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Mumz [18]
3 years ago
15

I need help solving the problem.

Mathematics
1 answer:
Slav-nsk [51]3 years ago
4 0

Answer:

You're trying to find the equation for a portion of a line. The first thing I notice is that the slope is negative. The slope is -1/2 and the y-intercept would be -3. That means the answer is either A or B. X is greater than -2, so the answer is B.

I hope that helps.

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The graph below belongs to which function family ?
Phoenix [80]
This is an absolute value function.
6 0
3 years ago
Read 2 more answers
Please help!!!!!!!!!!!!!!!!!!
vlabodo [156]

Answer:

57°

Step-by-step explanation:

The sum of both angles is 90°

3x+5x+2=90

8x+2=90

8x=88

x=11

Plug x into the equation

Larger Angle

5(11)+2=57

Smaller Angle

3(11)=33

4 0
3 years ago
Read 2 more answers
PLEASE HELP ASAP! BRAINLIEST TO BEST/RIGHT ANSWER
Alex
I think it is x to the 14th power

8 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
Please help.<br> Explain please.
Sunny_sXe [5.5K]
When n = 1, 2(1) + 1 = 3

When n = 2, 2(2) + 1 = 5

When n = 3, 2(3) + 1 = 7

When n = 4, 2(4) + 1 = 9

When n = 5, 2(5) + 1 = 11

Sum = 3 + 5 + 7 + 9 + 11 = 35

----------------------------------------------------
Answer: 35
----------------------------------------------------
4 0
3 years ago
Read 2 more answers
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