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kvv77 [185]
3 years ago
12

Find the inverse of (1/3y)-7 Please show work!

Mathematics
1 answer:
Pani-rosa [81]3 years ago
7 0

Answer:

Step-by-step explanation:

if additive inverse then

(1/3y)+7

if multiplicative inverse then

3y-7

if both then

3y+7

if in this way

f(x)=(1/3x)-7

f(-x)=(-1/3)-7

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The function a(1) =1/2I^2 describes the area of an isosceles right triangle with leg I. Make a table of values for I=1,2,3 and 4
prohojiy [21]

Functions can be represented using equations, graphs and tables.

The function is given as:

a(l) = \frac 12 l^2

When l = 1, we have:

a(1) = \frac 12 \times 1^2

a(1) = 0.5

When l = 2, we have:

a(2) = \frac 12 \times 2^2

a(2) = 2.0

When l = 3, we have:

a(3) = \frac 12 \times 3^2

a(3) = 4.5

When l = 4, we have:

a(4) = \frac 12 \times 4^2

a(4) = 8.0

Represent the above results as a table, we have:

<u>l          a(l)</u>

1          0.5

2         2.0

3          4.5

4          8.0

Read more about tables and functions at:

brainly.com/question/13136492

8 0
3 years ago
Estimate the quotient what is 467.6 divided by 8
lorasvet [3.4K]
I’m thinking 58.45? if that makes sense

7 0
3 years ago
36 appointments booked daily. 4 out of 24 are canceled. how many are canceled daily
andrew-mc [135]

Answer:

8

Step-by-step explanation:

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3 years ago
Can someone please help me please
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Step-by-step explanation: math

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3 years ago
Find the limit if it exists lim x→0 sqrtx+7-sqrt7 over x
Triss [41]

Answer:

\frac{1}{ 2\sqrt{7} }

Step-by-step explanation:

\lim_{x\to 0}  \frac{ \sqrt{x + 7}  -  \sqrt{7} }{x}  \\  \\  = \lim_{x\to 0}  \frac{( \sqrt{x + 7}  -  \sqrt{7}) }{x}  \times  \frac{( \sqrt{x + 7}   +   \sqrt{7}) }{( \sqrt{x + 7}   +  \sqrt{7}) }  \\  \\   = \lim_{x\to 0}  \frac{( \sqrt{x + 7} )^{2}  -  (\sqrt{7})^{2}  }{x( \sqrt{x + 7}   +  \sqrt{7})}  \\  \\   = \lim_{x\to 0}  \frac{( {x + 7}  -  {7}) }{x( \sqrt{x + 7}   +  \sqrt{7})}   \\  \\ = \lim_{x\to 0}  \frac{ {\cancel x}}{\cancel x( \sqrt{x + 7}   +  \sqrt{7})} \\  \\ = \lim_{x\to 0}  \frac{ {1}}{\sqrt{x + 7}   +  \sqrt{7}}  \\  \\  =  \frac{1}{ \sqrt{0 + 7} +  \sqrt{7}  } \\  \\  =  \frac{1}{ \sqrt{7} +  \sqrt{7}  }  \\  \\  =  \frac{1}{ 2\sqrt{7} }

6 0
3 years ago
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