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3 years ago

9

Frank invested $1000 in an account with a 6% interest rate compounded annually. How much is the account after 3 years?

1 answer:

Wewaii [24]3 years ago

4 0

I belive c is the corect awnser

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Cameron, Arthur, and Jamie are playing soccer. Their locations are recorded by a motion tracking system. The grid shows distanc

Luden [163]

Answer:

1. Arthur is about 54 meters away from Cameron.

2. Jamie is about 56 meters away from Cameron.

3. Arthur is closer to Cameron.

Step-by-step explanation:

We have been given the locations of Cameron (70,10), Arthur (20,30) and Jamie (45,60) on the the grid in meters.

1. To find the distances of Arthur and Jamie from Cameron we will use distance formula.

$\text{Distance}=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

$\text{Distance between Cameron and Arthur}=\sqrt{(70-20)^{2}+(10-30)^{2}}$

$\text{Distance between Cameron and Arthur}=\sqrt{(50)^{2}+(-20)^{2}}$

$\text{Distance between Cameron and Arthur}=\sqrt{2500+400}$

$\text{Distance between Cameron and Arthur}=\sqrt{2900}$

$\text{Distance between Cameron and Arthur}=53.8516480713450403\approx 54$

Therefore, the distance Arthur is about 54 meters away from Cameron.

2. Let us find the distance between Cameron and Jamie.

$\text{Distance between Cameron and Jamie}=\sqrt{(70-45)^{2}+(10-60)^{2}}$

$\text{Distance between Cameron and Jamie}=\sqrt{(25)^{2}+(-50)^{2}}$

$\text{Distance between Cameron and Jamie}=\sqrt{625+2500}$

$\text{Distance between Cameron and Jamie}=\sqrt{3125}$

$\text{Distance between Cameron and Jamie}=55.9016994374947424\approx 56$

Therefore, Jamie is about 56 meters away from Cameron.

3. We can see that 56 is greater than 54, therefore, Arthur is closer to Cameron.

4 0

3 years ago

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Find the product. (1.8 x 10^5)(3.1 x 10^2)

Stells [14]

Answer:

5.58*10^7

Step-by-step explanation:

Recall that 10^a*10^b = 10^(a+b). Thus, 10^5*10^2 = 10^7.

Thus, (1.8 x 10^5)(3.1 x 10^2) = (1.8)(3.1)*10^7 = 5.58*10^7

7 0

3 years ago

Read 2 more answers

Element X is a radioactive isotope such that its mass decreases by 26% every day. If an experiment starts out with 810 grams of

ioda

Answer:

The hourly decay rate is of 1.25%, so the hourly rate of change is of -1.25%.

The function to represent the mass of the sample after t days is $A(t) = 810(0.74)^t$

Step-by-step explanation:

Exponential equation of decay:

The exponential equation for the amount of a substance is given by:

$A(t) = A(0)(1-r)^t$

In which A(0) is the initial amount and r is the decay rate, as a decimal.

Hourly rate of change:

Decreases 26% by day. A day has 24 hours. This means that $A(24) = (1-0.26)A(0) = 0.74A(0)$ ; We use this to find r.

$A(t) = A(0)(1-r)^t$

$0.74A(0) = A(0)(1-r)^{24}$

$(1-r)^{24} = 0.74$

$\sqrt[24]{(1-r)^{24}} = \sqrt[24]{0.74}$

$1 - r = (0.74)^{\frac{1}{24}}$

$1 - r = 0.9875$

$r = 1 - 0.9875 = 0.0125$

The hourly decay rate is of 1.25%, so the hourly rate of change is of -1.25%.

Starts out with 810 grams of Element X

This means that $A(0) = 810$

Element X is a radioactive isotope such that its mass decreases by 26% every day.

This means that we use, for this equation, r = 0.26.

The equation is:

$A(t) = A(0)(1-r)^t$

$A(t) = 810(1 - 0.26)^t$

$A(t) = 810(0.74)^t$

The function to represent the mass of the sample after t days is $A(t) = 810(0.74)^t$

5 0

3 years ago

The amount Diego charges to detail an SUV is proportional to the time it takes him to detail the vehicle. Diego charges $37.50 t

prisoha [69]

Answer:

$d = 15h$

Step-by-step explanation:

Given

$Charges\ \alpha\ Time$

Represent charges with d and time with h

So:

$d = \$37.50$ when $t = 2.50\ hours$

Required

Determine the formula

$Charges\ \alpha\ Time$

In other words,

$d\ \alpha\ h$

Convert to an equation

$d = k * h$

Where

k = constant of proportion

Substitute values for Charges and Time

$37.50 = k * 2.5$

Solve for k

$k = 37.50/2.50$

$k = 15$

Substitute 15 for k in $d = k * h$

$d = 15 * h$

$d = 15h$

<em>The above formula models the situation</em>

8 0

3 years ago

Plzzz help ?? Don't know

tensa zangetsu [6.8K]

I think its c yo...............

8 0

3 years ago