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Blababa [14]
3 years ago
8

ABC DEF AB corresponds to DF. True False

Mathematics
2 answers:
Alik [6]3 years ago
8 0
True or it could be AC corresponding to DF
Alex Ar [27]3 years ago
6 0

it's false

I guess

have a nice day

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PLEASE HEELP AND THANK YOU
sasho [114]
Huh huh huh huh huh huh huh
8 0
3 years ago
Read 2 more answers
The denominator of a fraction is greater than the numerator by 8. If the numerator is increased by 17 and denominator is decreas
MA_775_DIABLO [31]

Answer:\frac{13}{21}

Step-by-step explanation:

2( N+ 17) = 3 (N +7)

2N+ 34= ·3N+ 21

N= 13

D= N+8 = 13+8 = 21

Fraction is \frac{13}{21}

When Im using N is equal to numerator and D for denominator, in order to get the fraction we substitute and the results combine in the fraction like this  \frac{N}{D}

3 0
1 year 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
Help find domain and range
kolezko [41]

Answer:

d: 4 r:5

Step-by-step explanation:

for domain the x coordinates doesn't go over -2 on the left or 2 on the right so if u add the units of the space in between it's 4 since 2- (-2)=4

for the range, the y coordinates never goes over 3 or -2 so range is 5 is u add the units in between

I hope that makes sense!

4 0
3 years ago
This is the last question plz help
kolezko [41]

Answer:

Skewed right as the values decrease to the right.

4 0
3 years ago
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