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Lindsey S.
3+ years of tutoring experience, majoring in Secondary Education-Mathematics
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Geometry
TutorMe
Question:

1.) What is the sum of the interior angles of a regular hexagon? 2.) If a 45-45-90 triangle has a hypotenuse with the length of 8, what are the measures of the two legs? 3.) If a circle has an inscribed angle of 48 degrees, what is the measure of the intercepted arc?

Lindsey S.
Answer:

1.) Use the formula $$ (N-2)(180) $$ to find the sum of the interior angles: $$ (6-2)(180) $$ $$ (4)(180)$$=$$720$$ 2. Step 1: Set $$L\sqrt{2}=hypotenuse$$ and solve/simplify $$L\sqrt{2}=8$$ Step 2: Solve for "$$L$$" $$L=\frac{8}{\sqrt{2}}$$ $$L=\frac{8}{\sqrt{2}}*\frac{\sqrt{2}}{\sqrt{2}}$$=$$\frac{8\sqrt{2}}{\sqrt{4}}$$=$$\frac{8\sqrt{2}}{2}$$=$$4\sqrt{2}$$ 3.) The measure of an intercepted arc is the measure of the inscribed angle times 2, therefore: $$48*2=96 degrees$$

English
TutorMe
Question:

1.) Define the prepositional phrase in the following sentence: The English notes in Lindsey's binder are outdated. 2.) Indicate whether "who" or "whom" should be inserted in the blank: The girl ______ you met yesterday is moving away. 3.) Indicate which word in the sentence is an adjective and which word is an adverb: Lindsey handled the breakable toy very carefully.

Lindsey S.
Answer:

1.) in Lindsey's binder (describes which notes we were talking about) 2.) whom 3.) "breakable" is the adjective and "very" is the adverb

Algebra
TutorMe
Question:

1.) Solve the equation: $$ 5(-3x-2)-(x-3)=-4(4x+5)+13 $$ 2.) Solve the equation: $$ 2x=\sqrt{x+3} $$ 3.) Find the slope: $$ (-2,4) , (1, 10) $$

Lindsey S.
Answer:

1.) Step 1: Distribute/multiply the factors $$ -15x-10-x+3=-16x-20+13 $$ Step 2: Group the like terms $$ -16x-7=-16x-7 $$ Step 3: Add $$ 16x+7 $$ to both sides of the equation, therefore $$ 0=0 $$ Therefore, the statement is true for all real number values of x 2.) Step 1: Square both sides of the equation $$ (2x)^2=(\sqrt{x+3})^2 $$ $$ 4x^2=x+3 $$ Step 2: Move everything to one side of the equation (equal sign) so 0= $$ 4x^2-x-3=0 $$ Step 3: Find factors for quadratic equation $$ (4x+3)(x-1)=0 $$ Step 4: Solve for $$ x $$ $$ 4x+3=0 ; x-1=0 $$ $$ 4x=-3 ; x=1 $$ $$ x=-\frac{3}{4} ; x=1 $$ 3.) Step 1: Use the slope formula to find the slope $$ m=\frac{y_2-y_1}{x_2-x_1} $$ $$ m=\frac{10-4}{1-(-2)} $$=$$ \frac{6}{3}$$=$$2$$

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