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maks197457 [2]
3 years ago
14

Write a word problem that is solved by 43 ÷ 5= 8 R3, in which the remainder is the only part needed to answer the question.

Mathematics
2 answers:
Sonbull [250]3 years ago
6 0
The answer should have been 8.6 that was wrong but it is okay you were close
Rashid [163]3 years ago
6 0
There are 5 children in a classroom and 1 teacher who oversees them. One day, the Principal brings in 43 M&M candies. He directs the students to split the candies evenly amongst themselves, and any candies leftover are to be the teacher's candies. How many candies will the teacher be enjoying on that day?
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Cos ( α ) = √ 6/ 6 and sin ( β ) = √ 2/4 . Find tan ( α − β )
Zina [86]

Answer:

\purple{ \bold{ \tan( \alpha  -  \beta ) = 1.00701798}}

Step-by-step explanation:

\cos( \alpha ) =  \frac{ \sqrt{6} }{6}  =  \frac{1}{ \sqrt{6} }  \\  \\  \therefore \:  \sin( \alpha )  =  \sqrt{1 -  { \cos}^{2} ( \alpha ) }  \\  \\  =  \sqrt{1 -  \bigg( {\frac{1}{ \sqrt{6} } \bigg )}^{2} }  \\  \\ =  \sqrt{1 -  {\frac{1}{ {6} }}}  \\  \\ =  \sqrt{ {\frac{6 - 1}{ {6} }}}   \\  \\  \red{\sin( \alpha ) =  \sqrt{ { \frac{5}{ {6} }}} } \\  \\  \tan( \alpha ) =  \frac{\sin( \alpha ) }{\cos( \alpha ) }  =  \sqrt{5}  \\  \\ \sin( \beta )  =  \frac{ \sqrt{2} }{4}  \\  \\  \implies \: \cos( \beta )  =   \sqrt{ \frac{7}{8} }  \\  \\ \tan( \beta )  =  \frac{\sin( \beta ) }{\cos( \beta ) } =  \frac{1}{ \sqrt{7} }   \\  \\  \tan( \alpha  -  \beta ) =  \frac{ \tan \alpha  -  \tan \beta }{1 +  \tan \alpha .  \tan \beta}  \\  \\  =  \frac{ \sqrt{5} -  \frac{1}{ \sqrt{7} }  }{1 +  \sqrt{5} . \frac{1}{ \sqrt{7} } }  \\  \\  =  \frac{ \sqrt{35} - 1 }{ \sqrt{7}  +  \sqrt{5} }  \\  \\  \purple{ \bold{ \tan( \alpha  -  \beta ) = 1.00701798}}

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Use the properties of 30-60-90 and 45-45-90 triangles to solve for x in each of the problems below. Then decode the secret messa
murzikaleks [220]

The trigonometric function gives the ratio of different sides of a right-angle triangle. The given problems can be solved as given below.

<h3>What are Trigonometric functions?</h3>

The trigonometric function gives the ratio of different sides of a right-angle triangle.

\rm Sin \theta=\dfrac{Perpendicular}{Hypotenuse}\\\\\\Cos \theta=\dfrac{Base}{Hypotenuse}\\\\\\Tan \theta=\dfrac{Perpendicular}{Base}\\\\\\Cosec \theta=\dfrac{Hypotenuse}{Perpendicular}\\\\\\Sec \theta=\dfrac{Hypotenuse}{Base}\\\\\\Cot \theta=\dfrac{Base}{Perpendicular}\\\\\\

where perpendicular is the side of the triangle which is opposite to the angle, and the hypotenuse is the longest side of the triangle which is opposite to the 90° angle.

1st.) x = 5 /Sin(30°)

x = 10

!) sin(45°) = 4/x

x = 4/sin(45°)

x = 4√2

I) Cos(45°) = √3 / x

x = √3 / Cos(45°)

x = √6

E) Tan(60°) = 3√3 / x

x = 3√3 / 3

W) For isosceles right-triangle, the angle made by the legs and the hypotenuse is always 45°.

x = 45°

N) x² + x² = (7√2)²

x = 7

V) Tan(60°) = 7 / x

x = 7√3/3

K) x² + x² = (9)²

x = 9/√2

Y) Sin(60°) = 7√3/x

x = 14

M) Sin(30°) = x/11

x = 11/2

T) Sin(45°) = x/√10

x = √5

A) x + 2x + 90° = 180°

x = 30°

O) Sin(45°) = √2 / x

x = 2

R) Tan(30°) = x / 4

x = 4/√3 = 4√3 / 3

S) Sin(60°) = x / (10/3)

x = 5√3 / 3

Learn more about Trigonometric functions:

brainly.com/question/6904750

#SPJ1

5 0
1 year ago
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