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Shkiper50 [21]
3 years ago
10

ormula1" title="\left \{ {{5x+3y=1} \atop {7x+3y=5}} \right." alt="\left \{ {{5x+3y=1} \atop {7x+3y=5}} \right." align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Elena-2011 [213]3 years ago
5 0

Answer:

(2, - 3)

Step-by-step explanation:

given the 2 equations

5x + 3y = 1 → (1)

7x + 3y = 5 → (2)

multiply all terms in (1) by - 1

- 5x - 3y = - 1 → (3)

add (2) and (3) term by term to eliminate the term in y

(7x - 5x) + (3y - 3y) = (5 - 1)

2x = 4 ( divide both sides by 2 )

x = 2

Substitute x = 2 in either (1) or (2) for corresponding value of y

(2) : 14 + 3y = 5 ( subtract 14 from both sides )

3y = - 9 ( divide both sides by 3 )

y = - 3

solution is (2, - 3)


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1) C. 4 - 3·i

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Step-by-step explanation:

1) Given that the real part of the complex number = 4

The imaginary of the complex number = -3

The general form of representing complex numbers is z = a + b·i, we have;

The binomial equivalent to the complex number is z = 4 - 3·i

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The second function, y = -(1/2)·x² also has a vertex (h, k) = (0, 0)

The coefficient, 'a' is positive, therefore, the graph opens up

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Therefore;

The second graph shares the same vertex, is inverted, and opens wider than the first graph

3) The given functions are;

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When x = 1, y = x² + 3 = 1 + 3 = 4

∴ When x = 1, y = 4

Second function;

When y = 4, y = 4 = (x - 2)² + 3

√(1) = x - 2

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∴ When x = 3, y = 4

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∴ When x = 2, y = 7

Second function;

When y = 7, y = 7 = (x - 2)² + 3

√4 = 2 = (x - 2)

x = 2 + 2 = 4

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∴ When x = 4, y = 7

Therefore, the second function, y = (x - 2)² + 3, has the x-value shifted 2 units to the right for a given value of 'y'

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The ratio of the corresponding length and width of figures B and B' are;

3/4.5 = 4/6

Therefore, figure B' is similar but not congruent to figure B

5) A rotation and a reflection are rigid transformations and therefore, the dimensions and measure of the original figure and the image are the same;

∴ m∠k' = m∠k.

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Step-by-step explanation:

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