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Margarita [4]
3 years ago
15

If (44)x = 432, what is the value of x? 4 7 8 28

Mathematics
2 answers:
inessss [21]3 years ago
6 0
28 would be your vaule for x
postnew [5]3 years ago
3 0

44 x 4 = 176

44 x 7 =308

44 x 8 = 352

44 x 28 =1232


 none of those will =432

 are you sure you have the right numbers?


(44)x=432

x=432/44 = 9.8181


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Helpppppppp? please thank you
Snezhnost [94]

The x values are the same for both green dots, so you are looking at a simplified version of the distance formula.

d = sqrt( (x2 - x1)^2 + (y2 - y1)^2 )

Since x1 = x2 = 5, the first set of brackets = (5 - 5) = 0

d = sqrt( y2 - y1)^2 )

d = y2 - y1

y2 = 5

y1 = - 1

d = 5 - - 1

d = 6    Answer


8 0
3 years ago
Read 2 more answers
Need this quick!!!
Leya [2.2K]

A function assigns the values. The correct option is graph X.

<h3>What is a Function?</h3>

A function assigns the value of each element of one set to the other specific element of another set.

The function which is giving a constant term is the graph on the right, while the function where the value of x increases to positive infinity is the upper graph.

Hence, the correct option is graph X.

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7 0
2 years ago
Cora is using successive approximations to estimate a positive solution to
Advocard [28]
Um...... what......?
7 0
3 years ago
consider a wire 2 ft long cut into two pieces. one piece forms a circle with radius r and the other forms a square with side len
makkiz [27]

The formula for the radius r in terms of x . and for the maximum areas is x=2/\pi+4

Given that,

y forms a circle of radius r

y=2\pir

r=y/2\pi

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(2-y) = 4x

x=(2-y)/4

Now Sum of Area's=Area of Square +Area of Circle

Sum = \pir² + x²

Substitute the r and x values in above equation,

A(y)= y²/4\pi+(y-2)²/ 16

To maximize Area A(y)

A'(y)= 0

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Y max will be max, x to be maximum.

for maximum sum of areas,

x=2/\pi+4

Hence,The formula for the radius r in terms of x . and for the maximum areas is x=2/\pi+4

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4 0
1 year ago
Add and simplify
Tanya [424]

Answer:

Your 1st answer is correct

Step-by-step explanation:

\frac{1}{(x - 1)}  -  \frac{3}{(1 - x)}   + 2

\frac{1}{(x - 1)}  -  \frac{3}{ - (x - 1)}  + 2

\frac{1}{(x - 1)}   +  \frac{3}{(x - 1)}  + 2

\frac{4}{(x - 1)}  + 2

\frac{4 + 2(x - 1)}{x - 1}

\frac{4 + 2x - 2}{x - 1}

= \frac{2x  + 2}{x - 1}

=  \frac{2(x + 1)}{x - 1}

6 0
2 years ago
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