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Setler [38]
3 years ago
9

Find the distance between the given points (-7,5) and (-8,4)

Mathematics
2 answers:
yulyashka [42]3 years ago
5 0

Answer:

1

Step-by-step explanation:

Rashid [163]3 years ago
4 0

Answer:

2 units

Step-by-step explanation:

5 and 4 are 1 unit away

-7 and -8 are also 1 unit away

1 + 1 = 2

Hope this helped!

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An 8 kg toddler is running at a speed of 10 m/s. How much energy does he have?
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6 0
3 years ago
Find the surface area of the prism.<br> 7 cm<br> 13 cm<br> 5 cm<br> 12 cm<br> The surface area is
Sveta_85 [38]

Answer:

please give more information

Step-by-step explanation:

8 0
2 years ago
Problem page rita drive 420 miles using 18 gallons of gas. at this rate, how many gallons of gas would she need to drive 357 mil
bulgar [2K]
The answer is 8,330 
because you have to take 420 and divied it by 18 and youll get 23.33333 then you times it to 357 

hope you are satisfied 
3 0
3 years ago
Compared to its parent function the transformation of the following : y= -3x - 3 + 4
astra-53 [7]
The answer is B
Hope it helps!!
6 0
3 years ago
If f(x)=2x+sinx and the function g is the inverse of f then g'(2)=
Alexxx [7]
\bf f(x)=y=2x+sin(x)&#10;\\\\\\&#10;inverse\implies x=2y+sin(y)\leftarrow f^{-1}(x)\leftarrow g(x)&#10;\\\\\\&#10;\textit{now, the "y" in the inverse, is really just g(x)}&#10;\\\\\\&#10;\textit{so, we can write it as }x=2g(x)+sin[g(x)]\\\\&#10;-----------------------------\\\\

\bf \textit{let's use implicit differentiation}\\\\&#10;1=2\cfrac{dg(x)}{dx}+cos[g(x)]\cdot \cfrac{dg(x)}{dx}\impliedby \textit{common factor}&#10;\\\\\\&#10;1=\cfrac{dg(x)}{dx}[2+cos[g(x)]]\implies \cfrac{1}{[2+cos[g(x)]]}=\cfrac{dg(x)}{dx}=g'(x)\\\\&#10;-----------------------------\\\\&#10;g'(2)=\cfrac{1}{2+cos[g(2)]}

now, if we just knew what g(2)  is, we'd be golden, however, we dunno

BUT, recall, g(x) is the inverse of f(x), meaning, all domain for f(x) is really the range of g(x) and, the range for f(x), is the domain for g(x)

for inverse expressions, the domain and range is the same as the original, just switched over

so, g(2) = some range value
that  means if we use that value in f(x),   f( some range value) = 2

so... in short, instead of getting the range from g(2), let's get the domain of f(x) IF the range is 2

thus    2 = 2x+sin(x)

\bf 2=2x+sin(x)\implies 0=2x+sin(x)-2&#10;\\\\\\&#10;-----------------------------\\\\&#10;g'(2)=\cfrac{1}{2+cos[g(2)]}\implies g'(2)=\cfrac{1}{2+cos[2x+sin(x)-2]}

hmmm I was looking for some constant value... but hmm, not sure there is one, so I think that'd be it
5 0
3 years ago
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