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mixer [17]
3 years ago
7

What is the slope of the line that passes through the points (4,9) and (6,10)

Mathematics
1 answer:
max2010maxim [7]3 years ago
4 0

Answer: 2

Step-by-step explanation:

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Please answer I will make brainlest:)
Georgia [21]

Answer:

(3,-4)

Step-by-step explanation:

7 0
3 years ago
Please Help!!
GarryVolchara [31]

Answer:

D:) (2,2,) is the Answer

Step-by-step explanation:

Solve the following system:


{X - 2 Y = -2 | (equation 1)


{3 X - 2 Y = 2 | (equation 2)



Swap equation 1 with equation 2:


{3 X - 2 Y = 2 | (equation 1)


{X - 2 Y = -2 | (equation 2)



Subtract 1/3 × (equation 1) from equation 2:


{3 X - 2 Y = 2 | (equation 1)


{0 X - (4 Y)/3 = (-8)/3 | (equation 2)



Multiply equation 2 by -3/4:


{3 X - 2 Y = 2 | (equation 1)


{0 X+Y = 2 | (equation 2)



Add 2 × (equation 2) to equation 1:


{3 X+0 Y = 6 | (equation 1)


{0 X+Y = 2 | (equation 2)



Divide equation 1 by 3:


{X+0 Y = 2 | (equation 1)


{0 X+Y = 2 | (equation 2)



Collect results:


Answer:  {X = 2 , Y = 2

4 0
3 years ago
Read 2 more answers
An electrician must use 15 miles of electrical cable between electric poles. How many yards of cable will be used?
Dmitry [639]

Answer:

The 3rd option or 39,600 would be the correct answer.

Step-by-step explanation:

Hope this helps:)

7 0
3 years ago
Hannah put $2,000 in a savings
Elden [556K]

Answer:2000/180=11.11 p

Step-by-step explanation:

mark brainliest

6 0
3 years ago
Use Newton’s Method to find the solution to x^3+1=2x+3 use x_1=2 and find x_4 accurate to six decimal places. Hint use x^3-2x-2=
luda_lava [24]

Let f(x) = x^3 - 2x - 2. Then differentiating, we get

f'(x) = 3x^2 - 2

We approximate f(x) at x_1=2 with the tangent line,

f(x) \approx f(x_1) + f'(x_1) (x - x_1) = 10x - 18

The x-intercept for this approximation will be our next approximation for the root,

10x - 18 = 0 \implies x_2 = \dfrac95

Repeat this process. Approximate f(x) at x_2 = \frac95.

f(x) \approx f(x_2) + f'(x_2) (x-x_2) = \dfrac{193}{25}x - \dfrac{1708}{125}

Then

\dfrac{193}{25}x - \dfrac{1708}{125} = 0 \implies x_3 = \dfrac{1708}{965}

Once more. Approximate f(x) at x_3.

f(x) \approx f(x_3) + f'(x_3) (x - x_3) = \dfrac{6,889,342}{931,225}x - \dfrac{11,762,638,074}{898,632,125}

Then

\dfrac{6,889,342}{931,225}x - \dfrac{11,762,638,074}{898,632,125} = 0 \\\\ \implies x_4 = \dfrac{5,881,319,037}{3,324,107,515} \approx 1.769292663 \approx \boxed{1.769293}

Compare this to the actual root of f(x), which is approximately <u>1.76929</u>2354, matching up to the first 5 digits after the decimal place.

4 0
2 years ago
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