for f(x)=x^2-2, find the rate of change on the interval [-2,4]​

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Explanation:

If angle A is 43 degrees, then minor arc BC is 2*43 = 86 degrees according to the inscribed angle theorem. The central angle is twice that of the inscribed angle. Both of these angles subtend the same minor arc.

When I say "minor arc BC", I mean that we go from B to C along the shortest path. Any minor arc is always less than 180 degrees.

Since minor arc AB is 69 degrees, and minor arc BC is 86 degrees, this means arc ABC is arcAB+arcBC = 69+86 = 155 degrees

Let's say point D is some point on the circle that isn't between A and B, and it's not between B and C either. Refer to the diagram below. The diagram is to scale. The diagram your teacher provided is not to scale because arc ABC is way too big (it appears to be over 180 degrees). Hopefully the diagram below gives you a better sense of what's going on.

Because arc ABC = 155 degrees, this means the remaining part of the circle, arc ADC, is 360-(arc ABC) = 360-155 = 205 degrees

Inscribed angle B subtends arc ADC. So we'll use the inscribed angle theorem again, but this time go in reverse from before. We'll cut that 205 degree angle in half to get 205/2 = 102.5 degrees which is the measure of angle B. This value is exact. In this case, we don't need to apply any rounding.

Answer:

\((C) \ a=\frac{1}{b}.\)

Step-by-step explanation:

(a+b)²=a²+b²+2;

a²+2ab+b²=a²+b²+2;

2ab=2;

ab=1

Solution to the initial-value problem with the given initial condition y(0) = 5 and differential equation \(dy/dt = e^{7t\).

We are given the following:

1. Initial condition: y(0) = 5
2. Differential equation: dy/dt = e^(7t)

Here's a step-by-step solution:

Step 1: Integrate both sides of the differential equation with respect to t.
∫(dy/dt) dt = ∫\(e^{7t\) dt

Step 2: Integrate the right side.
y(t) = (1/7)\(e^{7t\) + C, where C is the integration constant.

Step 3: Apply the initial condition, y(0) = 5.
5 = (1/7)\(e^{7*0\) + C

Step 4: Solve for the integration constant, C.
5 = (1/7)\(e^0\) + C
5 = (1/7)(1) + C
C = 5 - 1/7
C = 34/7

Step 5: Write the final solution for y(t).
y(t) = (1/7)\(e^{7t\) + 34/7

This is the solution to the initial-value problem with the given initial condition y(0) = 5 and differential equation \(dy/dt = e^{7t\).

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Answer:

32/0 or infinity

Step-by-step explanation:

Therefore, the answer is 32/0 or infinity.

From list of Al-alloy series, age-hardenable aluminum alloy series are 2000, 6000, and 7000. These makes precipitation hardening.Other alloy series are not considered age-hardenable and have different properties

A process that involves the formation of fine precipitates within the alloy matrix, resulting in increased strength and hardness. The 2000 series alloys are known for their high strength and excellent mechanical properties, making them suitable for aerospace and structural applications.

The 6000 series alloys are widely used due to their good combination of strength, formability, and corrosion resistance, and are commonly employed in automotive and architectural applications. The 7000 series alloys offer exceptional strength and toughness and are frequently used in high-performance aerospace and defense applications.

The other alloy series listed (1000, 3000, 4000, and 5000) are not typically considered age-hardenable and have different properties and applications.

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Answer:

a

Step-by-step explanation:

Answer:

Answer=(X + 5, Y - 4)

Step-by-step explanation:

X + 5 means you move to the right 5 places then Y - 4 means you move down 4 places.

Answer:

Look at the photo and goodluck

(i) When testing whether girls who attended girls-only high schools do better in math, it is important to control for various factors that could potentially influence math scores.

Some factors to consider are:

(ii) The equation relating the math score (score) to girlhs (dummy variable indicating girls-only high school attendance) and other factors can be written as:

score = β0 + β1 * girlhs + β2 * socioeconomic status + β3 * prior academic performance + β4 * school quality + β5 * peer effects + β6 * teacher quality + β7 * access to resources + ε

This equation represents the structural relationship between the math score and the factors being controlled for. The coefficients β1, β2, β3, β4, β5, β6, and β7 represent the respective effects of girlhs and the other factors on the math score.

(iii) Parental support and motivation, which are unmeasured factors, may be correlated with girlhs. This is because parents who choose to send their daughters to girls-only high schools might have certain preferences or beliefs regarding education, which could include providing higher levels of support and motivation. However, without directly measuring parental support and motivation, it is difficult to establish a definitive correlation.

(iv) To ensure that numghs (the number of girls-only high schools within a 20-mile radius of a girl's home) is a valid instrumental variable (IV) for girlhs, certain assumptions are needed:

(v) If the coefficient estimate on the chosen IV numghs is negative and statistically significant in the reduced form estimation, it suggests a strong relationship between the instrumental variable and the attendance at girls-only high schools. This provides some confidence in the validity of the IV. However, the decision to proceed with IV estimation should also consider other factors such as the strength of the instruments, the overall model fit, and the robustness of the results to alternative specifications.

It is important to carefully evaluate the assumptions and limitations of the IV estimation approach before drawing conclusions in math.

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Answer:

Representative.

Step-by-step explanation:

In Statistics, sampling can be defined as a process used to collect or select data (objects, observations, or individuals) from a larger statistical population using specific procedures.

A representative sample has characteristics similar to those of the entire population and, therefore, can be used to draw a conclusion about the (general) population. Thus, it is considered typically to be a subset of population and can be used to accurately make conclusions and reflect the characteristics of the larger sample (population).

Answer:

9 cubed is 729 if that's what your asking

The intercept is 3 and 5

The x-intercept is the point where a line crosses the x-axis, and the y-intercept is the point where a line crosses the y-axis.

To find x- intercept, y=0

x² + 2x -15=0

x² -5x + 3x -15=0

x(x-5) - 3(x-5) = 0

(x-5)(x-3)=0

Hence, the x- intercept is 3 and 5.

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Answer: x≥15

Step-by-step explanation: this means basically anything equal to or higher than 15

Answer:

its c

Step-by-step explanation:

i just too quiz and it siad this

Answer:

c

Step-by-step explanation:

took the quiz and got 100% on it

Answer:

Your answer is in the picture I have provided.

Step-by-step explanation:

PLEASE MARK BRAINLIEST

a) By strong induction, any amount of postage worth 8 cents or more can be made from 3-cent or 5-cent stamps.

b) By strong induction, any amount of postage worth 24 cents or more can be made from 7-cent or 5-cent stamps.

c) By strong induction, any amount of postage worth 12 cents or more can be made from 3-cent or 7-cent stamps.

a. Prove that any amount of postage worth 8 cents or more can be made from 3-cent or 5-cent stamps.

Base case: For postage worth 8 cents, we can use two 4-cent stamps, which can be made using a combination of one 3-cent stamp and one 5-cent stamp.

Induction hypothesis: Assume that any amount of postage worth k cents or less, where k is greater than or equal to 8, can be made from 3-cent or 5-cent stamps.

Induction step: Consider any amount of postage worth (k+1) cents. Since k is greater than or equal to 8, we can use the induction hypothesis to make k cents using 3-cent or 5-cent stamps. Then, we can add one more stamp to make (k+1) cents. If the last stamp we added was a 3-cent stamp, we can replace it with a 5-cent stamp to get the same value. If the last stamp we added was a 5-cent stamp, we can replace it with two 3-cent stamps to get the same value. Therefore, any amount of postage worth (k+1) cents can be made from 3-cent or 5-cent stamps.

b. Prove that any amount of postage worth 24 cents or more can be made from 7-cent or 5-cent stamps.

Base case: For postage worth 24 cents, we can use three 8-cent stamps, which can be made using a combination of one 7-cent stamp and one 5-cent stamp.

Induction hypothesis: Assume that any amount of postage worth k cents or less, where k is greater than or equal to 24, can be made from 7-cent or 5-cent stamps.

Induction step: Consider any amount of postage worth (k+1) cents. Since k is greater than or equal to 24, we can use the induction hypothesis to make k cents using 7-cent or 5-cent stamps. Then, we can add one more stamp to make (k+1) cents. If the last stamp we added was a 5-cent stamp, we can replace it with two 7-cent stamps to get the same value. If the last stamp we added was a 7-cent stamp, we can replace it with three 5-cent stamps to get the same value. Therefore, any amount of postage worth (k+1) cents can be made from 7-cent or 5-cent stamps.

c. Prove that any amount of postage worth 12 cents or more can be made from 3-cent or 7-cent stamps.

Base case: For postage worth 12 cents, we can use one 3-cent stamp and three 3-cent stamps, which can be made using a combination of two 7-cent stamps.

Induction hypothesis: Assume that any amount of postage worth k cents or less, where k is greater than or equal to 12, can be made from 3-cent or 7-cent stamps.

Induction step: Consider any amount of postage worth (k+1) cents. Since k is greater than or equal to 12, we can use the induction hypothesis to make k cents using 3-cent or 7-cent stamps. Then, we can add one more stamp to make (k+1) cents. If the last stamp we added was a 3-cent stamp, we can replace it with two 7-cent stamps to get the same value. If the last stamp we added was a 7-cent stamp, we can replace it with one 3-cent stamp and two 7-cent stamps to get the same value. Therefore, any amount of postage worth (k+1) cents can be made from 3

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Answer:

50cm or 500mm

Step-by-step explanation:

There are 4 base edges and 4 lateral edges, therefore, multiply the number for both edges by 4 and add.

Considering the fact that we were given two units of measurement, millimeters and centimeters, but we know that there are 10mm (millimeters) in a cm (centimeter) , we can just divide 55 by 10 to get 5.5.

¦

Therefore;

5.5 x 4 = 22

7 x 4 = 28

28 + 22 = 50cm

The diagram can be drawn in order to assist in answering the question.

The probability that the first 2 presentations will be by Todd and Garrett, in that order, is 1/342.

What is probability?

In mathematics, probability is a measure of the likelihood or chance of an event occurring. It is expressed as a number between 0 and 1, with 0 indicating an impossible event and 1 indicating a certain event.

There are 19 students in the class, so there are 19 possible students who could go first. Once the first student has been chosen, there are 18 students left who could go second. Since the order of the presentations matters, we can use the multiplication rule of probability to find the probability that Todd goes first and Garrett goes second:

P(Todd first, Garrett second) = P(Todd first) * P(Garrett second | Todd first)

P(Todd first) = 1/19

P(Garrett second | Todd first) = 1/18

Multiplying these probabilities, we get:

P(Todd first, Garrett second) = (1/19) * (1/18) = 1/342

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The height of the hot air ballon can be modeled with the linear equation:

y = 300 + 900*x

Remember that a linear equation is written as:

y = a*x + b

Where a is the slope and b is the y-intercept.

Let's define the variables that we need to use for this problem:

y = height of the hot air balloon.

x = time in minutes.

We know that initially, the height of the balloon is at a height of 300 feet, and then it ascends at a constant rate of 900 feet per minute,

So after x minutes, the height will be modeled by the linear equation:

y = 300 + 900*x

Where the y-intercept is 300, the initial height, and the slope is 900, the rate of change.

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We can use the distributive property and the greatest common factor to write 36 + 4 as 4(9 + 1) or 4(10).

Using the distributive property, we can write:

36 + 4 = 4(9 + 1)

Next, we can identify the greatest common factor (GCF) of 9 and 1, which is 1. Since there are no other common factors between 9 and 1, we can write:

9 + 1 = 10

Therefore, we can express 36 + 4 as:

36 + 4 = 4(9 + 1) = 4(1 × 9 + 1) = 4(10)

So, 36 + 4 can be written as the product of 4 and 10, where 4 is the GCF of 36 and 4, and 10 is the sum of the two terms divided by their GCF. This is an example of factoring by grouping, which can be a useful technique in algebra and other areas of mathematics.

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The distance from point A to point C and the bearing from point A to point C is 229.5 miles.

Given that:

AB = 370 miles

angle A = 24 degrees

BC = 240 miles

angle B = 37 degrees 

To determine the distance from point A to C, we can use the cosine law since we are given two angles and two sides of a regular triangle. 

The angle opposite side AC is angle B which is equal to 37 degrees. 

(AC)^2 =(AB)^2 + (BC)^2 - 2(AB)(BC)cos(B)

Now we have to solve for AC:

(AC)^2 =(370)^2 + (240)^2 - 2(370)(240)cos(37) 

AC = 229.48 miles.

Hence the answer is, the distance from point A to point C and the bearing from point A to point C is 229.5 miles. 

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The diagonals of a Rhombus are congruent and perpendicular bisectors of each other​

If a parallelogram contains a pair of consecutive sides that are called congruent, then it is considered a rhombus, or if a parallelogram bisects two angles of the same, then it is a rhombus.

If the diagonals of a quadrilateral are perpendicular bisectors to each other, then the quadrilateral is a rhombus.

There are  general properties of a Rhombus are:

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Answer:

Step-by-step explanation:

The total number of students,

→ 100 students

No. of students were absent & failed,

→ 7 + 10

→ 17 students

No. of students totally passed,

→ 100 - 17

→ 83 students

No. of students passed in Science,

→ 80 students

No. of students passed in Mathematics,

→ 71 students

Then add both science and math,

→ 80 + 71

→ 151 students

No. of students passed in both subjects,

→ 151 - 83

→ 68 students

Therefore, the number of students who passed in both subjects are 68 students.

Answer:

b) 12.6 inches

Step-by-step explanation:

tangent is the ratio of the side opposite over adjacent, so if we let x= length of the opposite side, 1.40=x/9, x=12.6 inches

b) 12.6 inches

Hope it helps!!

The f′(a) when a = −6 is -7. This means that the slope of the tangent line of the graph of f(x) at x = -6 is -7.

To compute f′(a) algebraically for the given value of a, we use the following differentiation rule which is known as the Power Rule.

This states that:If f(x) = xn, where n is any real number, then f′(x) = nxⁿ⁻¹This is valid for any value of x.

Therefore, we can differentiate f(x) = −7x + 5 with respect to x using the power rule as follows:

f(x) = −7x + 5

⇒ f′(x) = d/dx (−7x + 5)

⇒ f′(x) = d/dx (−7x) + d/dx(5)

⇒ f′(x) = −7(d/dx(x)) + 0

⇒ f′(x) = −7⋅1 = −7

Hence, the derivative of f(x) with respect to x is -7.Now, we evaluate f′(a) when a = −6 as follows:f′(x) = −7 evaluated at x = −6⇒ f′(−6) = −7

Therefore, f′(a) when a = −6 is -7. This means that the slope of the tangent line of the graph of f(x) at x = -6 is -7.

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Answer:

where is the problem

Step-by-step explanation:

The probability that you select a black card and a red card is 26/51 as per the condition mentioned "choosing two cards (without replacement )".

The area of mathematics known as probability studies potential outcomes of events as well as their relative probabilities and distributions. Simply put, probability is the likelihood that something will occur. When we don't know how an event will turn out, we can discuss the likelihood or likelihood of various outcomes. Statistics is the study of events that follow a probability distribution.

Here,

Since we have not mentioned the order,

so it can black-red or red-black,

=26/52*26/51+26/52*26/51

=2*26/52*26/51

=26/51

Assuming you choose two cards (without replacement), there is a 26/51 chance that you will choose a black card and a red card.

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Given the following question:

Clermont - Ferrand

15 letters in total

Three R's in the 15 letters

15 - 3 = 12

12 possible letters we can receive

Simplfied fraction = 4/5

Decimal = 0.8