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3 years ago

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5. I, II, III, IV, V, Conjecture:

1 answer:

NemiM [27]3 years ago

4 0

The conjecture is each term is the corresponding Roman numeral of the precious term

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(x + y + 2)(y + 1)<br> Solve

RideAnS [48]

Hello there! We are given the following equation and want to simplify it:

$(x+y+2)(y+1)$

First, let's use the Distributive Property:

$(a+b)(c)=ac+bc$

For this step, we are going to let:

$a=y$

$b=1$

$c=(x+y+2)$

After applying this version of the Distributive Property, we are given the equation:

$(y)(x+y+2)+(1)(x+y+2)$

We're going to use the distributive party again, but a slightly different version that's almost the same as first:

$(a)(b+c+d)=ab+ac+ad$

Using this on both terms of the equation changes the equation to:

$(xy+y^2+2y)+(x+y+2)$

Now, we just remove the parenthesis and combine like terms

$y^2+3y+xy+x+2$

This should be your answer. Let me know if you need any clarifications, thanks!

~ Padoru

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3 years ago

What is the inverse of y = x+3/4 ?<br><br> a. y=x-3/4<br> b. y=4x-3<br> c. y=4x+3<br> d. y= x-3/-4

viva [34]

Y = x + 3/4

x = y - 3/4

swap the variables:-

y = x - 3/4 (answer). Its a.

4 0

3 years ago

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Use linear approximation to approximate √25.3 as follows.

Sophie [7]

The idea is to use the tangent line to $f(x)=\sqrt x$ at $x=25$ in order to approximate $f(25.3)=\sqrt{25.3}$ .

We have

$f(x)=\sqrt x\implies f(25)=\sqrt{25}=5$

$f'(x)=\dfrac1{2\sqrt x}\implies f'(25)=\dfrac1{10}$

so the linear approximation to $f(x)$ is

$L(x)=f(5)+f'(5)(x-5)=5+\dfrac{x-5}{10}=\dfrac x{10}+\dfrac92$

Hence $m=\frac1{10}$ and $b=\frac92$ .

Then

$f(25.3)\approx L(25.3)=\dfrac{25.3}{10}+\dfrac92=\boxed{7.03}$

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3 years ago

PPPPPPPPPPLLLLLLLLLLZZZZZZ HELP!<br><br> A.<br> B.<br> C.<br> D.

BigorU [14]

The answer to the question is d

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3 years ago

PLEASE PLEASE PLEASE ANSWER ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

gtnhenbr [62]

The answer is times the 5500 adding the 6600 and as it adds up the box rate to 1.75 the answer is 1.75 insterest rate and 132000 dollars left

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3 years ago

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