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lozanna [386]
3 years ago
12

Select the three number bonds that represent factor pairs of the num Number bond: 1 times 56 = 56. Number bond: 4 times 14 = 56.

Number bond: 5 times 6 = 56. Number bond: 7 times 8 = 56.
Mathematics
1 answer:
MaRussiya [10]3 years ago
8 0

Answer:

Number bond: 1 times 56 = 56.

Number bond: 4 times 14 = 56.

Number bond: 7 times 8 = 56.

Step-by-step explanation:

As given,

Number bond: 1 times 56 = 56.             .......(1)

Number bond: 4 times 14 = 56.             .......(2)

Number bond: 5 times 6 = 56.              .......(3)

Number bond: 7 times 8 = 56.              .......(4)

Equation (1) represents factor pair of the number  because

1 times 56 = 1× 56 = 56

It satisfies

Equation (2) represents factor pair of the number  because

4 times 14 = 4× 14 = 56

It satisfies

Equation (3) does not represents factor pair of the number  because

5 times 6 = 5× 6 = 30

It does not satisfies

Equation (4) represents factor pair of the number  because

7 times 8 = 7× 8 = 56

It satisfies

So, the correct answer is -Number bond: 1 times 56 = 56.

                                           Number bond: 4 times 14 = 56.

                                            Number bond: 7 times 8 = 56.

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Which table represents the graph of a logarithmic function in the form y=log3x when b>1?
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Answer:

<u><em>The satisfied table of the given function</em></u>y = log_{b} (x)<u><em></em></u>

<em>x                    1/8            1/4             1/2              1             2</em>

<em>y                    -3                 -2            -1               0               1</em>

<em></em>

Step-by-step explanation:

<u><em>Explanation</em></u> :-

Given logarithmic function y = log_{b} (x)   if b >1

Given first table

i)

put x = \frac{1}{8}     given b > 1 so we can choose b = 2

y = log_{2} (\frac{1}{8} )

y = log_{2} (2^{-3}  )

we will apply logarithmic formula

log x ⁿ = n log (x)

y = log_{2} (2^{-3}  ) = -3 log_{2} (2) = -3 (1) = -3

<em>y = -3</em>

<em>ii)</em>

<em>put x = </em>\frac{1}{4}<em>     given b > 1 so we can choose b = 2</em>

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<em></em>y = log_{2} (2^{-2}  )<em></em>

we will apply logarithmic formula

log x ⁿ = n log (x)

y = log_{2} (2^{-2}  ) = -2 log_{2} (2) = -2 (1) = -2

<em>y = -2</em>

<em>iii) </em>

<em>put x = </em>\frac{1}{2}<em>     given b > 1 so we can choose b = 2</em>

<em></em>y = log_{2} (\frac{1}{2} )<em></em>

y = log_{2} (2^{-1}  )

<em>we will apply logarithmic formula </em>

<em>log x ⁿ = n log (x)</em>

y = log_{2} (2^{-1}  ) = -1 log_{2} (2) = - (1) = -1

<em>y = -1</em>

<em>iv) </em>

<em>put x = 1     given b > 1 so we can choose b = 2</em>

<em></em>y = log_{2} (1 )<em> = 0</em>

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<em>v) </em>

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y = log_{2} (2 )

<em>y = 1</em>

<em></em>

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<em>y                    -3                 -2            -1               0               1</em>

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