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4 years ago

14

Tiffany put a $550 item on layaway by making a 15% down payment and agreeing to pay $20 a month. How many months sooner would sh

e pay off the item if she increased her monthly payment to $80?

2 answers:

Harrizon [31]4 years ago

5 0

18 months or 4 times faster

mixer [17]4 years ago

3 0

Answer:3 months

Step-by-step explanation:

APEX

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What is the absolute value of Point Clabeled on the number line?

Yuki888 [10]

Answer:

2 1/4

Step-by-step explanation:

An absolute value cannot be negative, so the negative answer choices are eliminated. The point labeled C is 5 tick marks to the right of 1. We know from the other numbers on the line that 4 tick marks constitutes one unit, so C is 5/4 = 1 1/4 units to the right of 1. Its value is 2 1/4.

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3 years ago

If y = 7x – 5, which of the following sets represents possible inputs and outputs of the function, represented as ordered pairs?

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The third set, (0, -5) (2,9) (-3,-26)

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3 years ago

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The circle below is centered at the point (4,1) and has a radius of length 2. What is it’s equation?

vekshin1

Answer:

Option B:

$(x-4)^2+(y-1)^2 = 4$

Step-by-step explanation:

Given

Centre = (h,k) = (4,1)

Radius = r = 2

The standard equation of circle is:

$(x-h)^2+(y-k)^2=r^2$

Putting the values of h,k and r

$(x-4)^2+(y-1)^2 = (2)^2\\(x-4)^2+(y-1)^2 = 4$

Hence, the correct option is option B

$(x-4)^2+(y-1)^2 = 4$

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3 years ago

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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 $r = 0.16$

So

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

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

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

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.

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