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Kamila [148]
3 years ago
14

ASAP PLEASE !!! Subtract g(x) from f(x)

28x%29%3D%20log%28x%2B2%29" id="TexFormula1" title="f(x)=\sqrt{x+7\\} \\g(x)= log(x+2)" alt="f(x)=\sqrt{x+7\\} \\g(x)= log(x+2)" align="absmiddle" class="latex-formula">
Mathematics
2 answers:
avanturin [10]3 years ago
7 0

\boxed{\large{\bold{\textbf{\textsf{{\color{blue}{Answer}}}}}}:)}

f(x)=\sqrt{x+7} \\g(x)= log(x+2)

we have to subtract g(x)from f(x)

  • \sf{\sqrt{x+7}-log(x+2)   }
Alecsey [184]3 years ago
4 0

Answer:

sqrt(x+7) - log(x+2)

Step-by-step explanation:

f(x) = sqrt(x+7)

g(x) log(x+2)

f(x) - g(x) = sqrt(x+7) - log(x+2)

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Lana71 [14]
B should be correct
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Answer:

CE = 17

Step-by-step explanation:

∵ m∠D = 90

∵ DK ⊥ CE

∴ m∠KDE = m∠KCD⇒Complement angles to angle CDK

In the two Δ KDE and KCD:

∵ m∠KDE = m∠KCD

∵ m∠DKE = m∠CKD

∵ DK is a common side

∴ Δ KDE is similar to ΔKCD

∴ \frac{KD}{KC}=\frac{DE}{CD}=\frac{KE}{KD}

∵ DE : CD = 5 : 3

∴ \frac{KD}{KC}=\frac{5}{3}

∴ KD = 5/3 KC

∵ KE = KC + 8

∵ \frac{KE}{KD}=\frac{5}{3}

∴ \frac{KC+8}{\frac{5}{3}KC }=\frac{5}{3}

∴ KC + 8 = \frac{25}{9}KC

∴ \frac{25}{9}KC - KC=8

∴ \frac{16}{9}KC=8

∴ KC = (8 × 9) ÷ 16 = 4.5

∴ KE = 8 + 4.5 = 12.5

∴ CE = 12.5 + 4.5 = 17

7 0
3 years ago
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Answer:

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Step-by-step explanation:

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Answer:

Option A

Step-by-step explanation:

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Option B

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Statement is False.

Option C

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Given statement is False.

Option D

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Answer:

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