Ask question

Ask question

All categories

3 years ago

15

Ralph sold brownies for $0.66 a piece in order to earn money to buy baseball cards. If the cards cost $0.90 per pack, and if

Ralph had no money left over after buying them, what is the least number of brownies he must have sold?

1 answer:

Olegator [25]3 years ago

6 0

The least amount of brownies that he has to sell would be 15 brownies in order to have no money left after buying the cards.

Hope this Helps

You might be interested in

Solve the system of linear equations 3x-2y=1 9x-6y=1

erik [133]

solution:

3x-2y=1 , -9x-6y=-3.

solution system can be represent in (13,0) ( 1 3 , 0 ).

6 0

3 years ago

I need help 1.09 divided by 3.08 show your solution Thank you so much if u helpt me

gayaneshka [121]

Answer:

<h2 /><h2><em>answe</em><em>r</em><em> will</em><em> be</em><em> </em><em>0</em><em>.</em><em>3</em><em>5</em><em>3</em><em>8</em></h2>

Step-by-step explanation:

$\frac{1.09}{3.08} = 0.3538$

6 0

4 years ago

Which country are you from?

densk [106]

Answer:

Whom are you asking????

3 0

3 years ago

Read 2 more answers

John enlarged one side of his square flower bed by 8 ft to make it into a rectangle. The area of the new flower bed is 3 times t

kobusy [5.1K]

Answer:

4

Step-by-step explanation:

4*4=16

(4+8)4=48

16*3=16!

5 0

3 years ago

Read 2 more answers

Find x?<br> In 3x - In(x - 4) = ln(2x - 1) +ln3

earnstyle [38]

Answer:

$x = \displaystyle \frac{5 + \sqrt{17}}{2}$ .

Step-by-step explanation:

Because $3\, x$ is found in the input to a logarithm function in the original equation, it must be true that $3\, x > 0$ . Therefore, $x > 0$ .

Similarly, because $(x - 4)$ and $(2\, x - 1)$ are two other inputs to the logarithm function in the original equation, they should also be positive. Therefore, $x > 4$ .

Let $a$ and $b$ represent two positive numbers (that is: $a > 0$ and $b > 0$ .) The following are two properties of logarithm:

$\displaystyle \ln (a) + \ln(b) = \ln\left(a \cdot b\right)$ .

$\displaystyle \ln (a) - \ln(b) = \ln\left(\frac{a}{b}\right)$ .

Apply these two properties to rewrite the original equation.

Left-hand side of this equation:

$\begin{aligned}&\ln(3\, x) - \ln(x - 4)= \ln\left(\frac{3\, x}{x -4}\right)\end{aligned}$

Right-hand side of this equation:

$\ln(2\, x- 1) + \ln(3) = \ln\left(3 \left(2\, x - 1\right)\right)$ .

Equate these two expressions:

$\begin{aligned}\ln\left(\frac{3\, x}{x -4}\right) = \ln(3(2\, x - 1))\end{aligned}$ .

The natural logarithm function $\ln$ is one-to-one for all positive inputs. Therefore, for the equality $\begin{aligned}\ln\left(\frac{3\, x}{x -4}\right) = \ln(3(2\, x - 1))\end{aligned}$ to hold, the two inputs to the logarithm function have to be equal and positive. That is:

$\displaystyle \frac{3\ x}{x - 4} = 3\, (2\, x - 1)$ .

Simplify and solve this equation for $x$ :

$x^2 - 5\, x + 2 = 0$ .

There are two real (but not rational) solutions to this quadratic equation: $\displaystyle \frac{5 + \sqrt{17}}{2}$ and $\displaystyle \frac{5 - \sqrt{17}}{2}$ .

However, the second solution, $\displaystyle \frac{5 - \sqrt{17}}{2}$ , is not suitable. The reason is that if $x = \displaystyle \frac{5 - \sqrt{17}}{2}$ , then $(x - 4)$ , one of the inputs to the logarithm function in the original equation, would be smaller than zero. That is not acceptable because the inputs to logarithm functions should be greater than zero.

The only solution that satisfies the requirements would be $\displaystyle \frac{5 + \sqrt{17}}{2}$ .

Therefore, $x = \displaystyle \frac{5 + \sqrt{17}}{2}$ .

7 0

3 years ago