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jenyasd209 [6]
2 years ago
14

steve uses 8/9 gallons of paint to paint 4 bird houses. how many gallons of paint does he use for each birdhouse?

Mathematics
2 answers:
Kobotan [32]2 years ago
7 0
2/9 gallons each is your answer. This is because you have to divide the 8/9 between the four birdhouses, which is the same as dividing the 8 in the 8/9 by 4.
Hope this helps!
Lesechka [4]2 years ago
4 0
He uses 0.2 of the paint
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(a)Use differentiation to find the minimum value of f(x)= 3x^2 - 5x + 10 and the value(s) of x for which this minimum occurs.
elena-14-01-66 [18.8K]

Hi there!

Begin by differentiating f(x) using the power rule:

dy/dx = nxⁿ⁻¹

Therefore:

f(x) = 3x² - 5x + 10

f'(x) = 6x - 5

Set this equation equal to 0 to find the x-intercept:

0 = 6x - 5

5 = 6x

x = 5/6, which is where the graph goes from NEGATIVE to POSITIVE, so there is a MINIMUM at this value.

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In the diagram of O, m∠JOK = 60° and OJ = 6 in. What is the exact area of the shaded region?
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Answer:   6π

<u>Step-by-step explanation:</u>

Area of a circle is π r².

Area of a section of a circle is π r² × the section of the circle.

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2 years ago
Help me to prove the given below. Thank you in advance
sleet_krkn [62]

Answer:

Construct MO, where O is the midpoint of CG. (As shown in the attached figure)

(1) LM = MG //given, CM is the median to the hypotenuse      

(2) CO = OG // construction    

(3) MO is a mid-segment   //(1), (2) , Definition of a mid-segment

(4) MO // LC //triangle mid-segment theorem    

(5) ∠MOG ≅ ∠LCO // corresponding angles in two parallel lines intersected by a transversal line (CG)

(6) m∠LCO=90°  //given, ΔLCG is a right triangle

(7) m∠MOG=90° //(5), (6), definition of congruent angles

(8) m∠MOC=180° – 90° = 90° // linear pair

(9) ∠MOG ≅ ∠MOC //(7),(8), definition of congruent angles

(10) MO=MO //common side, reflexive property of equality  

(11) ΔMCO ≅ ΔMGO  //(2), (9), (10) Side-Angle-Side postulate

(12) MC = MG // Corresponding sides in congruent triangles (CPCTC)      

(13) MG= ½LG // Given, CM is the median to the hypotenuse    

(15) MC = ½LG //(12) , (13), substitution  

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