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never [62]
3 years ago
5

Is this right idk but i need work fir it

Mathematics
1 answer:
bija089 [108]3 years ago
7 0

Answer:

The answer is C 5/4

Step-by-step explanation:

When lines are parallel and you know there is a scalar factor, make the ratio by putting the original over the dilation 15:12 or 15/12 this is the ratio and after you need to reduce by dividing by the greatest common factor. 15 and 12's GCF is 3 so (15/3)/(12/3) = 5/4.

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More angels more help
hodyreva [135]
You mean angles? right?
4 0
3 years ago
For Cooper's lemonade recipe, 6 lemons are required to make 3 cups of lemonade. At
AnnZ [28]

Answer:

2 lemons per cup

Step-by-step explanation

If you have 6 lemons and you can make 3 cups you will need to divide

6/3 which is 2

so your answer is 2 lemons per cup

7 0
3 years ago
PLZ HELP MEE!!!<br> NO LINKS OR WEBSITES!!! YOU WILL GET REPORTED!!!
Mama L [17]

Answer:

90 in^{2}

Step-by-step explanation:

Base is an equilateral triangle

slant hight= 12 in

side of base= 5 in

so, perimeter of base = 3(5)=15 in

lateral area = 1/2 pl =1/2 (15)(12) in

=90 in²

--------

hope it helps..

have a great day!!

3 0
3 years ago
Read 2 more answers
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
Help.... Other question got deleted.
Helen [10]
-2/25 is the slope
x1=4
hope this helps!
7 0
3 years ago
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