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3 years ago

Can someone help me

Mathematics

2 answers:

lisabon 2012 [21]3 years ago

6 0

The correct answer I’d 110+95=205

nignag [31]3 years ago

4 0

The second answer, cause it best shows the equation, but they just moved the numbers around. It is also the only equation with all the same numbers used in the original equation, too

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A circular pond 24 yd in

Schach [20]

Answer:

The walk will cost $8164.

Step-by-step explanation:

Given:

Diameter of the circular pond (D) = 24 yd

Width of the gravel path (x) = 2 yd

Cost per yard of the path = $50

Now, radius of the circular pond is half of the diameter and is given as:

$Radius,R=\frac{D}{2}=\frac{24}{2}=12\ yd$

Now, area of the pond is given as:

$A_{pond}=\pi R^2=3.14\times (12)^2=3.14\times 144=452.16\ yd^2$

Area of the complete path including the pond area is given as:

$A_{outer}=\pi(R+x)^2=3.14\times(12+2)^2=3.14\times196=615.44\ yd^2$

Now, area of the gravel path can be obtained by subtracting the pond area from the total outer area. This gives,

$A_{path}=A_{outer}-A_{pond}\\\\A_{path}=615.44-452.16=163.28\ yd^2$

Now, using unitary method,

Cost of 1 square yard of path = $50

∴ Cost of 163.28 square yard of path = 50 × 163.28 = $8164

Hence, the walk will cost $8164.

5 0

3 years ago

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Alinara [238K]

<span>-10z5 is the answer. </span>

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3 years ago

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Is each statement true for parallelogram DEFG? Drag each statement into the correct box.

prisoha [69]

Yes all of them are true

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3 years ago

The volume of a solid is measured in units cubed. True False

weeeeeb [17]

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Ask your teacher find the limit. use l'hospital's rule where appropriate. if there is a more elementary method, consider using i

liberstina [14]

Answer:

ln(5/3)

Step-by-step explanation:

The desired limit represents the logarithm of an indeterminate form, so L'Hopital's rule could be applied. However, the logarithm can be simplified to a form that is not indeterminate.

<h3>Limit</h3>

We can cancel factors of (x-1), which are what make the expression indeterminate at x=1. Then the limit can be evaluated directly by substituting x=1.

$\diplaystyle \lim\limits_{x\to1}{(\ln(x^5-1)-\ln(x^3-1))}=\lim\limits_{x\to1}\ln{\left(\dfrac{x^5-1}{x^3-1}\right)}\\\\=\lim\limits_{x\to1}\ln\left(\dfrac{x^4+x^3+x^2+x+1}{x^2+x+1}\right)=\ln{\dfrac{5}{3}}$

8 0

1 year ago