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Tutor profile: Didarul I.

Inactive
Didarul I.
Studying Pharmacology and Clinical Pharmacy at North South University
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Questions

Subject: Geometry

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Question:

If one of the short sides of a 45-45-90 triangle equals 5, how long is the hypotenuse?

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Didarul I.
Answer:

Using the Pythagorean theorem, x2 + y2 = h2. And since it is a 45-45-90 triangle the two short sides are equal. Therefore 52 + 52 = h2 . Multiplied out 25 + 25 = h2. Therefore h2 = 50, so h = √50 = √2 * √25 or 5√2.

Subject: Basic Math

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Question:

In January, Widget Company made 180,129 widgets. In February, the same company made 200,382 widgets. By what percentage did the amount of widgets made increase?

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Didarul I.
Answer:

To find the percent increase, use the following formula: (New value-Old valueOld value)×100 In the case of Widget Company, (200382−180129180129)×100=11.2 Therefore, the percent increase is 11.2%

Subject: Algebra

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Question:

Use algebra to solve the following system of linear equations: x=1+3y 6y+3x=−27

Inactive
Didarul I.
Answer:

Substitution can be used by solving one of the equations for either x or y, and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the y= form, and then set both equations equal to each other. Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable. For this problem, substitution makes the most sense because the first equation is already solved for a variable. We can substitute the expression that is equal to x, into the x of our second equation: 6y+3(1+3y)=−27 Next, we need to distribute and combine like terms: 6y+3+9y=−27 15y+3=−27 We are solving for the value of y, which means we need to isolate the y to one side of the equation. We can subtract 3 from both sides: 15y+3=−27 −3 −315y=−30 Then divide both sides by 15 to solve for y: 15y15=−3015 y=−2 Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both x and y values. Now that we have the value of yy, we can plug that value into the yy variable in one of our given equations and solve for x\textup: x=1+3(−2) x=1+(−6) x=−5 Our point of intersection, and the solution to the system of linear equations is (−5,−2).

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