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natima [27]
2 years ago
6

What is the solution to the following equation? 4x - 9 - 35​

Mathematics
1 answer:
lorasvet [3.4K]2 years ago
4 0

Answer:

4x-9-35

4×=9+35

4x=44

4x/4=44/44

×=11

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Someone help please I really need the answer fast if that’s possible!!??
ycow [4]

Answer:

no yes no i dumb

Step-by-step explanation:

5 0
2 years ago
Solve the equation by using square roots. 2 x 2 − 9 = 11
VashaNatasha [74]

Answer:

x = ± \sqrt{10}

Step-by-step explanation:

Given

2x² - 9 = 11 ( add 9 to both sides )

2x² = 20 ( divide both sides by 2 )

x² = 10 ( take the square root of both sides )

x = ± \sqrt{10} ← exact solutions

3 0
3 years ago
Determine if the ordered pair (-5, -1) is a<br> solution of 2x - y =-11.
Marrrta [24]
No, when you plug is -5 for x and -1 for why, it ends up being -9 not -11. (-5,-1) is not a solution.
7 0
3 years ago
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WILL GIVE BRAINLIEST!!! Given tan(theta) = 4/3 and pi &lt; theta &lt; (3pi)/2, find cos(2theta).
babymother [125]

Answer:  \cos 2 \theta = \dfrac{-7}{25}

Step-by-step explanation:

Given: \tan \theta = \dfrac{4}{3}  and  \pi < \theta< \dfrac{3\pi }{2}

To find: \cos 2 \theta

Now as we know

\cos 2x = \dfrac{1- \tan ^2 x}{1+ \tan ^2 x}

So we have

\cos 2 \theta = \dfrac{1-(\dfrac{4}{3} )^2}{1+(\dfrac{4}{3} )^2} \\\\\Rightarrow \cos 2 \theta = \dfrac{1- \dfrac{16}{9} }{1+ \dfrac{16}{9} } = \dfrac{\dfrac{9-16}{9} }{\dfrac{9+16}{9}} = \dfrac{-7}{25}

Therefore d. is the correct option

Hence , \cos 2 \theta = \dfrac{-7}{25}

3 0
3 years ago
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Find the exact value of the following expression (without using a calculator): tan(Sin^-1 x/2)
Sati [7]

Answer:

tan(Sin^-1 x/2)=  \frac{x/2}{\sqrt{1-x^{2}/4 } }

Step-by-step explanation:

Let sin^-1 x/2= θ

then sinθ= x/2

on the basis of unit circle, we have a triangle with hypotenuse of length 1,   one side of length x/2 and opposite angle of θ.

          tan(Sin^-1 x/2) = tanθ

           tanθ= sinθ/cosθ

as per trigonometric identities cosθ= √(1-sin^2θ)

           tanθ= sinθ/ √(1-sin^2θ)

substituting the value sinθ=x/2 in the above equation

             tanθ= \frac{x/2}{\sqrt{1-x^{2}/4 } }

now substituting the value sin^-1 x/2= θ in above equation

             tan(sin^-1 x/2) =  \frac{x/2}{\sqrt{1-x^{2}/4 } }

!

3 0
3 years ago
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