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Mnenie [13.5K]
3 years ago
7

I think the answer is y=8x but I’m not sure and neither are my parents. My parents told me to look it up but I can’t find any he

lp other than this. Please answer soon!

Mathematics
1 answer:
Ugo [173]3 years ago
5 0
Y = 8x Yea that’s right
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At Kennedy school there are 37 girls and 36 boys in the third grade.How many students are in the third grade at Kennedy School?
zubka84 [21]
37 + 36 = 73

There are 73 students in the third grade at Kennedy School.
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A coordinate grid with one line labeled y equals 3 x plus StartFraction 3 over 4 EndFraction. The line passes through points at
Mademuasel [1]

Answer:

the  answer is c

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Step-by-step explanation:

6 0
3 years ago
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Rearrange x = 7y - 5 to make y the subject.
nalin [4]

By rearranging x = 7y - 5 to make y the subject, we get

x + 5 = 7y ( By transposition method )

or, \huge \purple {\tt {\frac{x + 5}{7}  = y}}

<h2>Answer:</h2>

\huge \pink {\tt {y =  \frac{x + 5}{7}}}

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If the perimeter is 102, how long is the longest side?
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8 0
2 years ago
If f(x)=x^3-x+2, then (f^-1)'(2)
yawa3891 [41]

Note that f(x) as given is <em>not</em> invertible. By definition of inverse function,

f\left(f^{-1}(x)\right) = x

\implies f^{-1}(x)^3 - f^{-1}(x) + 2 = x

which is a cubic polynomial in f^{-1}(x) with three distinct roots, so we could have three possible inverses, each valid over a subset of the domain of f(x).

Choose one of these inverses by restricting the domain of f(x) accordingly. Since a polynomial is monotonic between its extrema, we can determine where f(x) has its critical/turning points, then split the real line at these points.

f'(x) = 3x² - 1 = 0   ⇒   x = ±1/√3

So, we have three subsets over which f(x) can be considered invertible.

• (-∞, -1/√3)

• (-1/√3, 1/√3)

• (1/√3, ∞)

By the inverse function theorem,

\left(f^{-1}\right)'(b) = \dfrac1{f'(a)}

where f(a) = b.

Solve f(x) = 2 for x :

x³ - x + 2 = 2

x³ - x = 0

x (x² - 1) = 0

x (x - 1) (x + 1) = 0

x = 0   or   x = 1   or   x = -1

Then \left(f^{-1}\right)'(2) can be one of

• 1/f'(-1) = 1/2, if we restrict to (-∞, -1/√3);

• 1/f'(0) = -1, if we restrict to (-1/√3, 1/√3); or

• 1/f'(1) = 1/2, if we restrict to (1/√3, ∞)

6 0
2 years ago
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