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djyliett [7]
3 years ago
10

Find three consecutive odd integers such that the sum of the largest and twice the smallest is 25. If x represents the smallest

integer, then which equation could be used to solve the problem? 2x + x + 2 = 25 2x + x + 4 = 25 2x + 2x + 4 = 25
Mathematics
1 answer:
Murrr4er [49]3 years ago
5 0
1. x
2. x + 2
3. x + 4

x + 4 + 2x = 25. Second option is the answer.


Combine like terms
3x + 4 = 25
Subtract 4 from both sides
3x = 21
Divide both sides by 3
x = 7
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stellarik [79]

The simplified expression for 6(2( y + x)) is 12y + 12x.

Accordin to the given question.

We have an expression

6(2( y + x))

Therefore,

The simplified expression for 6(2( y + x)) is given by

6(2(y + x))

= 6(2y + 2x)          (by distributive law)

= 12y + 12x              (by distributive law)

Hence, the simplified expression for 6(2( y + x)) is 12y + 12x.

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5 0
1 year ago
30 POINTS AVAILABLE
kherson [118]

Answer:

\large\boxed{(x-2)^2+(y-1)^2=34}

Step-by-step explanation:

The equation of a circle in standard form:

(x-h)^2+(y-k)^2=r^2

(h, k) - center

r - radius

We have the endpoints of the diameter: (-1, 6) and (5, -4).

Midpoint of diameter is a center of a circle.

The formula of a midpoint:

\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)

Substitute:

h=\dfrac{-1+5}{2}=\dfrac{4}{2}=2\\\\k=\dfrac{6+(-4)}{2}=\dfrac{2}{2}=1

The center is in (2, 1).

The radius length is equal to the distance between the center of the circle and the endpoint of the diameter.

The formula of a distance between two points:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Substitute the coordinates of the points (2, 1) and (5, -4):

r=\sqrt{(5-2)^2+(-4-1)^2}=\sqrt{3^2+(-5)^2}=\sqrt{9+25}=\sqrt{34}

Finally we have:

(x-2)^2+(y-1)^2=(\sqrt{34})^2

3 0
3 years ago
Find the least number In the data set 62,58,56,61,59,57
arlik [135]
56 really to easy ...........
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IV
nikitadnepr [17]
The answer is all real numbers
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3 years ago
3 determine the highest real root of f (x) = x3− 6x2 + 11x − 6.1: (a) graphically. (b) using the newton-raphson method (three it
Juliette [100K]

(a) See the first attachment for a graph. This graphing calculator displays roots to 3 decimal places. (The third attachment shows a different graphing calculator and 10 significant digits.)

(b) In the table of the first attachment, the column headed by g(x) gives iterations of Newton's Method. (For Newton's method, it is convenient to let the calculator's derivative function compute the derivative f'(x) of the function f(x). We have defined g(x) = x - f(x)/f'(x).) The result of the 3rd iteration is ...

... x ≈ 3.0473167

(c) The function h(x₁, x₂) computes iterations using the secant method. The results for three iterations of that method are shown below the table in the attachment. The result of the 3rd iteration is ...

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(d) The function h(x, x+0.01) computes the modified secant method as required by the problem statement. The result of the 3rd iteration is ...

... x ≈ 3.0477377

(e) Using <em>Mathematica</em>, the roots are found to be as shown in the second attachment. The highest root is about ...

... x ≈ 3.0466805180

_____

<em>Comment on these methods</em>

Newton's method can have convergence problems if the starting point is not sufficiently close to the root. A graphing calculator that gives a 3-digit approximation (or better) can help avoid this issue. For the calculator used here, the output of "g(x)" is computed even as the input is typed, so one can simply copy the function output to the input to get a 12-significant digit approximation of the root as fast as you can type it.

The "modified" secant method is a variation of the secant method that does not require two values of the function to start with. Instead, it uses a value of x that is "close" to the one given. For our purpose here, we can use the same h(x1, x2) for both methods, with a different x2 for the modified method.

We have defined h(x1, x2) = x1 - f(x1)(f(x1)-f(x2))/(x1 -x2).

6 0
3 years ago
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