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2 years ago

Please help like I really need this and no one is answering like please I really need it

Mathematics

2 answers:

const2013 [10]2 years ago

6 0

What’s the question and answer choices

Slav-nsk [51]2 years ago

4 0

Answer:

Row 1:

56×2² = 224 in²

25×3² = 225 in²

15×4² = 240 in²

Row 2:

22×1.5² = 49.5 in²

35×6² = 1260 in²

18×3.5² = 220.5 in²

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Approximately 0.8 of middle school students have cell phones. Which of the following fractions describes how many middle school

lana [24]

Answer: E. 4/5

Step-by-step explanation: You have to put the decimal (0.8) over its place value. So in this case 0.8 turns into 8/10. Then you must simplify 8/10, by dividing the numerator (8) and the denominator (10) by the greatest common factor (2). Which gives you the answer 4/5.

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2 years ago

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Illusion [34]

Answer: 0.714286

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2 years ago

What is the answer to this??/

Alja [10]

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Step-by-step explanation:

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2 years ago

Find the limit if it exists lim x→0 sqrtx+7-sqrt7 over x

Triss [41]

Answer:

$\frac{1}{ 2\sqrt{7} }$

Step-by-step explanation:

$\lim_{x\to 0} \frac{ \sqrt{x + 7} - \sqrt{7} }{x} \\ \\ = \lim_{x\to 0} \frac{( \sqrt{x + 7} - \sqrt{7}) }{x} \times \frac{( \sqrt{x + 7} + \sqrt{7}) }{( \sqrt{x + 7} + \sqrt{7}) } \\ \\ = \lim_{x\to 0} \frac{( \sqrt{x + 7} )^{2} - (\sqrt{7})^{2} }{x( \sqrt{x + 7} + \sqrt{7})} \\ \\ = \lim_{x\to 0} \frac{( {x + 7} - {7}) }{x( \sqrt{x + 7} + \sqrt{7})} \\ \\ = \lim_{x\to 0} \frac{ {\cancel x}}{\cancel x( \sqrt{x + 7} + \sqrt{7})} \\ \\ = \lim_{x\to 0} \frac{ {1}}{\sqrt{x + 7} + \sqrt{7}} \\ \\ = \frac{1}{ \sqrt{0 + 7} + \sqrt{7} } \\ \\ = \frac{1}{ \sqrt{7} + \sqrt{7} } \\ \\ = \frac{1}{ 2\sqrt{7} }$

6 0

2 years ago

Tan inverse X + tan inverse Y + tan inverse z=pie prove that X+Y+Z=xyz<br>

vfiekz [6]

Answer:

see explanation

Step-by-step explanation:

Given

$tan^{-1}$ x + $tan^{-1}$ y + $tan^{-1}$ z = π

let

$tan^{-1}$ x = A , $tan^{-1}$ y = B , $tan^{-1}$ z = C , so

x = tanA, y = tanB , z = tanC

Substituting values

A + B + C = π ( subtract C from both sides )

A + B = π - C ( take tan of both sides )

tan(A + B) = tan(π - C) = - tanC ( expand left side using addition identity for tan )

$\frac{tanA+tanB}{1-tanAtanB}$ = - tanC ( multiply both sides by 1 - tanAtanB )

tanA + tanB = - tanC( 1 - tanAtanB) ← distribute

tanA+ tanB = - tanC + tanAtanBtanC ( add tanC to both sides )

tanA + tanB + tanC = tanAtanBtanC , that is

x + y + z = xyz

3 0

2 years ago