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GREYUIT [131]
2 years ago
15

The Hurry-Mart convenient store is re-stocking their shelves. They always stock the m&m to Hershey bars at a ratio of 10 to

8. How many Hershey bars do they need if they have 15 bags of m&ms ?
Mathematics
1 answer:
castortr0y [4]2 years ago
8 0

Answer:

12 Hershey bars

Step-by-step explanation:

if its 10 to 8 and you have 15 m&m then you would have to do the 10 which you would have to put 8 then 5 is half of 10 so you would just half 8 which is 4 so 8+4+12. 12 Hershey bars needed to restock the shelf.

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Step-by-step explanation:

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3 years ago
I got a screenshot, You got answer
liq [111]

Answer:

A. V = (pi)(3.25)^2(12)

Step-by-step explanation:

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A = (pi)(r)^2

Multiply that by the height to find the formula for the volume of a cylinder,

V=(pi)(r)^2(h)

Where (pi) represents the value (3.1415), (r) represents the radius, and (h) represents the height of the cylinder. Now substitute in the given values, remember, this problem gives the diameter of the cylinder, divide that value by two to find the radius.

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3 0
3 years ago
What is an equation of the line that passes through the points (1, 6)<br> and (1, 2)?
victus00 [196]

Answer:y = -5x + 11

Step-by-step explanation:

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m = (1-6) / (2-1) = -5/1 = -5

Then use the point slope formula to find the equation of the line:

m = (y-y0)/(x-x0)

-5 = (y-1)/(x-2)

-5(x-2) = y-1

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8 0
3 years ago
At the beginning of the year Jason had $80 in his savings account. Each month he added $15 to his account. Write in expression f
zmey [24]
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6 0
3 years ago
85.87 J or heat energy is added to a 34.8 g mass of substance the temperature rises from 21.76°C
Andru [333]

Answer:

The required specific heat is 196.94 joule per kg per °C  

Step-by-step explanation:

Given as :

The heat generated = Q = 85.87 J

Mass of substance  (m)= 34.8 gram = 0.0348 kg

Change in temperature = T2 - T1 = 34.29°C - 21.76°C = 12.53°C

Let the specific heat = S

Now we know that

Heat = Mass × specific heat × change in temperature

Or, Q = msΔt

Or, 85.87 = (0.0348 kg ) × S × 12.53°C

Or , 85.87 = 0.4360 × S

Or, S = \frac{85.87}{0.4360}

∴ S = 196.94 joule per kg per °C

Hence the required specific heat is 196.94 joule per kg per °C   Answer

7 0
3 years ago
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