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lukranit [14]
3 years ago
6

What is the method for this (X+1) squared

Mathematics
2 answers:
Alexeev081 [22]3 years ago
8 0
Distribution since you are squaring both
sladkih [1.3K]3 years ago
7 0
(X+1)^2
=(x+1)(x+1)
=(x*(x+1))+(x+1)
=x^2 +x+x+1
=x^2 +2x +1
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Find the perimeter of the shape. (Answer choice In pic)
Lana71 [14]
Answer: 49.2 cm



Explanation: I don’t know how to explain it without a visual so I hope you end up figuring it out! All I can say is: use Pythagorean’s theorem for that triangle to the left (if you know what I mean) to find the hypotenuse and then add all the sides up!

Have a great day!
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6 0
3 years ago
Can anyone find the distance?
BigorU [14]
Sqrt((y2-y1)^2)+(x2-x1)^2)
6 0
2 years ago
If limx→3f(x)=7, which of the following must be true? I. f is continuous at x = 3 II. f is differentiable at x = 3
Viktor [21]

Without knowing anything else about f(x), neither of these need be true.

Suppose

f(x)=|x-3|+7=\begin{cases}10-x&\text{for }x3\\0&\text{for }x=3\end{cases}

Then (I) isn't true because, while the limit exists as x\to3 and is equal to 7, we have f(3)=0\neq7, so f(x) is not continuous there.

(II) also true because f(x) is not differentiable at x=3; that is,

\displaystyle\lim_{x\to3^-}f'(x)=\lim_{x\to3}(10-x)'=\lim_{x\to3}(-1)=-1

but

\displaystyle\lim_{x\to3^+}f'(x)=\lim_{x\to3}(x+4)'=\lim_{x\to3}1=1

which means the derivative does not exist at x=3.

3 0
3 years ago
The height of a trapezoid can be expressed as x-4, while the bases can be expressed as x+4 and x+9. If the area of the trapezoid
hram777 [196]

Check the picture below.

\bf \textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h=height\\ a,b=\stackrel{bases}{parallel~sides}\\ \cline{1-1} h= x - 4\\ a = x+4\\ b=x+9\\ A=99 \end{cases}\implies 99=\cfrac{(x-4)[(x+4)+(x+9)]}{2} \\\\\\ 99=\cfrac{(x-4)[2x+13]}{2}\implies 198=\stackrel{\mathbb{FOIL}}{2x^2+5x-52}\implies 0=2x^2+5x-250

\bf 0=(2x+25)(x-10)\implies x= \begin{cases} ~~\begin{matrix} -25 \\[-0.6em]\cline{1-1}\\[-5pt]\end{matrix}~~\\ 10 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{larger base}}{x+9}\implies 10+9\implies 19

keeping in mind that "x" cannot be equal to -25, since that'd give us negative values on either base and the bases are a positive value.

7 0
2 years ago
Read 2 more answers
If you could count $1 per second, how many years would it take to count a 1 billion
pogonyaev
I do believe it's 273,972 years.
6 0
3 years ago
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