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Advocard [28]
3 years ago
12

How to solve -6x + 3y = 18 Y=2x - 2

Mathematics
2 answers:
posledela3 years ago
6 0
Let's solve this by substitution.  Note that y is defined:  y = 2x - 2.  Subst. 2x - 2 for y in the other equation:  -6x + 3(2x - 2) = 18.

Then -6x + 6x - 6 = 18.  Unfortunately, -6 does not equal 18, so there is no solution.  The graphs of these two lines are parallel.
ioda3 years ago
6 0
-6x + 3(2x-2)= 18

-6x + 6x - 6 = 18

-6 ¥ 18
no solution
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14 POINTS 14 Points 14 points
viva [34]

Answer:

r = 10 cm

Step-by-step explanation:

r² = 8² + 6²

r² = 64 + 36 = 100

r = 10 cm

8 0
3 years ago
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An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t
USPshnik [31]
The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

\displaystyle\int_a^bf(x)\,\mathrm dx

Split up the interval [a,b] into n equal subintervals,

[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]

where a=x_0 and b=x_n. Each subinterval has measure (width) \dfrac{a-b}n.

Now denote the left- and right-endpoint approximations by L and R, respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are \{x_0,x_1,\cdots,x_{n-1}\}. Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, \{x_1,x_2,\cdots,x_n\}.

So, you have

L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)
R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)

Now let T denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)

Factoring out \dfrac12 and regrouping the terms, you have

T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)

which is equivalent to

T=\dfrac12\left(L+R)

and is the average of L and R.

So the trapezoidal approximation for your problem should be \dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2
4 0
3 years ago
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fiasKO [112]
The correct answer should be A.
4 0
3 years ago
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Kobotan [32]

Answer:

x=5

Explanation:

We need to get x to stand by itself on one side of the equation.

4x-2 = 3x+3  ← add 2 to each side

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vladimir1956 [14]

Answer:

162/9 = 18 minutes

Step-by-step explanation:

5 0
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