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Sedbober [7]
4 years ago
14

If there is 360 grams of radioactive material with a half-life of 8 hours, how much of the radioactive material will be left aft

er 32 hours and is the radioactive decay modeled by a linear function or an exponential function?
A) 22.5 grams; linear
B) 22.5 grams; exponential
C) 45 grams; linear
D) 45 grams; exponential
Mathematics
1 answer:
AnnZ [28]4 years ago
7 0
22.5 grams : Exponential

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John ran the 100-m dash with a time of 9.81 sec. if this pace could be maintained for an entire 26-mi marathon, what would his t
Dmitry_Shevchenko [17]

Answer:

4104.79 Seconds or about 1.14 Hours

Step-by-step explanation:

26mi = 41842.9m

100/9.81 = 41842.9/x

41842.9 x 9.81 = 100x

410478.849 = 100x

x = 4104.78849

7 0
3 years ago
Find the value of 10! / (10-2) !
Gnesinka [82]

Answer:

90

Step-by-step explanation:

10! / (10 - 2)!

=10!/8!

=10 * 9 * 8! /8!

=90 * 8!/8!

90

Hope this helps. Please mark as brainliest if possible. Have a nice day

3 0
3 years ago
Polygon with only 2 congruent sides and 2 congruent angles
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A rectangle or a parallelogram is a polygon with only 2 congruent sides and 2 congruent angles.
5 0
3 years ago
Solve for possible values of x. 3x^2 = 18x
uysha [10]

(0,0) and (6,0)

First we subtract 18x from both sides

3x²-18x=0

Factor out a 3x

3x(x-6)=0

This means that x is either 0 or 6

7 0
3 years ago
Read 2 more answers
You invested $ 7000 between two accounts paying 3 % and 5 % annual? interest, respectively. If the total interest earned for the
natita [175]

Answer:

In the first account was invested \$3,000  at 3%

In the second account was invested \$4,000  at 5%

Step-by-step explanation:

we know that

The simple interest formula is equal to

I=P(rt)

where

I is the Interest Value

P is the Principal amount of money to be invested

r is the rate of interest  

t is Number of Time Periods

in this problem we have

First account

t=1 years\\ P=\$x\\ r=0.03

substitute in the formula above

I=x(0.03*1)

I=0.03x

Second account

t=1 years\\ P=\$(7,000-x)\\ r=0.05

substitute in the formula above

I=(7,000-x)(0.05*1)

I=350-0.05x

Remember that

The interest is equal to \$290

so

Adds the interest of both accounts

0.03x+350-0.05x=\$290

0.05x-0.03x=350-290

0.02x=60

x=\$3,000

therefore

In the first account was invested \$3,000  at 3%

In the second account was invested \$7,000-\$3,000=\$4,000  at 5%

7 0
3 years ago
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