Factor the following polynomial fully.

10x^2 + 16x – 8

A. 2(5x – 4)(x + 1)
B. (2x + 4)(5x – 2)
C.(10x – 8)(x + 1)
D.2(x + 2)(5x – 2)

Answer:

58π or about 182.12

Step-by-step explanation:

In order to find circumference with diameter, you need to use the formula:

C = πd

Multiply the diameter (58) by pi, and since pi is an irrational number, let's use 3.14.

3.14(58) = 182.12

58π would be the answer as well if you don't multiply pi.

Answer:

3

left

Step-by-step explanation:

1000/36=0.036

hope this helps

If X and Y are independent normal random variables with the same mean μ and the same variance σ^2. The random variables W = X + Y and U = 2X do not have same distribution because variance is different.

As X and Y are independent.

So W = X+Y is also a normal distribution with mean is equal to 0 and variance = 2σ^2.

U= 2X also normal because linear combination of independent variable or scaler multiplication is also a normal distribution.

So U is also normal with mean= 0 and variance 4σ^2

But W and U do not have same distribution because variance is different.

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The right question is:

Suppose X and Y are independent normal random variables with the same mean μ and the same variance σ^2. Do the random variables W = X + Y and U = 2X have the same distribution? Explain.

The starting temperature would be 29 degrees.

What are the initial value and final value?

Initial value refers to the starting value or the value of a variable at the beginning of a process or time period. Final value refers to the end value or the value of a variable at the end of a process or time period. In the context of a mathematical function, the initial value would be the value of the function at the beginning of the domain and the final value would be the value of the function at the end of the domain.

If the temperature outside dropped 23 degrees in 6 hours. The final temperature was -6,

The starting temperature would be 29 degrees (-6 + 23).

Hence, the starting temperature would be 29 degrees.

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Taylor polynomials of degree 2 and 3 centered at x = 0 for the function f(x) = 16tan(x) are:

P2(x) = 16x + 8x^2

P3(x) = 16x + 8x^2

To find the Taylor polynomials centered at x = 0 for the function f(x) = 16tan(x), we can use the Taylor series expansion for the tangent function and truncate it to the desired degree.

The Taylor series expansion for tangent function is:

tan(x) = x + (1/3)x^3 + (2/15)x^5 + (17/315)x^7 + ...

Using this expansion, we can find the Taylor polynomials of degree 2 and 3 centered at x = 0:

Degree 2 Taylor polynomial:

P2(x) = f(0) + f'(0)(x - 0) + (1/2!)f''(0)(x - 0)^2

= 16tan(0) + 16sec^2(0)(x - 0) + (1/2!)16sec^2(0)(x - 0)^2

= 0 + 16x + 8x^2

Degree 3 Taylor polynomial:

P3(x) = P2(x) + (1/3!)f'''(0)(x - 0)^3

= 0 + 16x + 8x^2 + (1/3!)(48sec^2(0)tan(0))(x - 0)^3

= 16x + 8x^2

Therefore, the Taylor polynomials of degree 2 and 3 centered at x = 0 for the function f(x) = 16tan(x) are:

P2(x) = 16x + 8x^2

P3(x) = 16x + 8x^2

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a. This is a voluntary response sampling strategy. Bias may occur because only those with strong opinions are likely to respond, which may not represent the views of all parents.

b. This is cluster sampling. Bias may occur if the selected school is not representative of the entire city's school system.

c. This is a census. Bias may occur if some parents do not return the survey, leading to nonresponse bias.

d. This is stratified random sampling. Bias may be minimized as each school is fairly represented, but nonresponse bias could still occur if some parents do not return the survey or respond to the phone call.

e. This is a voluntary response sampling strategy. Bias may occur as only those watching the news and with strong opinions are likely to participate, which may not represent the views of all parents.

f. This is convenience sampling. Bias may occur because only those attending the PTA meetings will have their opinions counted, which may not represent the views of all parents.

g. This is cluster sampling. Bias may occur if the selected classes are not representative of the entire city's school system.

h. This is systematic random sampling. Bias may be minimized as the sample is spread evenly throughout the enrollment records, but nonresponse bias could still occur if some parents are not available for the interview.

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Long-range artillery pieces are typically set at larger elevation angles, ranging from 55 to 65 degrees, rather than the maximum range angle of 45 degrees.

The angle of elevation determines the trajectory of a projectile fired from an artillery piece. When firing at maximum range, an angle of 45 degrees provides the farthest distance the projectile can travel. However, setting the elevation angle slightly higher, in the range of 55 to 65 degrees, offers certain advantages.

By increasing the elevation angle, the projectile follows a higher trajectory and spends more time in the air. This allows for a steeper angle of descent upon reaching the target, increasing the chances of hitting the desired point. Moreover, a higher angle of descent can enhance the effectiveness of certain types of munitions, such as high-explosive rounds.

While the exact elevation angle depends on factors like the projectile's speed, weight, and wind conditions, a range of 55 to 65 degrees is commonly employed for long-range artillery. These angles provide a balance between maximum range and improved accuracy, enabling the artillery piece to engage targets at extended distances effectively.

long-range artillery pieces are typically set at elevation angles between 55 and 65 degrees, surpassing the maximum range angle of 45 degrees. This adjustment in elevation offers improved accuracy and effectiveness when engaging targets over long distances. By utilizing a steeper trajectory and a higher angle of descent, artillery operators can increase the chances of hitting the intended target. These elevation angles provide a balance between maximizing range and achieving the desired level of accuracy, ensuring the artillery's effectiveness in long-range engagements.

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The value of the function 8s/ 2s^2−25s using inverse  Laplace transform  is equal to 4e^(25t/2).

Function is equal to,

8s/ 2s^2−25s

Value of 's' after factorizing the denominator we get,

2s^2−25s = 0

⇒ s( 2s -25 ) =0

⇒ s =0 or s =25/2

Now apply partial fraction decomposition we get,

8s/ 2s^2−25s = A/s + B /(2s -25)

Simplify it we get,

⇒ 8s = A(2s -25) + Bs

Now substitute s =0 we get,

⇒ 0 = A (-25) + 0

⇒ A =0

and s = 25/2

⇒8(25/2) = A(2×25/2 -25 ) + B(25/2)

⇒100 = B(25/2)

⇒B = 8

Now ,

8s/ 2s^2−25s = 0/s + 8 /(2s -25)

⇒ 8s/ 2s^2−25s = 8 /(2s -25)

Take inverse Laplace transform both the side we get,

L⁻¹ [8s / (2s^2 - 25s)] = L⁻¹ [8/(2s - 25)]

Apply , L⁻¹ [1/(as + b)] = (1/a)e^(-bt/a),

here,

a = 2 , b = -25

L⁻¹ [8s / (2s^2 - 25s)]

= L⁻¹ [8/(2s - 25)]

= (8/2) e^(25t/2)

= 4e^(25t/2)

Therefore, the value of  inverse Laplace transform for the given function is equal to 4e^(25t/2)

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The above question is incomplete, the complete question is:

Find the inverse Laplace transform of 8s/ 2s^2−25s.

(a) I have attached the graph.

(b) The function g(x) is given by g(x) = f(2x + 2)/3.  You can work this out algebraically.  You can also work this out by using the graph of the function.

(c) We can obtain the graph of y = g(x) from the graph of y = f(x) through the following transformations:

* stretch horizontally by a factor of 2

* stretch vertically by a factor of 3

* shift downwards by 2 units

Good for you mary.

Lol is there a question? I'll answer it if so

a) The solutions to the trigonometric equation 4 sin θ + 3 csc θ = 7 in the range of 0° to 360° are θ = 30° and θ = 150°.

b) In the range of 0° to 360°, the equation 2 csc θ = 6 sin θ + 1 has the solutions θ = 30°, θ = 150°, and θ = 270°.

c) The equation 5 sin θ = 4(1 + cos θ) has solutions θ = 60° and θ = 300° within the range of 0° to 360°.

d) Within the range of 0° to 360°, the equation 3 tan θ + 5 = 4 sec² θ has solutions θ = 67.4° and θ = 247.4°.

To solve the equation 4 sin θ + 3 csc θ = 7, we can simplify it to obtain a quadratic equation in terms of sin θ. By solving the quadratic equation, we find the values of θ that satisfy the equation within the given range.

For the equation 2 csc θ = 6 sin θ + 1, we can rearrange it and simplify to obtain a quadratic equation in terms of sin θ. Solving the quadratic equation yields the solutions for θ within the given range.

The equation 5 sin θ = 4(1 + cos θ) can be rewritten in terms of sin θ and cos θ. By manipulating the equation using trigonometric identities and simplifying, we arrive at a quadratic equation in terms of sin θ. Solving this equation provides the values of θ within the specified range.

To solve the equation 3 tan θ + 5 = 4 sec² θ, we can express tan θ in terms of sin θ and cos θ. By substituting the trigonometric identities and rearranging the equation, we obtain a quadratic equation in terms of sin θ. Solving this quadratic equation gives us the solutions for θ within the given range.

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Answer:

Step-by-step explanation:

Answer:

Step-by-step explanaton:

Answer:

31.8

Step-by-step explanation:

Answer:

Enrique spent D. $31.80 on soda.

Answer:

Faris' arrival time = 3:30 pm + 3 h 54 mnt = 7:24 pm

Answer:

19.24

Step-by-step explanation:

Answer:

120

Step-by-step explanation:

180 subtract 60 equals 120

Answer:

80% passed

Step-by-step explanation:

1. 24 out of 30 passed

2. In order to make it a % you have to multiply

3. 100 * 24/30

4. = 2400/30

5. Therefore the answer is 80

Answer:

The second choice. Connect the endpoint and the mark at the angle measure.

Step-by-step explanation:

Answer:

Connect the endpoint and the mark at the angle measure.

Step-by-step explanation:

Answer:

See below.

Step-by-step explanation:

5a)

In an isosceles triangle, the angles opposite the congruent sides are congruent.

m<A = 40 deg

m<B = 40 deg

m<A + m<B + m<C = 180

40 + 40 + m<C = 180

m<C = 100 deg

5b)

In a right triangle, one angle is a right angle.

m<A = 40 deg

m<B = 90 deg

m<A + m<B + m<C = 180

40 + 90 + m<C = 180

m<C = 50 deg

Answer:

A sum of RS 700 is to be used to give seven cash prizes to students of school for their overall academic performance. If each prize is RS 20 less then its preceding prize, find the value of each of the prizes.

dont know this stuff lol

Answer:

Horizontal shift of 4 units to the left.

Vertical translation of 8 units downward.  

Step-by-step explanation:

Given the quadratic function, y = (x + 4)² - 8, which represents the horizontal and vertical translations of the parent graph, y = x²:

The vertex form of the quadratic function is y = a(x - h)² + k

Where:

The vertex is (h , k), which is either the minimum (upward facing graph) or maximum (downward-facing graph).

The axis of symmetry occurs at x = h.

a =  determines whether the graph opens up or down, and makes the graph wider or narrower.

h = determines how far left or right the parent function is translated.

k = determines how far up or down the parent function is translated.

Going back to your quadratic function,

          y = (x + 4)² - 8

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Answer:

5 Bananas and 10 peaches.

Step-by-step explanation:

.4 x 5 = 2

.5 x 10 = 5

5 + 2 = 7

Answer: 126 in

I used a calculator. It matches.

hope this helps =)

Answer:

75.39 in^2

Step-by-step explanation:

circumference=2πr

30.8=2(3.14)r

30.8=6.28r

4.9=r

Now that we know the radius, we can plug it in for the area equation

area=πr^2

area=(3.14)(4.9)^2

area=(3.14)(24.01)

area=75.39 in^2

Answer:

the triangles are congruent

Step-by-step explanation:

method 1: Angle-Angle-Angle congruency test

angle BAC = angle DEC(given)

angle ABC = angle EDC

angle BCA = angle DCE

method 2: Side Angle Side Congruency test

BC = DC (both are 4 cm)

angle BCA = angle DCE (vertically opp. angles, straight lines intersect)

AC = EC (both are 6 cm)

side note:

1. when testing for congruency and giving the different congruency test, always ensure that the sides or angles mentioned are in corresponding order of the congruent triangles.

2. see that from the explanation given i do not anyhow mix up the sides with any sides or mess up points with any points(will show drawing of an example of what i mean)

pls dont mind the ugly handwriting as well since i dont have a stylus

a. the difference in the two data provided =31072940 - 24658823 = 6414117

b. the sum of the two data to the nearest thousand hundred = 31072940 - 24658823 = 55731763

nearest to Thousand hundred = 55730000