rewrite the exprossion in the form b^n b^15/b

Here, we use the property of multiplication of exponential expression which states when we multiply two exponential expressions with the same base, we keep the base and add the exponents.

Therefore,

\(x^(3+5) = x^8\)

Now,

\(x^(3+5) = x^8\)

is of the form:

\(x^b = x^p\)

When we have two equal expressions on either side of the equation, the power of the base remains the same. Therefore,

p = 8

There we have it. The value of p is 8. The full solution is shown below:

\(x^3 × x^5 \\= x^px^8\\ = x^p\)

We can see that the base of the exponential expression on either side is equal.

Therefore, the power of the base must be equal as well. In other words

,p = 8.

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The expression fraction with negative 4 times the quantity 2 minus the cube root of 8 times 6 end quantity as the numerator and 5 is simplified to give 8

The fraction: negative 4 times the quantity 2 minus the cube root of 8 times 6 end quantity as the numerator and 5 is written in words

Rewriting the fraction

-4( (2 - ∛8 * 6) / 5)

The equation is simplified as

= -4( (2 - ∛8 * 6) / 5)

= -4( (2 - 2 * 6) / 5)

= -4( (2 - 12) / 5)

= -4( (-10) / 5)

= -4( -2)

= 8

Option C, 8 is the correct answer

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Answer:

cab =2acdhuajsjsidbjxjdbiss

Answer:

all numbers to 36 except 1, 2, 3, 4, 6, 9, 12, 18, and 36.

Step-by-step explanation:

Answer:

\(\sf (a) \quad AC=2\sqrt{3}\:\:cm\)

\(\sf (b) \quad AB=2\sqrt{7}\:\:cm\)

\(\sf (c) \quad \angle ACB=90^{\circ}\)

Step-by-step explanation:

Cosine rule

\(\sf c^2=a^2+b^2-2ab \cos C\)

where:

Sketch the triangle using the given information (see attached).

Given:

Substitute the given values into the formula and solve for AC:

\(\implies \sf c^2=a^2+b^2-2ab \cos C\)

\(\implies \sf AC^2=2^2+4^2-2(2)(4) \cos 60^{\circ}\)

\(\implies \sf AC^2=4+16-16 \left(\dfrac{1}{2}\right)\)

\(\implies \sf AC^2=20-8\)

\(\implies \sf AC^2=12\)

\(\implies \sf AC=\sqrt{12}\)

\(\implies \sf AC=\sqrt{4 \cdot 3}\)

\(\implies \sf AC=\sqrt{4}\sqrt{3}\)

\(\implies \sf AC=2\sqrt{3}\:\:cm\)

Given:

Substitute the given values into the formula and solve for AB:

\(\implies \sf c^2=a^2+b^2-2ab \cos C\)

\(\implies \sf AB^2=2^2+4^2-2(2)(4) \cos 120^{\circ}\)

\(\implies \sf AB^2=4+16-16 \left(-\dfrac{1}{2}\right)\)

\(\implies \sf AB^2=20+8\)

\(\implies \sf AB^2=28\)

\(\implies \sf AB=\sqrt{28}\)

\(\implies \sf AB=\sqrt{4\cdot7}\)

\(\implies \sf AB=\sqrt{4}\sqrt{7}\)

\(\implies \sf AB=2\sqrt{7}\:\:cm\)

Given:

Substitute the given values into the formula and solve for ∠ACB:

\(\implies \sf c^2=a^2+b^2-2ab \cos C\)

\(\implies \sf \left(2\sqrt{7}\right)^2=\left(2\sqrt{3}\right)^2+4^2-2\left(2\sqrt{3}\right)(4) \cos ACB\)

\(\implies \sf 28=12+16-16\sqrt{3} \cos ACB\)

\(\implies \sf 28=28-16\sqrt{3} \cos ACB\)

\(\implies \sf 0=-16\sqrt{3} \cos ACB\)

\(\implies \sf \cos ACB=0\)

\(\implies \sf ACB=\cos^{-1}(0)\)

\(\implies \sf ACB=90^{\circ}\)

A) The random variable X is  The number of pages that advertise footwear

B) X = 1, 2, 3.....20

C) On average 3.02 pages expect to advertise footwear on them

Bean advertised footwear on 29 of its 192 catalog pages.

we randomly survey 20 pages

the random variable X is

X = The number of pages that advertise footwear

It is given that 20 pages have been surveyed

so, X = 1, 2, 3.....20

The average value for binomial distribution can be calculated as follows

μ = np

= 20 (29/192)

= 580/192

=3.02

On average 3.02 pages expect to advertise footwear on them

Therefore, A) The random variable X is  The number of pages that advertise footwear

B) X = 1, 2, 3.....20

C) On average 3.02 pages expect to advertise footwear on them

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The bearing of R from P is found using trigonometry and the angle addition formula. The bearing is 210 degrees, with a distance of 3 from Q and 5 from P.

To find the bearing of R from P, we can use the bearings and distances given in the problem. Let P be a point, and line segment from P to Q with length 5. Then, from Q to R with length 3.

The angle of bearing of Q from P is 150 degrees, which means that the angle between the line segment PQ and the North direction is 150 degrees. We can say this angle as ∠PQ North. Similarly, the bearing of R from Q is 60 degrees, so we can say the angle ∠Q R North as 60 degrees.

Use trigonometry to find the angle between PR and the North direction. We can use the Law of Cosines to find the angle between PR and PQ:

cos(∠PQR) = (5^2 + 3^2 - PR^2) / (2 * 5 * 3)

cos(∠PQR) = (25 + 9 - PR^2) / 30

cos(∠PQR) = (34 - PR^2) / 30

Solving for cos(∠PQR), we get:

cos(∠PQR) ≈ 0.388

Taking the inverse cosine, we get:

∠PQR ≈ 67.6 degrees

Use the angle addition formula to find the bearing of R from P. Since we know the bearings of Q from P and R from Q, we can use the angle addition formula to find the bearing of R from P:

∠P R North = ∠PQ North + ∠Q R North

∠P R North = 150 degrees + 60 degrees

∠P R North = 210 degrees

Therefore, the bearing of R from P is 210 degrees (correct to the nearest degree).

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Yes, with sides 8,9 and 12 triangles can be formed.

Triangle is a shape made of three sides in a two-dimensional plane. the sum of the three angles is 180 degrees.

From the triangle property, the sum of the two sides of the triangle should be greater than the third side.

From the given data the sum of the two sides is greater than the third side.

8 + 9 > 12

17 > 12

Hence, with sides 8,9 and 12 triangles can be formed.

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The paper clip costs $0.05 and the folder Costs $1.00 more, which is $1.05. Together, they add up to the total of $1.10.

Let's solve this problem step by step:

Let's assume the cost of the paper clip is x dollars.

According to the information given, the folder costs $1.00 more than the paper clip, so the cost of the folder would be (x + $1.00).

The total cost of the folder and the paper clip is $1.10, so we can write the equation:

x + (x + $1.00) = $1.10

Combining like terms, we have:

2x + $1.00 = $1.10

Subtracting $1.00 from both sides of the equation, we get:

2x = $0.10

Dividing both sides by 2, we find the value of x:

x = $0.10 / 2

x = $0.05

Therefore, the paper clip costs $0.05.

In summary, the paper clip costs $0.05 and the folder costs $1.00 more, which is $1.05. Together, they add up to the total of $1.10.

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Answer:

Q8 It is because the data collected are too scattered and cannot show an obvious trend.

if it's correct pls mark brainliest tks

Answer:

the answer Is 100........

Answer:

97

Step-by-step explanation:

y=mx+b

m= slope

y=97x+1

The correct answer is option D. Dean is incorrect because the graph is symmetrical.

A symmetrical graph is one that looks the same on both sides of a centre line. This indicates that the y-values of the points on the graph that are the same distance from the centre line are the same. This results in a symmetrical graph that is aesthetically consistent and easy to understand.

In the given question, Dean is mistaken since the graph is symmetrical. The graph depicts a bell-shaped distribution, with the highest frequency of values grouped around the centre and decreasing frequencies at both ends. Also, the intervals 0-9 and 80-89 have a 0 frequency. The graph, however, does not show a peak; instead, it displays a cluster of numbers ranging from 30 to 60.

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Answer:

40 feet

Step-by-step explanation:

We know that the area of a rectangle is represented as \(lw=a\), where l is the length and w is the width.

We already know the width, and we know the area, so we can plug these values into the equation.

\(l\cdot 6 = 240\)

Our goal is to now isolate the variable l, and to do this we can divide both sides by 6.

\((l\cdot6) \div6 = 240\div6\\\\l = 40\)

Hope this helped!

Answer:

l=a/w

Step-by-step explanation:

Length equals area divided by width.

The equation that models this situation is 11x + 7.5y = 99

Total money to buy stamps and envelopes = $99

Cost of a sheet of 20 stamps = $11

Cost of a box of 50 envelopes = $7.50

Let x represent the number of stamp sheets and y represent the number of boxes of envelopes

Formulating the equation we get the following:

Cost of a sheet of 20 stamps*Number of stamp sheets + Cost of a box of 50 envelopes*Number of boxes of envelopes = Total money available to buy stamps and envelopes

11x + 7.5y = 99

So, the equation representing this situation is 11x + 7.5y = 99

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Answer:

2545

Step-by-step explanation:

Step-by-step explanation:

We can use the kinematic equations of motion to solve this problem. Let's assume the initial velocity of the snowball is 18 feet per second and its initial height is 25 feet. Also, we know that the acceleration due to gravity is -32.2 feet per second squared (assuming downward direction as negative).

To find out when the snowball hits the ground, we can use the equation:

h = 25 + 18t - 16t^2

where h is the height of the snowball at time t. We want to find the value of t when h = 0 (since the snowball hits the ground at that point). Therefore, we can rewrite the equation as:

16t^2 - 18t - 25 = 0

Solving for t using the quadratic formula, we get:

t = (18 ± √(18^2 + 41625))/(2*16)

t = 2.25 seconds or -0.875 seconds

Since time cannot be negative, the snowball hits the ground after 2.25 seconds.

To find the maximum height the snowball reaches, we can use the fact that the maximum height occurs at the vertex of the parabolic trajectory. The x-coordinate of the vertex is given by:

t = -b/2a

where a and b are the coefficients of the quadratic equation. In this case, a = -16 and b = 18, so:

t = -18/(2*(-16)) = 0.5625 seconds

To find the corresponding height, we can substitute t = 0.5625 seconds into the equation for h:

h = 25 + 18(0.5625) - 16(0.5625)^2

h = 28.2656 feet

Therefore, the maximum height the snowball reaches is 28.2656 feet.

Answer:

Step-by-step explanation:

Area = length times width

w = width

length = w + 5

Area = 36

w(w +5) = 36

w² + 5w = 36

w² + 5w - 36 = 0

Factor

(w + 9)(w -4) = 0

solve each root

w + 9 = 0

w  + 9 - 9 = 0 - 9

w = -9

w - 4 = 0

w -4 +4 = 0 + 4

w = 4

The two roots to the equation are -9, 4.

-9 is not an answer to the problem, as it makes no sense.  But 4 is an actual root.

the length is 5 + 4 = 9

So the length is 9 units.

The linear equation that compares the population y of spoonerville as it relates to the number of years since 2000 be,

y - 150x

Given, In the year 2000, the state of Florida named an existing town after their great and brightest resident.

The town of Spoonerville was founded in 2000 but unfortunately the population decreased by 150 people per year. In 2007 the population was 75,450 people.

Let the population of Spoonerville be y,

and the number of years since 2000 be x

So, the equation be

Population = y - 150x

Hence, the linear equation that compares the population y of spoonerville as it relates to the number of years since 2000 be,

y - 150x

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When two angles add up to 180 degrees, they are supplementary (a Straight Angle). As long as the sum is 180, they don't have to be close to each other.

Given, a graph in which two lines g and f intersect each other at a point "Q" on a plane "V".

Two rays that have a common endpoint and are referred to as the angle's sides and vertices, respectively, form an angle. Two rays can form angles in the plane where they are positioned.

Angles with a sum of 180 degrees are referred to be complementary angles. Complementary angles are those formed by straight lines.

The 90-degree angles between the two lines are known as supplementary angles.

Therefore, WQ and QW are two other names for line g. R, Q, and S intersect at R. Point Q on Plane V is where two lines meet. WQ, RQ, and SQ are the three segments that have endpoint q. Every angle that has the vertex M is JMK and KML. KML and KLM are the two nearby angles. JMK and KML are the pair of supplementary angles.

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Justice is using a technique known as chunking.

Short-term memory can function more effectively thanks to chunking, however the quantity of chunks that can be stored efficiently declines as chunk size grows.

Large pieces of information may be divided into smaller bits and then grouped together using relevant information or attributes before being stored in order to improve retention of that information in the short term memory. To establish relational information or similarity between the information to be stored in order to improve retention and recall in the short term memory, information must be broken down before being regrouped. However, as chunk size grows, so does the efficiency of information chunking.

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The absolute minimum value of the function

\(f(x)=2x3 21x2−48x 6\)

,\(−8≤x≤2\) is -414 at

x = 6.

To find the absolute minimum value of the function

\(f(x) = 2x^3 - 21x^2 - 48x + 6\) in the interval

-8 ≤ x ≤ 2, follow these steps:

1. Determine the derivative of f(x) with respect to x to find critical points:

 \(f'(x) = 6x^2 - 42x - 48\)

2. Set f'(x) to 0 and solve for x to find critical points:

 x = 6, -2

3. Check the function's value at the critical points and the endpoints of the interval:

\(f(-8) = 2560,  f(-2) = 72,  f(6) = -414,  f(2) = -52\)

4. Compare the values and determine the absolute minimum value, Therefore,The absolute minimum value of the function is -414 at x = 6.

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On solving the provided question, we can say that - by doing the equation we have 70 adult tickets were sold. and 130 children's tickets were sold.

An algebraic equation of the form y=mx+b, where m denotes the slope and b the y-intercept, is known as a linear equation. When y and x are variables, the aforementioned is referred to as a "linear equation in two variables." Bivariate linear equations are defined as linear equations with two variables. 2x - 3 = 0, 2y = 8, m + 1 = 0, x/2 = 3, x + y = 2, and 3x - y + z = 3 are examples of linear equations.

Total: $610, quantity: 200, adult: $5, number of children: x, child: $2

Price:\(5x + 2y = 610\)

Quantity: 200 when x plus y.

\(2(x + y = 200) 2x - 2y = -400 1(5x + 2y = 610) 5x + 2y = 610\\3x + 0 = 210\\ x = 70.\)

In either the Cost or Quantity equation, substitute "x = 70":

\(x + y = 200 70 + y = 200 y = 130\)

70 adult tickets were sold.

130 children's tickets were sold.

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       For C1(Q) = 3 - 2Q.

        For C2(Q) =  4.

     2. The AVC function

           For C1(Q) = 5/Q + 3 + 20/Q - Q.

           For C2(Q) = 3/Q + 4 + 2/Q.

     3. The AFC function

              For C1(Q)= 5/Q - 20/(5 + 3Q + 20/Q - Q)

              For C2(Q) = 0.

     4.  To find the AC function

        For C1(Q) = (5 + 3Q + 202 - Q^2)/Q + 5/Q - 20/(5 + 3Q + 20/Q - Q).

         For C2(Q) = (3 + 4Q + 2)/Q + 3/Q + 4 + 2/Q.

    5.To find the ATC function

    For C1(Q)= 5/Q² + 3/Q + 20/Q² - Q/Q + 5/Q - 20/(5Q + 3Q² + 20 - Q²)

     For C2(Q)= 3/Q² + 4/Q + 2/Q² + 3/Q + 4/Q + 2/Q.

1. To find the MC function, we take the derivative of the cost functions with respect to Q.

For C1(Q) = 5 + 3Q + 202 - Q^2, MC1(Q) = 3 - 2Q.

For C2(Q) = 3 + 4Q + 2, MC2(Q) = 4.

2. To find the AVC function, we divide the cost functions by Q.

For C1(Q), AVC1(Q) = (5 + 3Q + 202 - Q^2)/Q = 5/Q + 3 + 20/Q - Q.

For C2(Q), AVC2(Q) = (3 + 4Q + 2)/Q = 3/Q + 4 + 2/Q.

3. To find the AFC function, we subtract the AVC function from the ATC function.

For C1(Q), AFC1(Q) = (5 + 3Q + 202 - Q^2)/Q - (5 + 3Q + 202 - Q^2)/(5 + 3Q + 20/Q - Q)

= 5/Q - 20/(5 + 3Q + 20/Q - Q).

For

C2(Q), AFC2(Q) = (3 + 4Q + 2)/Q - (3 + 4Q + 2)/(3/Q + 4 + 2/Q) = 0.

4. To find the AC function, we add the AVC function to the AFC function.

For

C1(Q), AC1(Q) = (5 + 3Q + 202 - Q^2)/Q + 5/Q - 20/(5 + 3Q + 20/Q - Q).

For

C2(Q), AC2(Q) = (3 + 4Q + 2)/Q + 3/Q + 4 + 2/Q.

5. To find the ATC function, we divide the AC function by Q.

For

C1(Q), ATC1(Q) = [(5 + 3Q + 202 - Q²)/Q + 5/Q - 20/(5 + 3Q + 20/Q - Q)]/Q

= 5/Q² + 3/Q + 20/Q² - Q/Q + 5/Q - 20/(5Q + 3Q² + 20 - Q²).

For

C2(Q), ATC2(Q) = [(3 + 4Q + 2)/Q + 3/Q + 4 + 2/Q]/Q

= 3/Q² + 4/Q + 2/Q² + 3/Q + 4/Q + 2/Q.

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Answer:

3

Step-by-step explanation:

3 is the coefficient and y is the term

Answer:

The first one: 3

Step-by-step explanation:

Answer:

solution,

\( (\frac{1}{7} ) ^{3a + 3} = {343}^{a - 1} \\ or \: ( {7}^{ - 1} ) ^{3a + 3} = {(7}^{3} ) ^{a - 1} \\ or \: (7) ^{ - 3a - 3} = {(7)}^{3a - 3} \\ or \: - 3a - 3 = 3a - 33 \\ or \: - 3a - 3a = - 3 + 3 \\ or \: - 6a = 0 \\ \: a = 0\)

hope this helps..

Good luck on your assignment..

Answer:

its 0

Step-by-step explanation:

trust

Step-by-step explanation:

Answer:

x = 6

Step-by-step explanation:

We solve that above question using the trigonometric function of Tangent

Tan theta = Opposite/Adjacent

Theta = 45°

Opposite = 6

Adjacent = x

tan 45° = 6/x

tan 45 in rational form = 1

1 = 6/x

Cross Multiply

x = 6

Interior angles play an important role in geometry, particularly in proving theorems related to parallel lines and angles. Understanding this concept can help you solve problems involving parallel lines and transversals.

Interior angles that lie on opposite sides of a transversal are called alternate interior angles. Alternate interior angles are formed when a transversal intersects two parallel lines. Here's how you can identify alternate interior angles:

1. Look for a transversal that intersects two parallel lines.

2. Identify any pair of interior angles that are on opposite sides of the transversal.

3. These angles are called alternate interior angles.

Alternate interior angles are congruent, which means they have the same measure. This property allows us to solve for unknown angles or prove certain geometric relationships.

For example, if we have two parallel lines cut by a transversal, and we know the measure of one alternate interior angle, we can use that information to find the measure of another alternate interior angle.

In the diagram, if angle 1 is 60 degrees, then angle 2 will also be 60 degrees. This is because alternate interior angles are congruent. Alternate interior angles play an important role in geometry, particularly in proving theorems related to parallel lines and angles. Understanding this concept can help you solve problems involving parallel lines and transversals.

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The phase diagrams provide a visual representation of the system's behavior by plotting the vector field and trajectories in the x-y plane.

The given system of differential equations is described by:

\(˙x = a - x^2˙y = x - y\)

To find the equilibria, we set ˙x and ˙y equal to zero:

\(a - x^2 = 0 -- > x^2 = a -- > x = ±√ax - y = 0 -- > y = x\)

So, the equilibria are (±√a, ±√a).

Now let's analyze the behavior of the system for different values of 'a'.

For a > 0:

In this case, there are two real equilibria, (√a, √a) and (-√a, -√a). We can observe that (√a, √a) is a stable node, as the eigenvalues of the linearized system around this point have negative real parts. On the other hand, (-√a, -√a) is a saddle point, as the eigenvalues have opposite signs (one positive and one negative).

For a < 0:

In this case, there are no real equilibria since √a and -√a are imaginary. Therefore, the system has no equilibrium points.

To visualize the phase diagrams for a = 1 and a = -1:

For a = 1:

The system has two real equilibria, (1, 1) and (-1, -1). The point (1, 1) is a stable node, and (-1, -1) is a saddle point. The phase diagram would show trajectories converging towards (1, 1).

For a = -1:

Since a < 0, there are no equilibrium points, and thus the phase diagram would show no fixed points or trajectories.

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