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coldgirl [10]
3 years ago
13

SOS!!

Mathematics
1 answer:
wlad13 [49]3 years ago
4 0

The only difference between the two following expressions is their exponent though it’s equal. If you would analyze carefully, the two expressions are equal since 2/4 is just equal to ½. The two expressions are equal in value so the presentation of the exponent is their only difference.

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One acute angle of a right triangle is 12 degrees more than twice the other acute angle. Fine the acute angles of the right tria
Svetlanka [38]
Let one acute angle be X and one be Y
X+Y=90  -------Eq.1
2X+12=Y
2X-Y=-12------Eq.2
solving eq 1&2 we get,
3x=78
∴X=26
substituting value X in equation.1
X+Y=90
Y=90-26
∴Y=64
⇒answer:- X=26°
                   Y=64°


8 0
3 years ago
Solve the equation<br> 42x + 7 = 16
ehidna [41]

Answer:

should get 0.2 hope this helps you!

Step-by-step explanation:

7 0
3 years ago
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-BARSIC- [3]

Answer:

Assuming you mean until the lunch money balance runs out,

$2.20x = $23

Step-by-step explanation:

You spend $2.20 every day.  If you are trying to find how many days until the lunch balance runs out, you need to put in X as your variable.

The last step is to set it equal to $23 dollars to be able to factor out the answer.


Hope this Helped!

4 0
3 years ago
Please help on this khan problem i can't word it right so there is attachments for the introductions and options for answers
Gwar [14]

Answer:

c

Step-by-step explanation:

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7 0
3 years ago
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Use the data in the table to answer the question. Citations are "speeding tickets." You may fill in the table to help you answer
Vikentia [17]

Answer:

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Step-by-step explanation:

We are given the following information in the question:

Number of citations:      5      7.5      10      15       20

 Outputs Residuals:       3       6       10       5          6

We have to find the linear approximation of the data that passes through the points (5, 3) and (20, 6).

Linear approximation is given by:

The equation of line is given by:

(y-y_1) = \displaystyle\frac{y_2-y_2}{x_2-x_1}(x-x_1)

where, (x_1,y_1), (x_2.y_2) is the point through which the line passes.

The equation of line is:

(y-3) = \displaystyle\frac{6-3}{20-5}(x-5)\\\\15(y-3)= 3(x-5)\\15y = 3x-15+45\\15y = 3x + 30\\y = \frac{x}{5}+2

The above equation is the required linear approximation.

8 0
3 years ago
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