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bulgar [2K]
3 years ago
6

(−2)(−2)+4 is equal to

Mathematics
1 answer:
abruzzese [7]3 years ago
8 0

Answer:

8

Step-by-step explanation:

[-2][-2]=-2 x-2=4

4+4=8

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A delivery service charges $1.25 for
choli [55]

Answer: 6\ \text{miles}

Step-by-step explanation:

Given

A delivery service charge \$1.25 for each delivery

and an additional \$0.75 for each mile traveled

Suppose, x miles are covered

The charge is given by

\Rightarrow 5.75=1.25+0.75x\\\Rightarrow 4.5=0.75x\\\Rightarrow x=6\ \text{miles}

Therefore, driver traveled 6 miles for delivery

8 0
3 years ago
Mia makes cupcakes and sells them to raise money for charity.
Doss [256]

Answer: 81.82%

Step-by-step explanation: what you made - the ingredients cost so 37.51 - 6.82 = 30.69 that is her profit now we put it in percentage form which is 81.82% if there is any part you need me to go deeper to explain please dont hesitate to ask!

Hope this helps!!! Good luck!!! ;)

6 0
3 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
2 years ago
PLASE HELP!!! I WILL GIVE BRAINLIEST!!!
zhannawk [14.2K]
0.43 = 

3/100 = 0.03

+

4/10 = 0.4

0.03 + 0.4 = 0.43.

Hope this helped☺☺
4 0
3 years ago
Read 2 more answers
PLS HELP!! A line passes through the point (0, -3/2) and has a slope of 1/2. what is the equation of the line?
vesna_86 [32]
Your answer is y=.5x-1.5
6 0
3 years ago
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