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Rashid [163]
3 years ago
6

A circle has radius 50 cm. Which number would be close to the area

Mathematics
1 answer:
ale4655 [162]3 years ago
7 0

Answer:

look do is gone eat

Step-by-step explanation:

ok

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What is the solution of the system for y-2x=8 and 16+4x2y
Leto [7]

Answer

y = 2x +8

Step-by-step explanation:

Add  2 x  to both sides of the first equation to get:

y = 8 + 2 x


Divide both sides of the second equation by  2  to get:

y = 8 + 2 x


The two equations determine the same conditions on  xy  so this system of equations has an infinite number of solutions, namely all points on the line:

y = 2 x + 8

6 0
4 years ago
Read 2 more answers
Find the slope between (3,7) and (−2,11)
gavmur [86]

Use Y2-y1/x2-x1

11-7/-2-3

4/-5

4 0
3 years ago
Convert 3/8 into percent show the work
Kipish [7]
To get the decimal think of 3/8 as a division question -3 divided by 8

3/8=0.375

now times the decimal by 100

0.375 * 100=37.5

so the answer is going to be 37.5%
4 0
3 years ago
Read 2 more answers
Mike and Beth both write the decimal 0.8 (0.88888...)as a fraction. Mike: 45 Beth: 89 Which student wrote the correct fraction?
fenix001 [56]

Answer:

see the explanation

Step-by-step explanation:

we have

0.888...

This is a <u>repeating decimal</u> (Is a decimal that has a digit, or a block of digits, that repeat over and over and over again without ever ending)

Convert to fraction number

Let

x=0.888...

10x=8.888...

Subtract 0.888... from 8.888... to remove the decimal

10x-x=8.888...-0.888...

9x=8

Solve for x

x=8/9

therefore

Mike fraction is incorrect

because 4/5=0.8

0.8 is a <u>terminating decimal </u>(It's a decimal with a finite number of digits)

Mike's mistake was considering the number as a terminating decimal instead of a repeating decimal

Beth is correct

because

If you divide 8/9

the result is 0.8888888...

3 0
3 years ago
Why are there two solutions for the equation |6 + y| = 2? Explain.
GaryK [48]

Solution, \left|6+y\right|=2\quad :\quad y=-8\quad \mathrm{or}\quad \:y=-4

Steps:

|f\left(y\right)|=a\quad \Rightarrow \:f\left(y\right)=-a\quad \mathrm{or}\quad \:f\left(y\right)=a, 6+y=-2\quad \quad \mathrm{or}\quad \:\quad \:6+y=2

6+y=-2\quad :\quad y=-8,\\6+y=-2,\\\mathrm{Subtract\:}6\mathrm{\:from\:both\:sides}, 6+y-6=-2-6,\\\mathrm{Simplify}, y=-8

6+y=2\quad :\quad y=-4,\\6+y=2,\\\mathrm{Subtract\:}6\mathrm{\:from\:both\:sides},6+y-6=2-6\\\mathrm{Simplify},y=-4

\mathrm{Combine\:the\:ranges}, y=-8\quad \mathrm{or}\quad \:y=-4

\mathrm{The\:Correct\:Answer\:is\:Because\:of\:the\:absolute\:value,\:It\:could\:be\:Positive\:or\:negative.}

\mathrm{Hope\:This\:Helps!!!}

\mathrm{-Austint1414}

7 0
4 years ago
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