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liberstina [14]
3 years ago
5

Mmmmmnmmmmzbejheheheiiwjehejri

Mathematics
2 answers:
elena-s [515]3 years ago
7 0

Are you okay? Why the spam?

Gnoma [55]3 years ago
7 0

Answer:

andcksjdbvi2efjvnkojedbciujefdmg

Step-by-step explanation:

wqdmbskjfbsdkmc oisdjcxnjwdcnefjnweijfbejfjef

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The last one I believe

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MAVERICK [17]

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8, 54

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2 years ago
Please help! I don’t know the answer
Maru [420]

a1 = 54

a2 = (1/3)a1 = (1/3) * 54 = 18

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etc.

y = 54(\dfrac{1}{3})^{x - 1}

7 0
3 years ago
Find the least common multiple of the pair of polynomials.<br> 3y2 - 48 and y + 4
alexdok [17]

Step-by-step explanation:

3y2 - 48

So the least common multiple is 3

3( y2 - 16)

y + 4 there cannot be a multiple of y

7 0
4 years ago
PLEASE HELP ME ANSWER. NEEDS TO HAVE AN EXPLANATION.
finlep [7]

Answer:

1) C. 4 - 3·i

2) D. The second graph shares the same vertex, is inverted, and opens wider than the first graph

3) C. y = (x - 2)² + 3 would shift right two units

4) B. Figure B' is congruent but not similar to figure B

5) A. m∠k' = m∠k

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

2) The first graph equation is y = 2·x²

When x = 1, y = 2 and when x = 2, y = 8

The vertex = (h, k)

Where;

h = -b/(2.a) and b = 0, a = 2

∴ h = 0/(2 × 2) = 0

h = 0

k = f(h) = f(0) = 2 × 0² = 0

k = 0

The vertex, (h, k) = (0, 0)

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

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

When x = 1, y = -1/2 and when x = 2, y = -2

Therefore, the second function is wider

Therefore;

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

3) The given functions are;

First function; y = x² + 3 and second function; y = (x - 2)² + 3

First function;

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

x = 3

∴ When x = 3, y = 4

First function;

When x = 2, y = x² + 3 = 4 + 3 = 7

∴ When x = 2, y = 7

Second function;

When y = 7, y = 7 = (x - 2)² + 3

√4 = 2 = (x - 2)

x = 2 + 2 = 4

x = 4

∴ 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'

4) The lengths of the sides of figure B are 3 by 4, the lengths of the sides of figure B' 4.5 by 6

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.

3 0
3 years ago
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