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N76 [4]
3 years ago
13

Phil has budgeted $120 to rent a car during his weeklong vacation. At the rate of $40 for the week plus $0.10 per mile, how many

miles can he drive the car and stay within his budget?​
Mathematics
2 answers:
Serga [27]3 years ago
8 0

Answer:

1000 miles

Step-by-step explanation:

Rashid [163]3 years ago
6 0

Answer: 800 miles

Step-by-step explanation: His budget is $120

He then spends $40 for the week

So that is $80 that he has left

Each mile is only $0.10, so you divide $80 by $0.10

After dividing you get 800

That means that he can drive for 800 miles and still stay within  his budget

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2.The number of grams A of a certain radioactive substance present at time, in yearsfrom the present, t is given by the formulaA
professor190 [17]

Answer:

Given that,

The number of grams A of a certain radioactive substance present at time, in years

from the present, t is given by the formula

A=45e^{-0.0045(t)}

a) To find the initial amount of this substance

At t=0, we get

A=45e^{-0.0045(0)}A=45e^0

We know that e^0=1 ( anything to the power zero is 1)

we get,

A=45

The initial amount of the substance is 45 grams

b)To find thehalf-life of this substance

To find t when the substance becames half the amount.

A=45/2

Substitute this we get,

\frac{45}{2}=45e^{-0.0045(t)}

\frac{1}{2}=e^{-0.0045(t)}

Taking natural logarithm on both sides we get,

\ln (\frac{1}{2})=-0.0045(t)^{}(-1)\ln (\frac{1}{2})=0.0045(t)\ln (\frac{1}{2})^{-1}=0.0045(t)\ln (2)=0.0045(t)0.6931=0.0045(t)t=\frac{0.6931}{0.0045}t=154.02

Half-life of this substance is 154.02

c) To find the amount of substance will be present around in 2500 years

Put t=2500

we get,

A=45e^{-0.0045(2500)}A=45e^{-11.25}A=45\times0.000013=0.000585A=0.000585

The amount of substance will be present around in 2500 years is 0.000585 grams

4 0
1 year ago
Rearrange the following group of numbers from smallest to largest. 2.7, 2.06, 2
allochka39001 [22]
2,2.06,2.7
Look at this
This is the reason
2, 2.06, 2.70
6 0
3 years ago
Given that f(x) = 2x + 1 and g(x) = −5x + 2, solve for f(g(x)) when x = 3.
ANTONII [103]
f(x)=2x+1 \\
g(x)=-5x+2 \\ \\
f(g(x))=2(-5x+2)+1=-10x+4+1=-10x+5 \\
f(g(3))=-10 \times 3+5=-30+5=-25 \\
\boxed{f(g(3))=-25}

The answer is C.
4 0
3 years ago
Choose the correct answer.
Ronch [10]
What figure?? Please elaborate
6 0
3 years ago
Write the equation in y intercept form
Vadim26 [7]

Answer:

y=-2x+4

Explanation:

Subtract 6x from both sides of the equation.

3y=12−6x

Divide each term by 3 and simplify.

Divide each term in 3y=12−6x by 3.

Cancel the common factor of 3.

Simplify

5 0
3 years ago
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