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Delvig [45]
3 years ago
6

[PLEASE HELP] in the function above, the slope of it will be multiplied by 1/2 and it’s y value of its y intercept will be incre

ased by 3 units, which of the graphs below best shows the new function???

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

Answer:

The graph at the bottom left in your group of possible answers.

Step-by-step explanation:

Notice that the original given graph corresponds to the equation:

y=2x+1

since the line's slope is 2/1 = 2 and the y-intercept is at the point (0, 1).

So if one modifies the equation multiplying the current slope by 1/2, and the y intercept increased by 3 units, Then the new function would be:

y=x+4

A line of slope 1 and y-intercept at (0, 4)

Notice that the graph at the bottom left in your possible answers is representing such function.

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How do I determine z ∈ C:
saw5 [17]

Simplify the coefficient of z on the left side. We do this by rationalizing the denominators and multiplying them by their complex conjugates:

\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{3-2i}{1+i}\cdot\dfrac{1-i}{1-i} - \dfrac{5+3i}{1+2i}\cdot\dfrac{1-2i}{1-2i}

\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{(3-2i)(1-i)}{1-i^2} - \dfrac{(5+3i)(1-2i)}{1-(2i)^2}

\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{3 - 2i - 3i + 2i^2}{1-(-1)} - \dfrac{5 + 3i - 10i - 6i^2}{1-4(-1)}

\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{3 - 5i + 2(-1)}2 - \dfrac{5 - 7i - 6(-1)}5

\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{1 - 5i}2 - \dfrac{11 - 7i}5

\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{1 - 5i}2\cdot\dfrac55 - \dfrac{11 - 7i}5\cdot\dfrac22

\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{5 - 25i - 22 + 14i}{10}

\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = -\dfrac{17 + 11i}{10}

So, the equation is simplified to

-\dfrac{17+11i}{10} z = \dfrac12 - \dfrac{2i}5

Let's combine the fractions on the right side:

\dfrac12 - \dfrac{2i}5 = \dfrac12\cdot\dfrac55 - \dfrac{2i}5\cdot\dfrac22

\dfrac12 - \dfrac{2i}5 = \dfrac{5-4i}{10}

Then

-\dfrac{17+11i}{10} z = \dfrac{5-4i}{10}

reduces to

-(17+11i) z = 5-4i

Multiply both sides by -1/(17 + 11i) :

\dfrac{-(17+11i)}{-(17+11i)} z = \dfrac{5-4i}{-(17+11i)}

z = -\dfrac{5-4i}{17+11i}

Finally, simplify the right side:

-\dfrac{5-4i}{17+11i} = -\dfrac{5-4i}{17+11i} \cdot \dfrac{17-11i}{17-11i}

-\dfrac{5-4i}{17+11i} = -\dfrac{(5-4i)(17-11i)}{17^2-(11i)^2}

-\dfrac{5-4i}{17+11i} = -\dfrac{85 - 68i - 55i + 44i^2}{289-121(-1)}

-\dfrac{5-4i}{17+11i} = -\dfrac{85 - 68i - 55i + 44(-1)}{410}

-\dfrac{5-4i}{17+11i} = -\dfrac{41 - 123i}{410}

-\dfrac{5-4i}{17+11i} = -\dfrac{41 - 41\cdot3i}{410}

-\dfrac{5-4i}{17+11i} = -\dfrac{1 - 3i}{10}

So, the solution to the equation is

z = -\dfrac{1-3i}{10} = \boxed{-\dfrac1{10} + \dfrac3{10}i}

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3n+3=234\\
3n=231\\
n=77\\
n+1=78\\
n+2=79

77,78,79
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