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Ugo [173]
3 years ago
10

Jeremy determines that \sqrt{9} = 9^{\frac{1}{2}} . Part of his work is shown. \sqrt{9} = 3 = 3^1 = 3^{\frac{1}{2}+\frac{1}{2}}

= _________ = 9^{\frac{1}{2}} Which expression or equation should be placed in the blank to correctly complete Jeremy's work?
Mathematics
1 answer:
daser333 [38]3 years ago
7 0

3¹

Step-by-step explanation:

Jeremy determines that \sqrt{9} =9^{\frac{1}{2} }

His work should show \sqrt{9} =3=3^{1} =3^{\frac{1}{2} +\frac{1}{2} } =3^{1} =9^{\frac{1}{2} }

The part that adds  \frac{1}{2} +\frac{1}{2} =1 is missing

Thus two halves equals a whole, then it will be 3¹

Learn More

Indices : brainly.com/question/12720046

Keyword : Expressions, indices

#LearnwithBrainly

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Brainly.com
Slav-nsk [51]
The answer is a, because the absolute value of x(the actual weight of the product) minus the target weight of 18.25 needs to be greater that the target weight difference of .36 so that it is outside the range.  So basically, any number outside of the range will be correct.You can plug in number that you know are in/ out of the acceptable range of numbers for this problem in.  
|18.62-18.25|>.36
.37>.36
18.62 is not in range.





7 0
3 years ago
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The point (-5, 3) is reflected across the x axis. What are the coordinates of this reflection? Group of answer choices
harina [27]

Answer:

( -5, -3)

Step-by-step explanation:

8 0
3 years ago
What value of x is in the solution set of 8x – 6 > 12 + 2x?
Harman [31]

Answer:

8x - 6 > 12 + 2x \\ 8x - 2x > 12 + 6 \\ 6x > 18 \\  \frac{6x}{6}  >  \frac{18}{6}  \\ x > 3

<em>hope</em><em> </em><em>this</em><em> </em><em>helps</em><em> </em><em>you</em><em>.</em><em>.</em><em>.</em><em>!</em>

3 0
3 years ago
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Round each number to the place of the underlined digit the 4 is underlined in 324,650
Otrada [13]

Answer: my bad 324,000 isnt the answer. Its 325,000


Step-by-step explanation:


4 0
3 years ago
Draw at least six different sized rectangles that have an area of 64 square units
olga_2 [115]

Let

x------> the length of the rectangle

y------> the width of the rectangle

we know that

The area of a rectangle is equal to

A=x*y

A=64\ units^{2}

so

x*y=64 --------> equation 1

let's assume different values of x to get the different values of y

<u>case 1)</u> For x=64 units

substitute in the equation 1

x*y=64

y=64/x

y=64/64

y=1\ units      

the dimensions of the rectangle are 64 units x 1 unit

see the draw in the attached figure N 1

<u>case 2)</u> For x=32 units

substitute in the equation 1

x*y=64

y=64/x

y=64/32

y=2\ units

the dimensions of the rectangle are 32 units x 2 units

see the draw in the attached figure N 2

<u>case 3)</u> For x=16 units  

substitute in the equation 1

x*y=64

y=64/x

y=64/16

y=4\ units

the dimensions of the rectangle are 16 units x 4 units

see the draw in the attached figure N 3

<u>case 4)</u> For x=30 units

substitute in the equation 1

x*y=64      

y=64/x

y=64/30

y=\frac{32}{15} =2\frac{2}{15} \ units

the dimensions of the rectangle are 30 units x 2 (2/15) units

see the draw in the attached figure N 4

<u>case 5)</u> For x=40 units

substitute in the equation 1

x*y=64    

y=64/x

y=64/40

y=1.60\ units

the dimensions of the rectangle are 40 units x 1.60 units

see the draw in the attached figure N 5

<u>case 6)</u> For x=60 units

substitute in the equation 1

x*y=64    

y=64/x

y=64/60

y=\frac{16}{15} =1\frac{1}{15} \ units

the dimensions of the rectangle are 60 units x 1 (1/15) units

see the draw in the attached figure N 6

3 0
3 years ago
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