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Neporo4naja [7]

2 years ago

15

The cost of a peanut butter bar is $0.07 more than the cost of a chocolate bar. If you buy 5 peanut butter bars and 6 chocolate

bars, the total cost is $6.40. How much does the chocolate bar cost?
$0.61

$0.55

$0.54

$0.62

1 answer:

Licemer1 [7]2 years ago

4 0

Hello!

To be quick and simple, your answer would be $0.55

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Hello? i have a question.

Vedmedyk [2.9K]

Answer:

whats your question

Step-by-step explanation:

4 0

3 years ago

1.Arsenic-74 is used to locate brain tumors. It has a half-life of 17.5 days. 90 mg wereused in a procedure. Write an equation t

Sophie [7]

$1)\text{ }N_t\text{ = 90\lparen}\frac{1}{2})^{\frac{t}{17.5}}$

2) 5.625 mg will be left

Explanation:

1) Half-life = 17.5 days

initial amount of Arsenic-74 = 90 mg

To get the equation, we will use the equation of half-life:

$\begin{gathered} N_t\text{ = N}_0(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}} \\ where\text{ N}_t\text{ = amount remaining} \\ N_0\text{ = initial amount} \\ t_{\frac{1}{2}\text{ }}\text{ = half-life} \end{gathered}$ $N_t\text{ = 90\lparen}\frac{1}{2})^{\frac{t}{17.5}}$

2) we need to find the remaining amount of Arsenic-74 after 70 days

t = 70

$\begin{gathered} N_t=\text{ 90\lparen}\frac{1}{2})^{\frac{70}{17.5}} \\ N_t\text{ = 5.625 mg} \end{gathered}$

So after 70 days, 5.625 mg will be left

4 0

1 year ago

C(2, 5), D(3, 1)<br><br> Find the slope of the line that passes through each pair of points.

Tpy6a [65]

Slope=-4 or -4/1

Work:

(2, 5) (3, 1)
x1 y1. x2 y2

M= y2-y1/x2-x1
M= 1-5/3-2
M= -4/1
M=-4

5 0

3 years ago

The points A (0,4), B (6, 4) and C (6, -2) are 3 corners of a square ABCD.

Brilliant_brown [7]

Answer: It would be (0,-2)

Step-by-step explanation:

6 0

2 years ago

A sum of money is invested at 12% compounded quarterly. About how long will it take for the amount of money to double?

kompoz [17]

Answer:

5.9 years

Step-by-step explanation:

We believe a better representation of the compound interest formula is ...

$V(t)=P(1+\dfrac{r}{n})^{nt}$

We want to find the value of t for P=1 and V(t)=2. We are told n=4, so the formula becomes ...

$2=(1+\dfrac{.12}{4})^{4t}\\\\\log{(2)}=4t\log{(1.03)}\quad\text{take logs}\\\\t=\dfrac{\log{2}}{4\log{1.03}}\approx\boxed{5.9\quad\text{years}}$

8 0

3 years ago