PLEASE HELP which is equivalent??

Mathematics

1 answer:

Zolol [24]3 years ago

6 0

Answer:

the 3 one

Step-by-step explanation:

the rest would be like 0.9721111111 repeating but the 3 one won't be so its 3 one hive me brainlest

You might be interested in

Given that (-4,3) and (6,-3) are opposites points on the circle (i.e diameter) write the equation of the circle

ElenaW [278]

The center of the circle is the midpoint of the points or ((-4+6)/2,(-3+3)/2)=(1,0). Now, the diameter is the distance between the points. Using distance formulas (or noticing that the y-values are the same so it is a horizontal line), the distance is 10. Therefore, the radius is 5. The equation of a circle is (x-a)^2+(y-b)^2=r^2 where (a,b) is the center. Therefore, the equation is (x-1)^2+(y-0)^2=5^2.

7 0

4 years ago

17=11y+12x in slope intercept form please explain

finlep [7]

The answer is attached

Step-by-step explanation:

If this answer helped you then please consider making this brainliest and thank this response :)

3 0

3 years ago

Read 2 more answers

Darla spins a spinner and it points to a 7.She spins it again.Are these events dependent of independent?

Anvisha [2.4K]

No, it is independant

8 0

3 years ago

Read 2 more answers

A student calculates that 52.5 g of product can be prepared from a given set of initial amounts of the reactants. the student th

Yuliya22 [10]

Yield = 48.2/52.5 * 100 = 91.8%

6 0

3 years ago

Q.04: (11 points) Given the polar curve r = e θ , where 0 ≤ θ ≤ 2π. Find points on the curve in the form (r, θ) where there is a

Citrus2011 [14]

I suppose the curve is $r(\theta)=e^\theta$ .

Tangent lines to the curve have slope $\frac{dy}{dx}$ ; use the chain rule to get this in polar coordinates.

$\dfrac{dy}{dx}=\dfrac{dy}{d\theta}\dfrac{d\theta}{dx}=\dfrac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$

We have

$y(\theta)=r(\theta)\sin\theta\implies\dfrac{dy}{d\theta}=\dfrac{dr}{d\theta}\sin\theta+r(\theta)\cos\theta$

$x(\theta)=r(\theta)\cos\theta\implies\dfrac{dx}{d\theta}=\dfrac{dr}{d\theta}\cos\theta-r(\theta)\sin\theta$

$r(\theta)=e^\theta\implies\dfrac{dr}{d\theta}=e^\theta$

$\implies\dfrac{dy}{dx}=\dfrac{e^\theta\sin\theta+e^\theta\cos\theta}{e^\theta\cos\theta-e^\theta\sin\theta}=\dfrac{\sin\theta+\cos\theta}{\cos\theta-\sin\theta}$