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seropon [69]
2 years ago
5

(-4,1)and(3,6) Find the slope of the line passing through the pair of points or state that the slope is undefined. Then indicate

whether the line through the poonts rises, falls, is horziontal, or is vertical.​
Mathematics
2 answers:
katen-ka-za [31]2 years ago
6 0

Answer:

Honestly, I never really figured how to do the slope but you can go to m a t h w a y . c o m   it should help you it helped me tons!!

Step-by-step explanation:

sorry if this is no help

Bezzdna [24]2 years ago
4 0

Answer:

The slope is 5/7 and it rises

Step-by-step explanation:

(-4,1) and (3,6)

6 - 1 / 3 - (-4)

5 / 3+4

= 5 / 7

and it rises since the answer is positive

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Clare spent $5.17 on apples. How many pounds of apples did Clare buy?
ZanzabumX [31]

Answer: complete the question please

Step-by-step explanation:

How much is a pound/lb? Then do 5.17 divided by (however much is a pound/lb)

6 0
3 years ago
If RG = 4X+12 and HI = 10x-15, then x =​
Zinaida [17]
T and g are the answer because you can’t do 10x and 15
6 0
2 years ago
Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer
agasfer [191]

Answer:

No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming:  the function is f(x)=x^{2} in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if  both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

g'(0)=g'(1)

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

g(x)=x^{2}\\g'(x)=2x\\ g'(0)=2(0) \Rightarrow g'(0)=0

g(x)=x^{2}\\g'(x)=2x\\ g'(1)=2(1) \Rightarrow g'(1)=2

g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)

3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Because this is the same as to calculate the limit from the left and right side, of g(x).

f'(c)=\lim_{x\rightarrow c}\left [\frac{f(b)-f(a)}{b-a} \right ]\\\\g'(0)=\lim_{x\rightarrow 0}\left [\frac{g(b)-g(a)}{b-a} \right ]\\\\g'(1)=\lim_{x\rightarrow 1}\left [\frac{g(b)-g(a)}{b-a} \right ]

This is what the Bilateral Theorem says:

\lim_{x\rightarrow c^{-}}f(x)=L\Leftrightarrow \lim_{x\rightarrow c^{+}}f(x)=L\:and\:\lim_{x\rightarrow c^{-}}f(x)=L

4 0
3 years ago
Simplify<br> -4 1/5 (-13 1/10)<br> pls help
oksian1 [2.3K]

Answer:

55 1/50 or 2751/50

Step-by-step explanation:

Convert both expressions to improper form

Make their denominators the same

Multiply the numerators

Simplify the fraction if needed

4 0
3 years ago
How many more poodles did ava observe than bulldogs?
Andreas93 [3]
You need to put the Answers and picture/problem before asking someone to answer it.
8 0
3 years ago
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