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kherson [118]
3 years ago
7

Enter the factor under the radical

ula1" title="(a - b) \sqrt{a - b} " alt="(a - b) \sqrt{a - b} " align="absmiddle" class="latex-formula">
​
Mathematics
2 answers:
kvasek [131]3 years ago
3 0

\\ \rm\longmapsto (a-b)\sqrt{a-b}

\\ \rm\longmapsto (a-b)(a-b)^{\frac{1}{2}}

\\ \rm\longmapsto (a-b)^{1+\dfrac{1}{2}}

\\ \rm\longmapsto (a-b)^{\dfrac{3}{2}}

irga5000 [103]3 years ago
3 0

Answer:

\dashrightarrow \: { \tt{(a - b) \sqrt{a - b} }} \\  \\ \dashrightarrow \: { \tt{ {(a - b)}^{1}  {(a - b)}^{ \frac{1}{2} } }}

• from law of indices:

{ \boxed{ \rm{ ({x}^{n} )(  {x}^{m} ) =  {x}^{(n + m)} }}}

therefore:

\dashrightarrow \: { \tt{ {(a - b)}^{(1 +  \frac{1}{2} )} }} \\  \\ \dashrightarrow \: { \tt{ {(a - b)}^{ \frac{3}{2} } }}

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Answer:

Part a) The temperature at noon was -3°F

Part b) The temperature at 3:00 pm was +2°F

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Step-by-step explanation:

The complete question is

When Leo woke up he saw that the temperature was -8°F. By noon the temperature has increased 5°F. Part a) What was the temperature at noon?

Part b) By 3:00 pm, the temperature increased by another 5 degrees Fahrenheit was the temperature at 3:00 pm positive or negative?

Part c) By 11:00 pm, the temperature had dropped 8°F. Was the

temperature at 11:00 PM positive or negative? Explain.​

Part a) we know that

You can use a number line to adds the numbers

-8+5=-3°F

therefore

The temperature at noon was -3°F

Part b) we know that

the temperature increased by another 5 degrees

You can use a number line to adds the numbers

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therefore

The temperature at 3:00 pm was +2°F

Part c) we know that

the temperature had dropped 8°F

You can use a number line to adds the numbers

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Hope this helped!! Have an amazing day :D

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