How can you use transformations to graph this function?

y=3x[tex] 7^{-x} [/tex]+2

Answer:

Its -0.2 Because time cannot be a negative value.

Step-by-step explanation:

The correct value would be the 2.7 sec. one.

An algebraic equation with the second degree of the variable is called an Quadratic equation.

Here, Indicated when h(t) = 0, we are to find for the time where the football has reached a certain height.

⇒ - 16t² + 40t + 7 = 0

Then, the values of t would be -0.2 and 2.7.

Since,  we are to look for the time, it is impossible for us to have an answer that has a negative value.

So based from our answer, the solution that we will eliminate is the -0.2 one. Since its value is negative.

Thus, the correct value would be the 2.7 sec. one.

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Answer:

c

Step-by-step explanation:

Answer:

The sum is fraction therefore rational

Step-by-step explanation:

Given two numbers

[tex]-2\frac{3}{4}\text{ and }\frac{5}{9}[/tex]

we have to tell that how their sum is rational.

we add the above two number

[tex]-2\frac{3}{4}+\frac{5}{9}[/tex]

[tex]\frac{-11}{4}+\frac{5}{9}[/tex]

[tex]\frac{-99+20}{36}=\frac{-79}{36}=-2.194444...[/tex]

Hence, the sum is we can say fraction or non-terminating but repeating which is the definition of a rational number.

∴ from the given options answer C is correct.

The sum is fraction therefore rational.

Answer:

The LA theorem is a special case of the AAS theorem and the ASA postulate.

Step-by-step explanation:

The LA theorem is a special case of the AAS theorem and the ASA postulate.

LA(Leg - Acute) theorem states that if the leg and one acute angle of one right triangle are congruent to the corresponding leg and acute angle of another right triangle, then the triangles are congruent.

AAS (Angle Angle Side) states that if two angles and the non included side of one triangle are congruent to two angles and the non included side of another triangle, then these two triangles are congruent.

ASA (Angle Side Angle) postulate states that if two angles and one included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent.

The number in which the 6 has a value that is one-tenth the value of the 6 in 34,761 is 3,476.1

In the number 34,761, the digit 6 is in the tens place, giving it a value of 60. To find a number where the value of 6 is one-tenth of 60, we look for a place value that is one position to the right.

In 3,476.1, the 6 is in the ones place, making its value 6, which is one-tenth of 60. This shift from the tens place to the ones place decreases the value by a factor of ten.

Answer:

Approximately 83 bags each carrying 50 pounds of soil will carry 1 cubic yard of soil or 4,128 pounds of soil.

Step-by-step explanation:

We are given the following information in the question:

1 cubic yards is equal to 4128 pounds that is

[tex]1~yard^3 = 4,128~pounds[/tex]

Amount of soil 1 bag can hold = 50 pounds

Number of bags to hold 1 cubic yard of soil or 4,128 pounds of soil:

Formula:

[tex]\text{Number of bags} = \displaystyle\frac{\text{Total amount of soil}}{\text{Amount of soil 1 bag can hold}}[/tex]

[tex]= \displaystyle\frac{4128}{50} = 82.56[/tex]

Thus, approximately 83 bags each carrying 50 pounds of soil will carry 1 cubic yard of soil or 4,128 pounds of soil.

The required measure of the length of the given line is 8.6 units.

Distance is defined as the object traveling at a particular speed in time, from one point to another.

here,
Locate the end point of line in the graph,
end point of the line from the graph is given as (0, 8) and (7, 3)

Applying the distance formula,
D = √[[x₂ - x₁]² + [y₂ - y₁]²]

Substitute the point in the above equation,

D = √[[7]² + [3 - 8]²]
D = √49 + 25
D = √74
D = 8.6

Thus, the required measure of the length of the given line is 8.6 units.

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The common speed of the blocks just after the collision is 0.4 times the initial velocity of the first block (v₁).

To determine the common speed of the blocks just after the completely inelastic collision, we can use the principle of conservation of linear momentum.

According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

Let's denote the velocity of the first block (m = 0.050 kg) before the collision as v₁ and the velocity of the second block (m = 0.075 kg) before the collision as v₂. Since the second block is stationary, its initial velocity, v₂, is 0.

1. Before the collision, the total momentum is given by:

Initial momentum = m₁v₁ + m₂v₂

where m₁ and m₂ are the masses of the first and second blocks, respectively.

Plugging in the values, we have:

Initial momentum = (0.050 kg) v₁ + (0.075 kg) 0

Initial momentum = (0.050 kg)v₁

2. After the collision, the two blocks stick together and move as a single unit with a common velocity, denoted as vf.

The final momentum is given by:

Final momentum = (m₁ + m₂) vf

where (m₁ + m₂) is the total mass of the system.

Plugging in the values, we have:

Final momentum = (0.050 kg + 0.075 kg) × vf

Final momentum = 0.125 kg × vf

According to the principle of conservation of linear momentum, the total momentum before the collision is equal to the total momentum after the collision:

0.050 kg × v₁ = 0.125 kg × vf

To determine the common speed of the blocks just after the collision, we need to solve for vf:

vf = (0.050 kg × v₁) / (0.125 kg)

Calculating the numerical value, vf = 0.4 ×  v1

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Priscilla and Herb would take approximately 44.44 hours to address 10,000 envelopes together, as Priscilla can address 100 envelopes per hour and Herb can address 125 envelopes per hour.

The question asks about the combined work rate of Priscilla and Herb in addressing 10,000 envelopes. To solve this, we calculate each of their rates of work and then combine them to find the total rate. Priscilla completes 10,000 envelopes in 100 hours, so her rate is 100 envelopes per hour. Herb completes 10,000 envelopes in 80 hours, so his rate is 125 envelopes per hour. Working together, their combined rate is 100 + 125 = 225 envelopes per hour. To address 10,000 envelopes at this rate, we divide 10,000 by 225, which equals approximately 44.44 hours.

Using simple mathematical operations, we know that 201 bolts are produced in 21 minutes.

So, the bolts produced in 21 minutes are:

Calculate as follows:

Therefore, using simple mathematical operations, we know that 201 bolts are produced in 21 minutes.

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Final answer:

To find the number of bolts produced in 21 minutes, set up a proportion using the given information: 268 bolts / 28 minutes = x bolts / 21 minutes. Simplifying, we find that x = 201 bolts.

Explanation:

To find the number of bolts produced in 21 minutes, we can set up a proportion using the given information. We know that the machine produces 268 bolts in 28 minutes. Let x be the number of bolts produced in 21 minutes.

Using the proportion:

268 bolts / 28 minutes = x bolts / 21 minutes

Multiplying both sides by 21, we get:

(268 bolts / 28 minutes) * 21 minutes = x bolts

Simplifying, we find that x = 201 bolts.

A double integral of Sin(x^2 + y^2) over a region between two circles with radii 2 and 5 is transformed into polar coordinates for simplification. The limits of the polar coordinates correspond to this region, and the integral becomes ∫ ∫ rSin(r^2) dr dθ, with r ranging from 2 to 5, and θ from 0 to π/2.

The student asked about the calculation of a double integral over a region between two circles, presented in polar coordinates, where the function is Sin(x^2 + y^2). For simplification, we can change to polar coordinates. In polar coordinates, x = rcosθ and y = rsinθ. Thus, x^2 + y^2 becomes r^2, and the function becomes Sin(r^2).

For a region between two circles of radii 2 and 5 in the first quadrant, we can describe the limits of the polar coordinates. r will range from 2 to 5, and θ from 0 to π/2. Thus, the integral will be ∫(from 0 to π/2) ∫(from 2 to 5) rSin(r^2) dr dθ. Evaluation of the integral will require techniques of polar integration.

It's important to note that this is a theoretical discussion and the actual integration may have complex results depending on the function to integrate.

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We need solve the given integral, convert coordinates to polar form and integrate step by step. First we will solve the inner integral from 2 to 5 and then the outer integral from 0 to π/2

We need to evaluate the double integral of Sin(x² + y²) over the region between the circles with radii 2 and 5.

In polar coordinates, where x = r cos(θ) and y = r sin(θ), this integral becomes:

∫(from 0 to π/2) ∫(from 2 to 5) sin(r²) r dr dθ

Let's solve the inner integral first, with respect to r:

Firstly, set up the integral: ∫(from 2 to 5) sin(r²) r dr.

Then Use the substitution u = r², then du = 2r dr.

Now Rewriting the integral: (1/2) ∫(from 4 to 25) sin(u) du = (1/2) [-cos(u)] (from 4 to 25).

and evaluating this: (1/2) [-cos(25) + cos(4)].

Next, integrate the result with respect to θ:

Set up the outer integral: (1/2) ∫(from 0 to π/2) [-cos(25) + cos(4)] dθ.

Now since [-cos(25) + cos(4)] is constant, the integral is straightforward: (1/2) [-cos(25) + cos(4)] * (π/2).

Hence the final answer: (π/4) [-cos(25) + cos(4)].

Therefore, the integral evaluates to (π/4) [-cos(25) + cos(4)].

EDITED QUESTION

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Find the double integral  Sin(x² + y²) over the region between the circles with radii 2 and 5.