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3 years ago

PLEASE HELP. DUE IN 10 MINUTES. 20 POINTS!!

Mathematics

1 answer:

Kryger [21]3 years ago

7 0

Answer:

y=2+3x

Step-by-step explanation:

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Ok last question for today <br> Will mark brainliest

Pepsi [2]

Answer:9

Step-by-step explanation:

8 * 9 = 72

8, 16, 24, 32, 40, 48, 56, 64, 72

1, 2, 3, 4, 5, 6, 7, 8, 9

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3 years ago

Edward's school is 132 miles away from his home . He goes by car to school and back. The car consumed 4 gallons of gas taking hi

mixer [17]

Total distance is 132 miles x 2 times/day x 5 days = 1320 miles
That used 4 gallons of fuel, so 1320 miles/4 gallons = 330 miles per gallon

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3 years ago

Monthly payment of 1,000 where interest is 6% compounded monthly

inessss [21]

Answer: you would have 1,060 in a month if that’s what you are asking

Step-by-step explanation:

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3 years ago

The value of a certain car decreases by 16% each year. What is the 1⁄2-life of the car?

svet-max [94.6K]

Answer:

The half life of the car is 3.98 years.

Step-by-step explanation:

The value of the car after t years is given by the following equation:

$V(t) = V(0)(1-r)^{t}$

In which V(0) is the initial value and r is the constant decay rate, as a decimal.

The value of a certain car decreases by 16% each year.

This means that $r = 0.16$

$V(t) = V(0)(1-r)^{t}$

$V(t) = V(0)(1-0.16)^{t}$

$V(t) = V(0)(0.84)^{t}$

What is the 1⁄2-life of the car?

This is t for which V(t) = 0.5V(0). So

$V(t) = V(0)(0.84)^{t}$

$0.5V(0) = V(0)(0.84)^{t}$

$(0.84)^{t} = 0.5$

$\log{(0.84)^{t}} = \log{0.5}$

$t\log{0.84} = \log{0.5}$

$t = \frac{\log{0.5}}{\log{0.84}}$

$t = 3.98$

The half life of the car is 3.98 years.

7 0

3 years ago

What is the gcf of 23 and 50

valkas [14]

Answer:

The greatest common factor of 23 and 50 is 1

Step-by-step explanation:

7 0

3 years ago

￼PLEASE HELP. DUE IN 10 MINUTES. 20 POINTS!!

PLEASE HELP. DUE IN 10 MINUTES. 20 POINTS!!