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Feliz [49]
3 years ago
6

Jim has ten more dollars than Gail. Together they have at most twenty-seven dollars. A. Determine the possible amounts of money

Gail could have. B. Sketch the solution on a number line.

Mathematics
2 answers:
Irina18 [472]3 years ago
4 0
Gail could have up to 17 dollars. 
Len [333]3 years ago
3 0

Answer: Gail could have from 0 to 8.5 and Jim could have 10 to 18.5

Step-by-step explanation:

Gail = G

Jim = G + 10

J + G <= 27 ∴

G+10+G <=27

2G + 10 <= 27

2G <= 27 - 10

2G <= 17

G <= 17/2

G <= 8.5

As it is money, let's not say that they have negative money.

So, Gail has from 0 to 8.5. As Jim has 10 more than Gail, he has from 10 to 18.5. At the most : 8.5 + 18.5 = 27.

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s2008m [1.1K]

Answer:

The expression in the form 7^n

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Step-by-step explanation:

7^2\:\times \:7^6

\mathrm{Apply\:exponent\:rule}:\quad \:a^b\times \:a^c=a^{b+c}

7^2\times \:7^6=7^{2+6}

So the expression becomes

7^2\:\times \:7^6=7^{2+6}

              =7^{2+6}

Add the numbers: 2+6=8

              =7^8

Therefore, the expression in the form 7^n

 7^2\times \:\:7^6=7^8                        

7 0
3 years ago
What is the value of x in the equation 2^12 = 2^15x?
Radda [10]

Answer:

X= 1/8

Step-by-step explanation:

i got u

4 0
3 years ago
Radium-226 has a half-life of 1620 years. for a 150 gram sample, how many grams will remain after 200 years?
VashaNatasha [74]
Ah, yes

if you start with P amount of something and the half life is h units (in years) and the time elapsed is t (in years) then the amount left is A

the equation is
A=P(\frac{1}{2})^\frac{t}{h}


so
given
h=1620
P=150
t=200
so
A=150(\frac{1}{2})^\frac{200}{1620}
A=150(\frac{1}{2})^\frac{10}{81}
use your calculator
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3 years ago
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Natalka [10]

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Step-by-step explanation:

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JulijaS [17]

Answer:

y=25,700(0.85)^{x}

Step-by-step explanation:

we know that

In this problem we have a exponential function of the form

y=a(b)^{x}

where

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x ----> the number of days

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a=25,700 gal

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b=(1-r)=1-0.15=0.85

The function is equal to

y=25,700(0.85)^{x}

8 0
3 years ago
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