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never [62]
3 years ago
11

answer as a mixed number if possible. A bowl of cereal had 3 3/6 grams of sugar in it. If Victor ate 2 bowl8a week, how many gra

ms of sugar would he have eaten?
Mathematics
1 answer:
GrogVix [38]3 years ago
7 0

<em>7 grams of sugar</em>

<h2>Explanation:</h2>

________________________________________

I'll assume your answer is as follows:

<em>A bowl of cereal had 3 3/6 grams of sugar in it. If Victor ate 2 bowl a week, how many grams of sugar would he have eaten?</em>

________________________________________

If so, then we know that:

  • A bowl of cereal had 3 3/6 grams of sugar in it.
  • Victor ate 2 bowl a week

In other words, the total of sugar he ate can be calculated as:

3\frac{3}{6}\times2 \\ \\ Converting \ mixed \ fraction \ into \ improper \ fraction: \\ \\ =(3+\frac{3}{6})\times2 \\ \\ =(\frac{3(6)+3}{6})\times 2 \\ \\ =(\frac{21}{6})\times 2 \\ \\ =(\frac{7}{2})\times 2 \\ \\ =7

So <em>he'd have eaten 7 grams of sugar</em>

<h2>Learn more:</h2>

Mixed numbers: brainly.com/question/383603

#LearnWithBrainly

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What is the volume of a hollow ball whose other radius is 200m and inner radius of 100m? What is the mass if it is iron?
kati45 [8]

Solution :

Volume of ball = Volume of outer ball - Volume of inner ball

V = \dfrac{4\pi R^3}{3} - \dfrac{4 \pi r^3}{3}\\\\V = \dfrac{4\pi}{3}( R^3 - r^3 )\\\\V = \dfrac{4\pi}{3}( 200^3 - 100^3 )\\\\V = \dfrac{4\pi}{3}(2^3 - 1^3) \times 10^6\\\\V = 2.93 \times 10^7 \ m^3

Now, density of iron is, d = 7300 kg/m³.

So, mass of iron ball is :

m = d\times V\\\\m = 7300\times 2.93 \times 10^7 \ kg\\\\m =  2.14 \times 10^{11}\ kg

Hence, this is the required solution.

5 0
3 years ago
Find the Taylor series for f(x)=sin(x) centered at c=π/2.sin(x)=∑ n=0 [infinity]On what interval is the expansion valid? Give yo
agasfer [191]

Answer:

sinx =1-\frac{1}{2} (x-\frac{\pi}{2} )^2+\frac{1}{24} (x-\frac{\pi}{2} )^4-\frac{1}{720} (x-\frac{\pi}{2} )^6+\frac{1}{40320} (x-\frac{\pi}{2} )^8+...

Step-by-step explanation:

given that f(x) = sin x

we have to find the Taylor series for that

f(x) = sin x   : f( = 1\\f'(x) = cos x :(f'\frac{\pi}{2})=0\\f"(x) = -sinx :f" (\frac{\pi}{2}) =-1\\f^4 (x) = -cosx : f^4 (\frac{\pi}{2}) =0

and so on.

i.e. 2nd, 4th, 6th terms would be 0

and also 1st, 5th, 9th terms would be positive for f value and 3rd, 9th,... would be negative

Using the above we can write Taylor series as

f(x) = f(a)+\frac{f'(a)}{1!} (x-\frac{\pi}{2}) +...+f^n(a) /n! (x- \frac{\pi}{2})^n+...

sinx =1-\frac{1}{2} (x-\frac{\pi}{2} )^2+\frac{1}{24} (x-\frac{\pi}{2} )^4-\frac{1}{720} (x-\frac{\pi}{2} )^6+\frac{1}{40320} (x-\frac{\pi}{2} )^8+...

This is valid for all real values of x.

x ∈(-\infty, infty)

3 0
3 years ago
Please help ? Thanks
rjkz [21]
Try answer C or B because either one could be the right answer
8 0
3 years ago
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Masja [62]

Answer:

1 milimeter

Step-by-step explanation:

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6 0
3 years ago
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When a force of 30 N acts on a certain object the acceleration of the object is 5m/s2 if the force is changed to 54 N what will
PtichkaEL [24]

Answer:

The correct answer is 9 m per second^{2}

Step-by-step explanation:

Force is defined by mass of an object times it's acceleration.

Let the mass of the object be m kilograms.

When a force of 30 N acts on a certain object the acceleration of the object is 5 m per second^{2}

∴ 30 = m × 5

⇒ m = 6.

If the force is changed to 54 N, let the acceleration of the object be x  m per second^{2}.

∴ 54 = m × x

⇒ x = \frac{54}{6} = 9

Thus when the force is 54 N, the acceleration of the object is 9 m per second^{2}.

6 0
4 years ago
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