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aalyn [17]
3 years ago
7

I got to have some help on this question...

Mathematics
2 answers:
dimaraw [331]3 years ago
4 0
12 inches is the length of the larger square's side
a_sh-v [17]3 years ago
4 0
I was going to try to do it some fancy equation way, but i couldn't seem to manipulate any of my equations just the right way. 


But.....I did come across the answer by chance while testing something in my calculator.

so, if y is the length of the original side of the square and x is the new length:
x+2=y

4x=perimeter of original square
4x=10+y
or
4x=10+(x+2)

I plugged 4 into this by chance and look what we have from that:
4x=10+(x+2)
4•4=10+(4+2)
16=10+6
16=16
Bingo!!! it's a match!

so the new length of one side is 6 inches.
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Use the alternative form of the derivative to find the derivative at x = c (if it exists). (If the derivative does not exist at
KIM [24]

Answer:

The derivative of the function does not exist.

Step-by-step explanation:

The alternative form of a derivative is given by:

f'(c)= \lim_{x \to c} \dfrac{f(x)-f(c)}{x-c}

Our function is defined as:

h(x)=|x+8|

i.e.  h(x)=   -(x+8) when x+8<0

           and   x+8  when x+8≥0

i.e. h(x)=  -x-8   when x<-8

      and   x+8  when x≥-8

Hence now we find the derivative of the function at c=-8

i.e. we need to find the Left hand derivative (L.H.D.) and Right hand derivative (R.H.D) of the function.

The L.H.D at a point 'a' is calculated as:

\lim_{x \to a^-} \dfrac{f(x)-f(a)}{x-a}\\\\=\lim_{h \to0} \dfrac{f(a-h)-f(a)}{a-h-a}= \lim_{h\to 0}  \dfrac{f(a-h)-f(a)}{-h}

Similarly R.H.D is given by:

\lim_{x \to a^+} \dfrac{f(x)-f(a)}{x+a}\\\\=\lim_{h \to 0} \dfrac{f(a+h)-f(a)}{a+h-a}= \lim_{h\to 0} \dfrac{f(a+h)-f(a)}{h}

Now for L.H.D we have to use the function h(x) =-x-8

and for R.H.D. we have to use the function h(x)=x+8

L.H.D.

we have a=-8

\lim_{x \to (-8)^-} \dfrac{h(x)-h(-8)}{x-(-8)}\\\\= \lim_{h \to0} \dfrac{h(-8-h)-h(-8)}{-8-h-(-8)}= \lim_{h\to 0} \dfrac{h(-8-h)-h(-8)}{-h}

= \lim_{h \to 0} \dfrac{8+h-8-0}{-h}= \lim_{h \to 0}\dfrac{h}{-h}=-1

similarly for R.H.D.

\lim_{x \to (-8)^+} \dfrac{h(x)-h(-8)}{x-(-8)}\\\\=\lim_{h \to 0} \dfrac{h(-8+h)-h(-8)}{-8+h-(-8)}= \lim_{h\to 0} \dfrac{h(-8+h)-h(-8)}{h}

\lim_{h \to 0} \dfrac{-8+h+8-0}{h}=\lim_{h \to 0}\dfrac{h}{h}=1

Now as L.H.D≠R.H.D.

Hence, the function is not differentiable.



4 0
2 years ago
What is the value of x?<br><br><br><br> x over 4 equals 14 over 16<br><br><br><br> x =
FinnZ [79.3K]
The answer is 3.5. use cross product multiplication
8 0
3 years ago
Read 2 more answers
What are the coordinates of the endpoints A'B after a reflection of segment AB over the line y = - x?
kupik [55]

Answer:

by reflection, if you mean swapped, then the answers are A'(1,-6) and B'(4,2)

i hope these are the answers you are looking for

Step-by-step explanation:

6 0
3 years ago
Can you help me please
Nitella [24]

Answer:

i cant read that

Step-by-step explanation:

8 0
2 years ago
Does anyone know how to solve with elimination and if so can you do this problem
Kryger [21]
Ok I will show my work in the comments
3 0
3 years ago
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