approx \frac{2000}{0.381} \approx 5243.87 \text{ feet} \][/tex]

So, the horizontal distance between the balloon and the ship is approximately 5243.87 feet.

Hence, the sample standard deviation is:

                         1.1832

The data is given by:

8, 8, 8, 5, 9, 8, 7, 6, 6, 7, 7, 7, 7, 9, 9.

The mean of these data points is given by:

[tex]Mean(x')=\dfrac{8+8+8+5+9+8+7+6+6+7+7+7+7+9+9}{15}\\\\i.e.\\\\Mean(x')=\dfrac{111}{15}\\\\i.e.\\\\Mean(x')=7.4[/tex]

Now,

x                x-x'                   (x-x')²

8            8-7.4=0.6              0.36

8            8-7.4=0.6              0.36

8            8-7.4=0.6              0.36    

5            5-7.4= -2.4             5.76

9            9-7.4=1.6                2.56

8            8-7.4=0.6              0.36

7            7-7.4= -0.4             0.16

6            6-7.4= -1.4              1.96

6            6-7.4= -1.4              1.96

7            7-7.4= -0.4             0.16

7            7-7.4= -0.4             0.16

7            7-7.4= -0.4             0.16

7            7-7.4= -0.4             0.16

9            9-7.4=1.6                2.56

9            9-7.4=1.6                2.56

                             ∑(x-x')²=19.69          

Now the variance of the sample population is given by:

[tex]Variance=\dfrac{\sum (x-x')^2}{n-1}[/tex]

where  n is the number of data points.

Here n= 15

Hence, n-1=14

Hence, we get:

[tex]Variance=\dfrac{19.6}{14}\\\\i.e.\\\\Variance=1.4[/tex]

We know that the standard deviation is the square root of the variance.

i.e.

[tex]Standard\ deviation=\sqrt{1.4}\\\\i.e.\\\\Standard\ deviation=1.1832[/tex]

Answer:

-5,5

Step-by-step explanation:

Answer:

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Answer:

Step-by-step explanation:

"A system of linear equations" covers a lot of territory. In Algebra 1, it usually means two linear equations in two unknowns. Each of those equations will graph as a line on a coordinate plane.

A solution is a point that satisfies all the equations. That is, it is a point that is on all the lines described by the system of equations.

The geometry of lines on a plane comes into play with regard to solutions.

MAD tells us how far, on average, all values are from the middle. So, in the example weight of elk antler pairs at Mount St Helens NVM are, on average, 2 away from the middle. On the other hand, weight of elk antler pairs at Rocky Mountain NP are, on average, 1.5 away from the middle. So, we can assure that elk antler pairs at Rocky Mountain NP weighs more than elk antler pairs at Mount St Helens NVM.