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2 years ago

Find the two x-intercepts of the function f and show that f '(x) = 0 at some point between the two x-intercepts.

Mathematics

1 answer:

suter [353]2 years ago

7 0

Step-by-step explanation:

f(x) = x√(x+8)

When f(x) = 0,

0 = x√(x+8)

x = 0, x + 8 = 0

x = -8

f'(x) = (1)√(x+8) + x • ½(x+8)^(-½) = 0

0 = √(x+8) + 1/[2√(x+8)] (x)

-2(x+8) = x

-2x - 16 = x

x = -16/3

(-8 < -16/3 < 0)

Therefore, x-intercept of f'(x) = 0 is somewhere between the two x-intercepts, ranging from -8 to 0.

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A box contains 6 dozens noodles.Each packetcontains 175gof the of the noodles and the packet itself has weight of 10g . If the f

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Answer:

Weight of empty box = 680 gram

Step-by-step explanation:

Given:

Number of noodle = 6 dozen = 12(6) = 72

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Total weight = 14 kg = 14,000 gram

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2 years ago

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Answer:

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2 years ago

good evening! Can someone please answer this, ill give you brainliest and your earning 50 points. Would be very appreciated.

AVprozaik [17]

Answer:

Part 1

Set B

Part 2

We can use the FOIL method to multiply these expressions:

$\textsf{FOIL}: \quad(a+b)(c+d)=ac+ad+bc+bd$

However, as all the expression are in the format $(a+b)^2$ , we can use the shortcut: $(a+b)^2=a^2+2ab+b^2$

Part 3

Using the shortcut to find the products:

$\begin{aligned}\implies (x+5)(x+5)& =(x+5)^2\\ & =x^2+2(x)(5)+5^2\\ & =x^2+10x+25\end{aligned}$

$\begin{aligned}\implies (2x+9)(2x+9) & = (2x+9)^2\\ & = (2x)^2+2(2x)(9)+9^2\\ & = 4x^2+36x+81\end{aligned}$

$\begin{aligned}\implies (x+1)^2 & = x^2+2(x)(1)+1^2\\ & = x^2+2x+1\end{aligned}$

Part 4

A new example of a multiplication problem based on the structure of Set B is: $(3x+2)(3x+2)$

$\begin{aligned}\implies (3x+2)(3x+2) & = (3x+2)^2\\ & = (3x)^2+2(3x)(2)+2^2\\ & = 9x^2+12x+4\end{aligned}$

Part 5

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