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DanielleElmas [232]
3 years ago
10

A sundae at an ice-cream shop costs $3, and each extra ounce of toppings costs

Mathematics
1 answer:
anyanavicka [17]3 years ago
7 0

Answer:

The number of ounce of topping that can be get is 4 .

Step-by-step explanation:

Given as :

The cost of a sundae = $3

The cost for each extra ounce of toppings = $0.50

Let The number of ounce of topping = n

The total spending cost on sundae with n extra topping = $5

Now, According to question

The total spending cost on sundae with n extra topping = The cost of a sundae + The number of ounce of topping × The cost for each extra ounce of toppings

Or , $3 + n × $0.50 = $5

or, 3 + 0.50 n = 5

or , 0.50 n = 5 - 3

or, 0.50 n = 2

∴  n = \dfrac{2}{0.50}

I.e n = 4

So, The number of ounce of topping = n = 4

Hence The number of ounce of topping that can be get is 4 . Answer

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1. Given any triangle ABC with sides BC=a, AC=b and AB=c, the following are true :

     i) the larger the angle, the larger the side in front of it, and the other way around as well. (Sine Law) Let a=20 in, then the largest angle is angle A.

      ii) Given the measures of the sides of a triangle. Then the cosines of any of the angles can be found by the following formula:

       a^{2}=b ^{2}+c ^{2}-2bc(cosA)

2.

20^{2}=9 ^{2}+13 ^{2}-2*9*13(cosA)

400=81+169-234(cosA)   150=-234(cosA)

cosA=150/-234= -0.641

3. m(A) = Arccos(-0.641)≈130°,

4. Remark: We calculate Arccos with a scientific calculator or computer software unless it is one of the well known values, ex Arccos(0.5)=60°, Arccos(-0.5)=120° etc

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The population of city B was approximately 7,150,000 and it increased by 0.8% in one year.what is the new population
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Sela's piano lesson starts at 4:45pm. The lesson is 30 minutes long. When does her lesson end
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What is the true solution to the equation below? 2 in e in2×-in e in 10×= in 30 A x=30 B x=75 C x=150 D x=300
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Answer:

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Step-by-step explanation:

Let as consider the given equation:

2\ln e^{\ln 2x}-\ln e^{\ln 10x}=\ln 30

It can be written as

2(\ln 2x)-(\ln 10x)=\ln 30         [\because \ln e^a=a]

\ln (2x)^2-(\ln 10x)=\ln 30        [\because \ln a^b=b\ln a]

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On comparing both sides, we get

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