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amm1812
3 years ago
14

A function is represented by these ordered pairs.

Mathematics
2 answers:
kobusy [5.1K]3 years ago
8 0

Answer:

literally just 4

Step-by-step explanation:

Crazy boy [7]3 years ago
4 0

answer: 4

Step-by-step explanation:

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A pencil cost $1.20 and a pen cost $2.90. Jasmiah bought an equal number of such pencils and pens. She spent $32.30 more on the
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If(√14/√7-2)-(√14/√7+2)=a√7+b√2 find the values of a and b where a and b are rational numbers​
seraphim [82]

Answer:

  • a = 4/3 and b = 0

============================

<h2>Given expression:</h2>

\dfrac{\sqrt{14} }{\sqrt{7}-2} -\dfrac{\sqrt{14} }{\sqrt{7}+2}

<h2>Simplify it in steps:</h2>

<h3>Step 1</h3>

Bring both fractions into common denominator:

\dfrac{\sqrt{14} (\sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)} - \dfrac{\sqrt{14} (\sqrt{7}-2)}{(\sqrt{7}-2)(\sqrt{7}+2)}

<h3>Step 2</h3>

Simplify:

\dfrac{\sqrt{14} ((\sqrt{7}+2) - (\sqrt{7}-2))}{(\sqrt{7}-2)(\sqrt{7}+2)} =

\dfrac{\sqrt{14} (\sqrt{7}+2 - \sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)} =

\dfrac{4\sqrt{14} }{(\sqrt{7}-2)(\sqrt{7}+2)} =

\dfrac{4\sqrt{14} }{(\sqrt{7})^2-2^2} =

\dfrac{4\sqrt{14} }{7-4} =

\dfrac{4}{3}  \sqrt{14} }

<h3>Step 3</h3>

Compare the result with given expression to get:

  • a = 4/3 and b = 0

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Step-by-step explanation:

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e

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