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Georgia [21]
3 years ago
11

Solve for x

Mathematics
2 answers:
elena-s [515]3 years ago
6 0

Answer: X=-4

Mark me as brainlest please, have a great day! and if you need more help comment on here!

Olenka [21]3 years ago
5 0
X is equal to -4. Hope I helped. If you want an explanation too I would be glad to provide that .
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Give the quotient in simplest form: 5/9 divided by 1/3
elena55 [62]

Answer:

5/9÷⅓

reciprocate ⅓ into 3/1

=5/9×3/1

=15/9

simplify

=5/3

6 0
3 years ago
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Find x in the given circle where o is the centre of the circle?​
artcher [175]

Answer:

X as an angle will be equal to 140°

Step-by-step explanation:

As AO = BO, AOB is an Isosceles triangle which has 2 equal angles OAB=OBA=20°, therefore BOA is equal to 140°

x is the angle opposite to BOA which will have the same angle measure.

5 0
3 years ago
What is an outcome please help I need this 100
Y_Kistochka [10]

Answer:

the way a thing turns out; a consequence

Step-by-step explanation:

I looked it up

6 0
3 years ago
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Let f (x) = -3x + 4. Describe the transformations from the graph of f to the graphs of g(x) = f (x) + 3 and h(x) = f (x - 2).
iris [78.8K]

Answer:

Answer:

g(x)=-6/5x+1/2

h(x)=-6/5x-1/2

Step-by-step explanation:

1).   g(x)=−f(x) ?

f(x)=6/5x−1/2

g(x)=−(6/5x−1/2)

g(x)=-6/5x+1/2

2). h(x)=f(−x) ?

f(-x)=6/5(-x)−1/2

f(-x)=-6/5x-1/2

h(x)=-6/5x-1/2

Step-by-step explanation:

5 0
2 years ago
Calculate two iterations of Newton's Method for the function using the given initial guess. (Round your answers to four decimal
pickupchik [31]

Answer:

The first and second iteration of Newton's Method are 3 and \frac{11}{6}.

Step-by-step explanation:

The Newton's Method is a multi-step numerical method for continuous diffentiable function of the form f(x) = 0 based on the following formula:

x_{i+1} = x_{i} -\frac{f(x_{i})}{f'(x_{i})}

Where:

x_{i} - i-th Approximation, dimensionless.

x_{i+1} - (i+1)-th Approximation, dimensionless.

f(x_{i}) - Function evaluated at i-th Approximation, dimensionless.

f'(x_{i}) - First derivative evaluated at (i+1)-th Approximation, dimensionless.

Let be f(x) = x^{2}-8 and f'(x) = 2\cdot x, the resultant expression is:

x_{i+1} = x_{i} -\frac{x_{i}^{2}-8}{2\cdot x_{i}}

First iteration: (x_{1} = 2)

x_{2} = 2-\frac{2^{2}-8}{2\cdot (2)}

x_{2} = 2 + \frac{4}{4}

x_{2} = 3

Second iteration: (x_{2} = 3)

x_{3} = 3-\frac{3^{2}-8}{2\cdot (3)}

x_{3} = 2 - \frac{1}{6}

x_{3} = \frac{11}{6}

7 0
4 years ago
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