What is 2/15 in simplest form

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 zero with a multiplicity of 1 is x = 1, and the zero x = -2 has a multiplicity of 2.

For a general polynomial like:

[tex]p(x) = (x - x_1)^n*(x - x_2)^m*...[/tex]

The zero x₁ has a multiplicity of n (same as the exponent in the term where the zero appears), while the zero x₂ has a multiplicity m.

Now let's go to our polynomial:

[tex]f(x) = (x - 3)^2(x + 2)^2(x - 1)[/tex]

Here we can see that the only zero with a multiplicity of 1, is x = 1, while the zero x = -2 has a multiplicity of 2.

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Answer:

first one is 1 and second is 2

Step-by-step explanation:

just got it right

Answer:

b

Step-by-step explanation:

The outlier affects the median of the data sets collected by Ivan by reducing the median.

An outlier is a number that is way smaller or way larger than that of other numbers in a data set. The outlier in the data set is 62. Median is the number at the center of a data set.

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Answer: 200 ft per hour

Step-by-step explanation:

Given : You are hiking on a trail that leads to the top of a mountain.

At 12 P.M. you reach an elevation of 3700 feet. At 4 P.M. you reach an elevation of 4500 feet.

The formula for the mean hourly change in elevation is :-

[tex]\dfrac{\text{Change in elevation}}{\text{Time in hours}}[/tex]

From the given information, the change in elevation=4500 ft-3700 ft= 800 ft

Time taken = 4 hours   [From 12 pm to 4 pm]

Now, Mean hourly change in elevation=[tex]\dfrac{800}{4}=200[/tex]

Hence, the mean hourly change in elevation = 200 ft

The total number of cases I should have is 42.

Multiplication is the mathematical operation that is used to determine the product of two or more numbers.

Total number of cases = number of layers x cases per layer

14 x 3 = 42

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Answer:

The temperature is -12.5 degrees Celsius.

Step-by-step explanation:

The daily maximum temperature, or high, describes how warm you can expect the air to be, usually from 7 a.m. to 7 p.m. The daily minimum temperature, or low, tells how much the air is expected to cool, usually overnight from 7 p.m. to 7 a.m.

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:

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

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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The value of Car 1 decreases linearly and can be described by a linear function. Without an exact exponential function for Car 2, we'll assume it may have a slower depreciation rate compared to Car 1. Autumn should consider Car 2 to likely have greater value after 6 years.

Part A: Identifying the Type of Function

To determine which type of function best describes the value of each car after a fixed number of years, we must look at the rate at which the car's value decreases. For Car 1, the value decreases by a constant amount each year ($6,000), which suggests a linear function. Conversely, Car 2 does not decrease by the same amount each year, but rather by amounts that seem to be getting progressively larger, hinting at an exponential function.

Part B: Writing the Functions

The linear function for Car 1 can be represented as f(x) = -6,000x + 44,000, since we start at $44,000 and decrease by $6,000 each year. For Car 2, an exponential decay function may fit the data; however, with only three points provided, determining the exact exponential function would require more complex regression analysis which we do not perform here. Assuming the rate of depreciation remains similar, we might estimate the function for Car 2 using a linear approximation for simplicity.

Part C: Future Car Value Comparison

Extending the linear depreciation model for Car 1, its value after 6 years would be f(x) = -6,000(6) + 44,000 = $8,000. A precise prediction for Car 2 after 6 years cannot be determined without an accurate exponential function, but it's apparent that Car 2 depreciates less rapidly than Car 1. Therefore, Autumn would likely find that Car 2 retains more value over 6 years.