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Charra [1.4K]
3 years ago
13

How do I solve this?

Mathematics
1 answer:
Helga [31]3 years ago
6 0
Look it up hope this helps
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How did you solve for mc?
natima [27]
You know that e= mc^2. In order to find the Value of mc, you have to take squareroot on both sides. So,
squareroot of e= mc
5 0
3 years ago
Duncan takes a break from studying and goes to the gym to swim laps. If swimming burns 9.05×105 cal per hour, how many kilojoule
fgiga [73]

For this case we have to by definition:

1 kilojoule equals 0.239006 kilocalorie.

1 kilocalorie equals 1000 calories

Then, if we have:9.05 * 10 ^ 5 \frac {calories} {hour} *\frac {1} {1000} \frac {kilocaloria} {calorie} = 9.05 * 10 ^ 2 \frac {kilocaloria} {hour}

Now:

9.05 * 10 ^ 2 \frac {kilocaloria} {hour} * \frac {1} {0.239006} \frac {kilojoule} {kilocaloria} = 3786.52 \frac {kilojoule} {hour}

Answer:

3786.52 \frac {kilojoule} {hour}

3 0
3 years ago
25 is what percent of 29?
kolezko [41]

Answer:

86.2%

Step-by-step explanation:

25/29=0.862

7 0
4 years ago
Read 2 more answers
For fun question
ZanzabumX [31]
Consider the function f(x)=x^{1/3}, which has derivative f'(x)=\dfrac13x^{-2/3}.

The linear approximation of f(x) for some value x within a neighborhood of x=c is given by

f(x)\approx f'(c)(x-c)+f(c)

Let c=64. Then (63.97)^{1/3} can be estimated to be

f(63.97)\approxf'(64)(63.97-64)+f(64)
\sqrt[3]{63.97}\approx4-\dfrac{0.03}{48}=3.999375

Since f'(x)>0 for x>0, it follows that f(x) must be strictly increasing over that part of its domain, which means the linear approximation lies strictly above the function f(x). This means the estimated value is an overestimation.

Indeed, the actual value is closer to the number 3.999374902...
4 0
3 years ago
3) A box of chocolates contains five milk chocolates, five dark chocolates, and five white
nevsk [136]
Dependenttttttttttttttttttttttt
7 0
3 years ago
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