Ask question

Ask question

All categories

3 years ago

12

While playing games on a particular smart phone, 1% of the battery is drained every five minutes of game play. The battery start

s off at 80% full . This can be represented by b(m)=80-m/5. What does B(75) represent?

1 answer:

Nostrana [21]3 years ago

8 0

Answer:

B(75) represents the percent of battery full after 75 minutes of game play

Step-by-step explanation:

Let m represent the number of minutes of game play. If 1% of the battery is drained every five minutes of game play, then after m minutes of game play $\dfrac{m}{5}\%$ was drained.

The battery starts off at 80% full. After m minutes the battery will have

$B(m)=80-\dfrac{m}{5}$

percent full.

When $m=75$ minutes, then

$B(75)=80-\dfrac{75}{5}=80-15=65\%.$

You might be interested in

Help with this PLease

vekshin1

Since it asks you to estimate, you would round the 4.6 up to 5. The estimate volume would be 125 inches^3; answer C.

4 0

3 years ago

Is 4x=2(2x-2) A no solution or one solution or infinite solution

maxonik [38]

Answer:

no solution

Step-by-step explanation:

2 x 2x, 2 x -2, you now have 4x=4x-4, subtract 4x and 4x and you are left with -4

3 0

2 years ago

The given matrix is the augmented matrix for a linear system. Use technology to perform the row operations needed to transform t

shtirl [24]

Answer:

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

Step-by-step explanation:

As the given Augmented matrix is

$\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 1 :

$r_{1}$ ↔ $r_{1} - r_{2}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 2 :

$r_{3}$ ↔ $r_{3} - 8r_{1}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\0&124&-54&77&:-82\end{array}\right]$

Step 3 :

$r_{2}$ ↔ $\frac{r_{2}}{7}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&124&-54&77&:-82\end{array}\right]$

Step 4 :

$r_{1}$ ↔ $r_{1} + 14r_{2}$ , $r_{3}$ ↔ $r_{3} - 124r_{2}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&- \frac{254}{7} &\frac{663}{7} &:-\frac{1690}{7} \end{array}\right]$

Step 5 :

$r_{3}$ ↔ $\frac{r_{3}. 7}{254}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right]$

Step 6 :

$r_{1}$ ↔ $r_{1} - 4r_{3}$ , $r_{2}$ ↔ $r_{2} + \frac{1}{7} r_{3}$

$\left[\begin{array}{ccccc}1&0&0&-\frac{71}{127} &:\frac{176}{127} \\0&1&0&-\frac{131}{254} &:\frac{284}{127} \\0&0&1&-\frac{663}{254} &:\frac{845}{127} \end{array}\right]$

∴ we get

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

6 0

2 years ago

Question number 4 please help

Korvikt [17]

$\frac{x^2+x-30}{x^2-3x+2}\\\\x^2-3x+2=0\\\\x^2-x-2x+2=0\\\\x(x-1)-2(x-1)=0\\\\(x-1)(x-2)=0\iff x-1=0\ \vee\ x-2=0\\\\x=1\ \vee\ x=2\\\\Answer:For\ x=1\ or\ x=2.$

8 0

2 years ago

If the product of a number and -4 is subtracted from the number, the result is 3 more than the number. Find the number

lesya [120]

Answer: The number is 0.75 or 3/4

Step-by-step explanation:

let x = the number.

Now set up an equation: The product of a number (x) and -4 = -4x. So -4x is subtracted from the number - x. x-(-4x)

Then, that equals 3 more than the number (x) = x+3

So the equation is x-(-4x)=x+3.

Then, solve the equation!

x-(-4x)=x+3

x+4x=x+3 (Distribute the negative to the parentheses)

5x=x+3 (Combine like terms)

4x=3 (Get the Xs on one side by subtracting x from both sides)

x=0.75 or 3/4 (divide by the coefficiant, 4, on both sides)

3 0

3 years ago

Read 2 more answers