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egoroff_w [7]
3 years ago
11

X + 6 = x – 1 Find x.

Mathematics
2 answers:
Lady bird [3.3K]3 years ago
6 0
I think it’s no solution
Inessa05 [86]3 years ago
3 0
The answer is indeed No solution due to it just not being able to solve. No number can be plugged in for that to be true
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What is the length of the hypotenuse
Viefleur [7K]

Answer:

15ft

Step-by-step explanation:

a^2 + b^2 = c^2

12 or 9 can equal to b or a

12^2 + 9^2 = c^2

144 + 81 = c ^2

225 = c^2

Square root of 225

15 = c

4 0
3 years ago
Read 2 more answers
Simplify 3 over 2 x 3 over 2 exact answer
nikitadnepr [17]

ANSWER : 9 over 4

or 2 and 1/4

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

*3 over 2 x 3 over 2 is 3/2 x 3/2* << this is for me to remember

Okay so multiplying fractions means for us to multiply across. That means if we line the fractions, we multiply. Like this

 3     x     3

-----         -----                3 times 3 is = 9

 2     x     2                  2 times 2 is = 4

9 over 4 is the same thing as 2 and 1/4  << FRACTION    

                                                ^^      

                                           WHOLE

Hope I helped (:

8 0
2 years ago
Read 2 more answers
Evaluate the following expression. Express your answer in simplest form. 3/16 + 5/8 × 1/2​
miss Akunina [59]

Answer:

1/2

Step-by-step explanation:

We have to follow the order of operations. Since there are no parenthesis or exponents, we do multiplication first. to multiple fractions, we multiply the two numerators (top parts), which in this situation would get us 5*1=5. Then, we multiply the two denominators (bottom parts), which would get us 8*2=16. then we put the first result over the second, and we get 5/16. since the two denominators are the same, we don't need to change anything. we can just add the numerator, and get 8/16, which simplifies to 1/2.

4 0
3 years ago
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}}&#10;\\\\\\&#10;\textit{using the pythagorean theorem}\\\\&#10;c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;\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}&#10;\qquad\qquad &#10;cos(\theta)=\cfrac{adjacent}{hypotenuse}&#10;\\\\\\&#10;% tangent&#10;tan(\theta)=\cfrac{opposite}{adjacent}&#10;\qquad \qquad &#10;% cotangent&#10;cot(\theta)=\cfrac{adjacent}{opposite}&#10;\\\\\\&#10;% cosecant&#10;csc(\theta)=\cfrac{hypotenuse}{opposite}&#10;\qquad \qquad &#10;% secant&#10;sec(\theta)=\cfrac{hypotenuse}{adjacent}

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

\bf sin(\theta)=\cfrac{-1}{6}&#10;\qquad\qquad &#10;cos(\theta)=\cfrac{\sqrt{35}}{6}&#10;\\\\\\&#10;% tangent&#10;tan(\theta)=\cfrac{-1}{\sqrt{35}}&#10;\qquad \qquad &#10;% cotangent&#10;cot(\theta)=\cfrac{\sqrt{35}}{1}&#10;\\\\\\&#10;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}&#10;\\\\\\&#10;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
Show that <br><br> Sin60 / cos60 = tan 60
Lerok [7]

Answer:

  • <em><u>sin60</u></em><em><u>=</u></em><em><u>opp\hyp</u></em>
  • <em><u>cos60</u></em><em><u>=</u></em><em><u>adja\hyp</u></em>
  • <em><u>tan60</u></em><em><u>=</u></em><em><u>opp\adj</u></em>
  • <em><u>sin60\cos60</u></em><em><u>=</u></em><em><u>tan60</u></em>
6 0
1 year ago
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