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nika2105 [10]
3 years ago
14

What percent is... 50% of 150% of x?

Mathematics
2 answers:
Varvara68 [4.7K]3 years ago
4 0

Answer:

75%

Step-by-step explanation:

faust18 [17]3 years ago
3 0
Convert both percents into decimals. To do that, move the percent sign two places to the left then multiply.

50% => 0.50
150% => 1.50
1.50 * 0.50 * x ; Start
0.75 * x            ; 1.50 * 0.50 is 0.75

The final answer is 0.75 * x. Which can also be expressed as 3/4x.
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If dy/dx equals cosine squared of the quantity pi times y over 2 and y = 0.5 when x = 0, then find the value of x when y = 2.5.
Ray Of Light [21]

Answer: 2 over pi

Step-by-step explanation: Bear with me, this is a hard one to explain it'll take 40 minutes to type out my explanation because my computer is not taking the photo, I could send it to you over snap or instagram immediately

7 0
3 years ago
Which equation represents the graph?
klio [65]
The second option.

That is because
y -3 = - \frac{1}{2}(x + 1) \\ (y -3)+3 = (- \frac{1}{2} x - \frac{1}{2}) + 3 \\ y = - \frac{1}{2}x + 2 \frac{1}{2}
6 0
3 years ago
The library is halfway between your house and your school. If your house is located at (-4,-3) and the library is at (2,-1), wha
aleksandrvk [35]
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6 0
4 years ago
the half-life of chromium-51 is 38 days. If the sample contained 510 grams. How much would remain after 1 year?​
madam [21]

Answer:

About 0.6548 grams will be remaining.  

Step-by-step explanation:

We can write an exponential function to model the situation. The standard exponential function is:

f(t)=a(r)^t

The original sample contained 510 grams. So, a = 510.

Each half-life, the amount decreases by half. So, r = 1/2.

For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.

Therefore, our function is:

\displaystyle f(t)=510\Big(\frac{1}{2}\Big)^{t/38}

One year has 365 days.

Therefore, the amount remaining after one year will be:

\displaystyle f(365)=510\Big(\frac{1}{2}\Big)^{365/38}\approx0.6548

About 0.6548 grams will be remaining.  

Alternatively, we can use the standard exponential growth/decay function modeled by:

f(t)=Ce^{kt}

The starting sample is 510. So, C = 510.

After one half-life (38 days), the remaining amount will be 255. Therefore:

255=510e^{38k}

Solving for k:

\displaystyle \frac{1}{2}=e^{38k}\Rightarrow k=\frac{1}{38}\ln\Big(\frac{1}{2}\Big)

Thus, our function is:

f(t)=510e^{t\ln(.5)/38}

Then after one year or 365 days, the amount remaining will be about:

f(365)=510e^{365\ln(.5)/38}\approx 0.6548

5 0
3 years ago
Write a mixed number and an improper fraction for the model below ​
Mariulka [41]

Answer:

2 and 3/4

11/4

Step-by-step explanation:

an improper fraction is any fraction with a numerator bigger then the denominator.

a mixed number is the "proper" version of an improper fraction.

the fraction 11/4 as a mixed number is 2 and 3/4.

to make an improper fraction a mixed number, find out how many times the denominator goes into the numerator. whatever you have left that didn't make 1 whole, turns into the fraction .

hope this helped

4 0
3 years ago
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