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natali 33 [55]
3 years ago
14

a vendor bought n dozen eggs at the rate of 22.6 of the rotating he sold the remaining acts at the rate of 2.50 per act find his

profit or loss​
Mathematics
1 answer:
babymother [125]3 years ago
7 0

Answer:

Your question was a bit confusing so I found another related question. You can reference it with your own details.

A vendor bought 10 dozen eggs at the rate of Rs 22 per dozen. Six of the eggs were rotten. He sold the remaining eggs at the rate of Rs 2.50 per egg . Find his profit or loss

Vendor bought 10 dozen eggs which means the vendor bought:

= 10 * 12 eggs

= 120 eggs

6 were rotten and he sold the rest for Rs. 2.50 per egg.

= (120 - 6) * 2.5

= Rs. 285

The amount he spent on the eggs was:

= 10 dozen * 22

= Rs. 220

Profit = 285 - 220

= Rs. 65

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Geometry math question no Guessing and Please show work thank you
nasty-shy [4]

We can name an angle either by naming its vertex or by three letters , keeping the letter of vertex in between.

so here vertex is F

we can name it as ∠EFG, ∠F , ∠GFE

This strikes out the option B that is ∠G

so option B is the answer

6 0
3 years ago
Gohan earned scores of 95, 100, 65, 95, 75, 98, 100, 55, 100, 45 on his last ten quizzes. What is the mean of his quiz scores? W
irinina [24]

Answer: Mean= 82.8

Median= 95

Mode=100

Step-by-step explanation:

Mean is the average of the scores

Mean= 95+100+65+95+75+98+100+55+100+45 =828/10= 82.8

To get the median, we rearrange the numbers in either ascending or descending order.

45,55,65,75,95,95,98,100,100,100

Median= 95

Mode is the number that occurs most which is 100

8 0
4 years ago
A rectangular banner covers an area of 2 7/8ft2. The length of the banner is 3/4ft. What is the width of the banner in ft?
hichkok12 [17]

Answer:

Width of the rectangular banner is 3\frac{5}{6} ft.

Step-by-step explanation:

Area of the rectangular banner = 2\frac{7}{8} square feet

                                                    = \frac{23}{8} square feet

Area of rectangle is given by,

Area = Length × Width

Length of the banner = \frac{3}{4} feet

From the formula,

\frac{23}{8}=\frac{3}{4}\times W

W = \frac{\frac{23}{8} }{\frac{3}{4} }

    = \frac{23}{8}\times \frac{4}{3}

    = \frac{23}{6}

     = 3\frac{5}{6} ft

Therefore, width of the rectangular banner is 3\frac{5}{6} ft

7 0
3 years ago
Complete the two-column proof using the answers in the bank. Given: V is the midpoint of (SU) ̅ .∠RSV ≅ ∠TUV Prove: ∆VUT ≅∆VSR
ivann1987 [24]

Step-by-step explanation:

1.  V is the midpoint of  (SU) ̅    1.  Given

2. UV ≅ VS           2.  Definition of midpoint

3. ∠RSV ≅ ∠TUV            3. Given

4.  ∠SVR ≅∠UVT           4.  Vertical Angle Theorem

5.  ∆VUT ≅ ∆VSR          5.  ASA Congruence  


8 0
3 years ago
∫<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bcosxdx%7D%7B%5Csqrt%7B1%2Bsin%5E%7B2%7Dx%20%7D%20%7D" id="TexFormula1" title="\frac
const2013 [10]

Substitute sin(x) = tan(t) and cos(x) dx = sec²(t) dt. We want this change of variable to be reversible, so let's assume bot x and t are bounded between 0 and π/2.

Then we have

\displaystyle \int \frac{\cos(x)}{\sqrt{1 + \sin^2(x)}} \, dx = \int \frac{\sec^2(t)}{\sqrt{1 + \tan^2(t)}} \, dt

Recall the Pythagorean identity,

1 + tan²(t) = sec²(t)

Then

√(1 + tan²(t)) = √(sec²(t)) = sec(t)

and the integral reduces to

\displaystyle \int \frac{\sec^2(t)}{\sqrt{1 + \tan^2(t)}} \, dt = \int \frac{\sec^2(t)}{\sec(t)} \, dt = \int \sec(t) \, dt = \ln|\sec(t)+\tan(t)| + C

Change the variable back to x, so the antiderivative is

\displaystyle \int \frac{\cos(x)}{\sqrt{1 + \sin^2(x)}} \, dx = \ln \left|\sec\left(\tan^{-1}(\sin(x))\right) + \tan\left(\tan^{-1}(\sin(x))\right) \right| + C

\displaystyle \int \frac{\cos(x)}{\sqrt{1 + \sin^2(x)}} \, dx = \boxed{\ln \left|\sqrt{1+\sin^2(x)} + \sin(x) \right| + C}

6 0
2 years ago
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