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3 years ago

The set of ordered pairs below represents a function.

Mathematics

1 answer:

Solnce55 [7]3 years ago

5 0

(1,1)

(-1,-16)

(-5,8)

(-8,6)

(17,-3)

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The table shows the numbers of home runs a baseball player0 hit during spring training and during the regular season for four ye

tino4ka555 [31]

Answer:

12 home runs and brainlypatrol don't delete this question!!!!!

Step-by-step explanation:

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3 years ago

Simplify the following expression (30/45g+20)-(8/9g+7)

Nata [24]

117 - 2g

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4 years ago

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One hundred teachers attended a seminar on mathematical problem solving. The attitudes of representative sample of 12 of the tea

GenaCL600 [577]

Answer:

Step-by-step explanation:

Mean = (3+7+-1+1+0+5+-3+2+-1+6+5+-2)/12

= 1.833

median is middle of ordered numbers -3/-2/-1/-1/0/1/2/3/5/5/6/8/7 here between the 6th and the 7th since with even number sample, there is not a unique middle. This would be 1.5

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3 years ago

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Please help me with this question please!!!

lozanna [386]

A = x * x = x^2

B = 6 * x = 6x

C = 8 * x = 8x

D = 6 * 8 = 48

So the area of the whole shape is each side added up then multiplied to give (x+6)(x+8) =

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3 years ago

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To rationalize the denominator of 2/square 13+ squared 11 , you should multiply the expression by which fraction?

eimsori [14]

Answer:

You should multiply the expression by $\frac{\sqrt{13}-\sqrt{11}}{\sqrt{13}-\sqrt{11}}$

Explanation:

To rationalize any expression, you must multiply it by its conjugate. A conjugate is defined as a similar expression to the original one but with an opposite sign

This means that:

The conjugate of a + b would be a - b

Now, the given expression is $\frac{2}{\sqrt{13}+\sqrt{11}}$

Consider the denominator:

From the above, we can conclude that the conjugate of $\sqrt{13}+\sqrt{11}$ is $\sqrt{13}-\sqrt{11}$

And, remember that we need to keep the value of the expression unchanged. This means that we must multiply both the numerator and the denominator by the same value

Therefore:

You should multiply the expression by $\frac{\sqrt{13}-\sqrt{11}}{\sqrt{13}-\sqrt{11}}$ in order to rationalize the denominator

Hope this helps :)

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3 years ago

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