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2 years ago

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(3-4^2)^2+8 Bdhdhdhsjsjsj

2 answers:

mestny [16]2 years ago

3 0

Answer: 177

Step-by-step explanation:

1. 4x4=16

(3-16)^2+8

2. 3-16=-13

(-13)^2+8

3. -13x-13=169

169+8

4.169+8=177

177

skelet666 [1.2K]2 years ago

3 0

Answer:

177

Step-by-step explanation:

<u>PEMDAS</u>

The PEMDAS rule is an acronym representing the order of operations in math:

Parentheses
Exponents
Multiplication and Division (from left to right)
Addition and Subtraction (from left to right)

Given expression:

$(3-4^2)^2+8$

<u>Parentheses</u>

Begin by calculating the part inside the parentheses.

Following PEMDAS, calculate the <u>exponent</u> first:

$\implies (3-4^2)^2+8=(3-16)^2+8$

Then carry out the <u>subtraction</u>:

$\implies (3-16)^2+8=(-13)^2+8$

<u>Exponents</u>

<u /> $\textsf{Apply the exponent rule}: \quad (-a)^n=a^n, \quad \textsf{if }n \textsf{ is even}$

$\implies (-13)^2+8=169+8$

<u>Addition</u>

$\implies 169+8=177$

Therefore:

$\implies (3-4^2)^2+8=177$

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PIT_PIT [208]

There are 2 tangent lines that pass through the point

$y=\frac{1}{(-1+\sqrt{3)^2} } (x-1)+2$

and

$y=\frac{1}{(-1-\sqrt{3)^2} } (x-1)+2$

Explanation:

Given:

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The point-slope form of the equation of a line tells us that the form of the tangent lines must be:

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For the lines to be tangent to the curve, we must substitute the first derivative of the curve for $m$ :

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$\frac{dy}{dx} =\frac{x+1-x}{(x+1)^2}$

$\frac{dy}{dx}= \frac{1}{(x+1)^2}$

$m=\frac{1}{(x+1)^2}$ $[2]$

Substitute equation [2] into equation [1]:

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Because the line must touch the curve, we may substitute $y=\frac{x}{x+1}:$

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Solve for x:

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$x=-2$ ± $\sqrt{3}$

$x=-2$ ± $\sqrt{3}$ <em> </em> $and$ $x=-2-\sqrt{3}$

There are 2 tangent lines.

$y=\frac{1}{(-1+\sqrt{3)^2} } (x-1)+2$

and

$y=\frac{1}{(-1-\sqrt{3)^2} } (x-1)+2$

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