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Alexeev081 [22]
3 years ago
14

Helpppppppppppppppppppppp i don't understand this at allllllll

Mathematics
2 answers:
yarga [219]3 years ago
8 0
The answer to this question would be the first box (left hand side). Hope that helps :)
balu736 [363]3 years ago
7 0

Answer:

2^3 (3 x 5^2 + 2^2) = 600x+32

 is the correct equation

Step-by-step explanation:

2^3 (3 x 5^2 + 2^2) = 600x+32\\

2^4 (3 x 5^5 +2^2)= 150000x+64\\

2^3 (3 x 5^2 +2^5) = 600x+256\\

2^4 (3 x 5^5 +2^1^6)= 150000x+1048576

Hope this helps! Brainliest would be much appreciated!

Have a great day! : )

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Factorise x^3+3x^2y+3xy^2+y^3
Nesterboy [21]
  • Answer:

<em>(x + y)³</em>

  • Step-by-step explanation:

<em>   use (a + b)³ = a³ + 3a²b + 3ab² + b³</em>

<em>x³ + 3x²y + 3xy² + y³ =</em>

<em>= (x + y)³</em>

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3 years ago
gracie bought a book for $8.98, a pen for $3.27, and a ruler for $1.50. if the sales tax is 8%, what did she spend?
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Answer: $14.85

Step-by-step explanation: 8.98+3.27+1.50=13.75

13.75 x 0.08=1.1

13.75 + 1.1= 14.85

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4 years ago
How can you prove that csc^2(θ)tan^2(θ)-1=tan^2(θ)
Oxana [17]

Answer:

Make use of the fact that as long as \sin(\theta) \ne 0 and \cos(\theta) \ne 0:

\displaystyle \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}.

\displaystyle \csc(\theta) = \frac{1}{\sin(\theta)}.

\sin^{2}(\theta) + \cos^{2}(\theta) = 1.

Step-by-step explanation:

Assume that \sin(\theta) \ne 0 and \cos(\theta) \ne 0.

Make use of the fact that \tan(\theta) = (\sin(\theta)) / (\cos(\theta)) and \csc(\theta) = (1) / (\sin(\theta)) to rewrite the given expression as a combination of \sin(\theta) and \cos(\theta).

\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \left(\frac{1}{\sin(\theta)}\right)^{2} \, \left(\frac{\sin(\theta)}{\cos(\theta)}\right)^{2} - 1 \\ =\; & \frac{\sin^{2}(\theta)}{\sin^{2}(\theta)\, \cos^{2}(\theta)} - 1\\ =\; & \frac{1}{\cos^{2}(\theta)} - 1\end{aligned}.

Since \cos(\theta) \ne 0:

\displaystyle 1 = \frac{\cos^{2}(\theta)}{\cos^{2}(\theta)}.

Substitute this equality into the expression:

\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \cdots\\ =\; & \frac{1}{\cos^{2}(\theta)} - 1 \\ =\; & \frac{1}{\cos^{2}(\theta)} - \frac{\cos^{2}(\theta)}{\cos^{2}(\theta)} \\ =\; & \frac{1 - \cos^{2}(\theta)}{\cos^{2}(\theta)}\end{aligned}.

By the Pythagorean identity, \sin^{2}(\theta) + \cos^{2}(\theta) = 1. Rearrange this identity to obtain:

\sin^{2}(\theta) = 1 - \cos^{2}(\theta).

Substitute this equality into the expression:

\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \cdots \\ =\; & \frac{1 - \cos^{2}(\theta)}{\cos^{2}(\theta)} \\ =\; & \frac{\sin^{2}(\theta)}{\cos^{2}(\theta)}\end{aligned}.

Again, make use of the fact that \tan(\theta) = (\sin(\theta)) / (\cos(\theta)) to obtain the desired result:

\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \cdots \\ =\; & \frac{\sin^{2}(\theta)}{\cos^{2}(\theta)}\\ =\; & \left(\frac{\sin(\theta)}{\cos(\theta)}\right)^{2} \\ =\; & \tan^{2}(\theta)\end{aligned}.

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