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2 years ago

What is the card balance

Mathematics

1 answer:

GaryK [48]2 years ago

8 0

Answer:

A card balance is the total amount of money that you currently owe on your credit card. The balance increases when purchases are made and decreases when payments are made. Purchases, balance transfers, foreign exchange, fees, and interest all factor into your credit card balance.

please brainlest and like

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I have $1000 invested in a coin worth $43000. If the coin reaches $100000 how much will I have invested?

Pavel [41]

Answer:

$2326

Step-by-step explanation:

I have $1000 invested in a coin worth $43000. If the coin reaches $100000 how much will I have invested?

To solve the above question, we have:

$43,000 = $1000

$100,000 = x

Cross Multiply

43000 × x = 100,000 × 1000

x = 100000 × 1000/43000

x = $2325.5813953

Approximately = $2326

Therefore, If the coin reaches $100000 you would have invested $2326

8 0

3 years ago

The slope of (17,2) and (18,-17)

Ahat [919]

<u>m= -19/1</u>

We need to use the slope equation

$\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

We are working with the points,

(17, 2) and (18, -17)

x1 y1 x2 y2

$\frac{-17-2}{18-17}$

<u>m= -19/1</u>

4 0

3 years ago

A bag cost 4 times as much as a dress if the bag cost $276 how much will Toni speed on the bag and 3 such dresses

timama [110]

276 divides by 4 = 69
So 3x69 = 207 + 276 = $483

6 0

3 years ago

What are the solutions of the equation 2x2 = 2?

kkurt [141]

$2x^2=2\ \ \ |divide\ both\ sides\ by\ 2\\\\x^2=1\to \boxed{x=-1\ or\ x=1}$

3 0

3 years ago

We take a milk carton from the refrigerator and put it on the table at room temperature. We assume that the temperature of the c

Kisachek [45]

We can solve this problem using separation of variables.

Then apply the initial conditions

EXPLANATION

We were given the first order differential equation

$\frac{dT}{dt}=k(T-a)$

We now separate the time and the temperature variables as follows,

$\frac{dT}{T-a}=kdt$

Integrating both sides of the differential equation, we obtain;

$ln(T-a)=kt +c$

This natural logarithmic equation can be rewritten as;

$T-a=e^{kt +c}$

Applying the laws of exponents, we obtain,

$T-a=e^{kt}\times e^{c}$

$T-a=e^{c}e^{kt}$

We were given the initial conditions,

$T(0)=4$

Let us apply this condition to obtain;

$4-20=e^{c}e^{k(0)}$

$-16=e^{c}$

Now our equation, becomes

$T-a=-16e^{kt}$

or

$T=a-16e^{kt}$

When we substitute a=20,
we obtain,

$T=20-16e^{kt}$

b) We were also given that,

$T(5)=8$

Let us apply this condition again to find k.

$8=20-16e^{5k}$

This implied

$-12=-16e^{5k}$

$\frac{-12}{-16}=e^{5k}$

$\frac{3}{4}=e^{5k}$

We take logarithm to base e of both sides,

$ln(\frac{3}{4})=5k$

This implies that,

$\frac{ln(\frac{3}{4})}{5}=k$

$k=-0.2877$

After 15 minutes, the temperature will be,

$T=20-16e^{-0.2877\times 15}$

$T=20-0.21376$

$T=19.786$

After 15 minutes, the temperature is approximately 20°C

6 0

3 years ago