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3 years ago

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:

(x + y)³

Step-by-step explanation:

use (a + b)³ = a³ + 3a²b + 3ab² + b³

x³ + 3x²y + 3xy² + y³ =

= (x + y)³

7 0

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?

Tamiku [17]

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

7 0

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:

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

5 0

2 years ago

Rachel is a nurse she earns 12 vacation days after working 768 hours how many hours does Rachel need to work to earn one vacatio

Deffense [45]

Rachel needs to work 64 hours to earn 1 vacati on day

6 0

3 years ago

Differentiate 4(e^4)^7*ln(e^4)

ohaa [14]

83847478/829298:74755

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3 years ago