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3 years ago

Which graph represents a function with the initial value of 1/2 Graph 2

Mathematics

1 answer:

viktelen [127]3 years ago

5 0

Graph 2 because It’s looks like it’s 1/2

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Rewrite as a product of two factors: 3 + 18d + 21y

Vlada [557]

Answer:

the Expression is :

(3) × (1 + 6d + 7y)

Step-by-step explanation:

here's the solution : -

=》

$3 + 18d + 21y$

=》

$(3 \times 1 )+ (3 \times 6d) + (3 \times 7y)$

now, since 3 is common in all terms let's take it out.

=》

$3 \times (1 + 6d + 7y)$

=》so, ,we got the above expression as product of 3 and ( 1 + 6d + 7y ) as factors.

3 0

3 years ago

Read 2 more answers

Please help me answer it

Neko [114]

So for number 3, so for every 15 he earns, he saves 6 so
15:6=5:2

then he earns 75 and saves x
5 times what=75
5 times 15=75

2 times 15=30
so the ration is 75:30
he saved $30

then

write a real life proportion
so rob has a pattern to make,
for every 5 blue tiles, he puts down, he must put down 7 green tiles,
he needs to fill a sqare area that is 12 feet in legnth and width
each tile is 1 foot
how many blue tiles will he need
so 5+7=12
area=12 times 12
area=144
so the number of tiles is 144
so blue=5/12
so 144 times 5/12=60 blue tiles

3 0

3 years ago

Please need help !!!!!!

faltersainse [42]

Answer:

I think its C .

6 0

3 years ago

Please help asap !!!

pashok25 [27]

She would need to but 48 videos to get the cash prize. Hope this helps :)

7 0

3 years ago

ALGEBRA 2!!!!!!!!! SHOW YOUR WORK!!!!!!!!!!!!!<br> Do f(g(x)) and g(f(x))

bezimeni [28]

$\bf f(x)=\cfrac{2x-3}{x+1}~\hspace{10em}g(x)=\cfrac{x+3}{2-x}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
f(~~g(x)~~)\implies \cfrac{2[g(x)]-3}{[g(x)]+1}\implies \cfrac{2\left( \frac{x+3}{2-x} \right)-3}{\left( \frac{x+3}{2-x} \right)+1}\implies
\cfrac{\frac{2x+6}{2-x}-3}{\frac{x+3}{2-x}+1}
\\\\\\
\cfrac{\frac{2x+6-6+3x}{2-x}}{\frac{x+3+2-x}{2-x}}\implies \cfrac{2x+6-6+3x}{2-x}\cdot \cfrac{2-x}{x+3+2-x}\implies \cfrac{5x}{5}\implies x$

$\bf \rule{34em}{0.25pt}\\\\
g(~~f(x)~~)\implies \cfrac{[f(x)]+3}{2-[f(x)]}\implies \cfrac{\frac{2x-3}{x+1}+3}{2-\frac{2x-3}{x+1}}\implies \cfrac{\frac{2x-3+3x+3}{x+1}}{\frac{2x+2-(2x-3)}{x+1}}
\\\\\\
\cfrac{2x-3+3x+3}{x+1}\cdot \cfrac{x+1}{2x+2-(2x-3)}\implies \cfrac{2x-3+3x+3}{x+1}\cdot \cfrac{x+1}{2x+2-2x+3}
\\\\\\
\cfrac{5x}{5}\implies x$

and in case you recall your inverses, when f( g(x) ) = x, or g( f(x) ) = x, simply means, they're inverse of each other.

4 0

4 years ago