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mestny [16]
3 years ago
14

Warren’s tape measure is marked in increments of 1/16 inch, not in decimal numbers. To measure the tiles, he needs to know the s

ide lengths to the nearest 1/16 inch. In this activity, you will convert decimal values to fractional values and then calculate the area of the tiles. The length of tile A is about 2.23607 inches. Which mark on the tape measure is closest to 2.23607 inches? Remember, the tape measure is marked in sixteenths of an inch.
Mathematics
1 answer:
Bezzdna [24]3 years ago
3 0

Answe

Step-by-step explanation:

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Find the remaining trigonometric ratios of θ if csc(θ) = -6 and cos(θ) is positive
VikaD [51]
Now, the cosecant of θ is -6, or namely -6/1.

however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.

we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now

\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a\implies \stackrel{IV~quadrant}{+\sqrt{35}=a}

recall that 

\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad\qquad 
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\\\\\\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\\\\\\
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
\qquad \qquad 
% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}

therefore, let's just plug that on the remaining ones,

\bf sin(\theta)=\cfrac{-1}{6}
\qquad\qquad 
cos(\theta)=\cfrac{\sqrt{35}}{6}
\\\\\\
% tangent
tan(\theta)=\cfrac{-1}{\sqrt{35}}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{\sqrt{35}}{1}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}

now, let's rationalize the denominator on tangent and secant,

\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}
3 0
3 years ago
How do you Solve d+7/−3=4?
vova2212 [387]

Answer:

One solution was found :

                  d = 19/3 = 6.333

Step-by-step explanation:

Step by step solution :

Step  1  :

            7

Simplify   ——

           -3

Equation at the end of step  1  :

        7    

 (d +  ——) -  4  = 0

       -3    

Step  2  :

Rewriting the whole as an Equivalent Fraction :

2.1   Adding a fraction to a whole

Rewrite the whole as a fraction using  -3  as the denominator :

         d     d • -3

    d =  —  =  ——————

         1       -3  

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

2.2       Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

d • 3 + 7 • -1     3d - 7

——————————————  =  ——————

      3              3  

Equation at the end of step  2  :

 (3d - 7)    

 ———————— -  4  = 0

    3        

Step  3  :

Rewriting the whole as an Equivalent Fraction :

3.1   Subtracting a whole from a fraction

Rewrite the whole as a fraction using  3  as the denominator :

        4     4 • 3

   4 =  —  =  —————

        1       3  

Adding fractions that have a common denominator :

3.2       Adding up the two equivalent fractions

(3d-7) - (4 • 3)     3d - 19

————————————————  =  ———————

       3                3  

Equation at the end of step  3  :

 3d - 19

 ———————  = 0

    3  

Step  4  :

When a fraction equals zero :

4.1    When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

 3d-19

 ————— • 3 = 0 • 3

   3  

Now, on the left hand side, the  3  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :

  3d-19  = 0

Solving a Single Variable Equation :

4.2      Solve  :    3d-19 = 0

Add  19  to both sides of the equation :

                     3d = 19

Divide both sides of the equation by 3:

                    d = 19/3 = 6.333

One solution was found :

                  d = 19/3 = 6.333

6 0
3 years ago
Find the measure of each<br> angle in the isosceles<br> trapezoid.<br> D<br> A<br> 136⁰<br> B<br> с
icang [17]

Answer:

b=136, c=44, d=44

Step-by-step explanation:

B is the same measure as A. C = D and A+B+C+D=360; C+D=88; C=44; D=44

3 0
2 years ago
Help. Will mark brainiest
Nikitich [7]

Answer:

Step-by-step explanation:

The third, fifth and sixth expressions (from the top) are NOT polynomials, and the reason in each case is that the expression has one or more negative powers of x in it.

8 0
3 years ago
Darrien read 97 pages last week.Evan read pages last week.How many 8 my 6. pages did, the boys read?
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I’m really confused about this
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