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2 years ago

Solve by completing the square with working out -6x-x^2-8

Mathematics

1 answer:

Goryan [66]2 years ago

7 0

-6x-x^2-8 = -(x+2) x (x+4)

(This is if you are factoring the expression)

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A flash disk has the capacity of storing 1 megabyte of data. How many kilobytes of data can be stored in the same flash disk

Luden [163]

For this sort of problem, you need to be familiar with prefixes.

So:

megabyte = $1 *10^6$ bytes

kilobyte = $1*10^3$ bytes

and a regular byte would just be $1$

Now, you will need to do a conversion:

$1megabyte*\frac{1*10^6bytes}{1megabyte}*\frac{1*10^{-3}kilobyte}{1byte}$

I'm going to explain this just in case, but we convert megabytes to regular bytes in the first half by using the information I gave you above. (In one megabyte, there are $1*10^6$ bytes!)

In the second fraction thought, remember one kilobyte is $1*10^3$ bytes. However, you usually only see one of each thing on the bottom of fractions, so you need to add a negative sign to the 3. (technically you could just do $\frac{1kilobyte}{1*10^3bytes}$ , but I think it is more correct to do write it out how I did).

*** They do equal the same thing though! Do whichever way is easier for you!

Now, your answer should be:

$10^3$ kilobytes (1000 kilobytes)

Hope this helped!

6 0

2 years ago

12 is 90% of what number?<br><br><br> with steps

a_sh-v [17]

Answer:

<h3><u>Let's</u><u> </u><u>understand the concept</u><u>:</u><u>-</u></h3>

Here 12 is 90% of a number.
So we can say that 90% of That number is 12 .
In this way we can get the the number

<h3><u>Solution</u><u>:</u><u>-</u></h3>

Let,

the number=x

<h3>According to the question ,</h3>

${:}\longrightarrow$ $\sf 90\% \:of \;x=12$

We know 90%=90/100
so add it

${:}\longrightarrow$ $\sf {\dfrac {90}{100}}\times x=12$

${:}\longrightarrow$ $\sf {\dfrac {90x}{100}}=12$

Now use cross multiplication method

${:}\longrightarrow$ $\sf 90x=12\times 100$

${:}\longrightarrow$ $\sf 90x=1200$

Now interchange sides .

${:}\longrightarrow$ $\sf x={\dfrac {1200}{90}}$

${:}\longrightarrow$ $\sf x={\dfrac {200}{15}}$

Answer will come in decimals so we take approximate value.

${:}\longrightarrow$ $\sf x=13.3$

${:}\longrightarrow$ $\sf x=approximately \:13$

$\therefore$ The number is 13.

6 0

2 years ago

A cafe sells cups of coffee and tea. There were 6 times as many cups of coffee sold as cups of tea within an hour, for a total o

Maurinko [17]

9 cups of tea were sold.

For each 6 cups of coffee, 1 cup of tea was sold. So 54/6 = 9.

4 0

2 years ago

The demand for coffee is given by the following equation, where QD stands for the quantity demanded and P stands for price.

mart [117]

Using linear function concepts, it is found that the slopes and intercepts of the functions are given as follows:

a) -4.

b) P = 25.

c) 2.

d) P = 5.

<h3>What is a linear function?</h3>

A linear function is modeled by:

y = mx + b

In which:

m is the slope, which is the rate of change, that is, by how much y changes when x changes by 1.

b is the y-intercept, which is the value of y when x = 0, and can also be interpreted as the initial value of the function.

The demand is given by:

D = 100 - 4P.

Hence the slope is of -4. The demand is equals to zero when:

100 - 4P = 0 -> P = 25.

The supply is given by:

QS = -10 + 2P.

Hence the slope is of 2. The supply is equals to zero when:

-10 + 2P = 0 -> P = 5.

More can be learned about linear function concepts at brainly.com/question/24808124

#SPJ1

4 0

1 year ago

Unsure how to do this calculus, the book isn't explaining it well. Thanks

krok68 [10]

One way to capture the domain of integration is with the set

$D = \left\{(x,y) \mid 0 \le x \le 1 \text{ and } -x \le y \le 0\right\}$

Then we can write the double integral as the iterated integral

$\displaystyle \iint_D \cos(y+x) \, dA = \int_0^1 \int_{-x}^0 \cos(y+x) \, dy \, dx$

Compute the integral with respect to $y$ .

$\displaystyle \int_{-x}^0 \cos(y+x) \, dy = \sin(y+x)\bigg|_{y=-x}^{y=0} = \sin(0+x) - \sin(-x+x) = \sin(x)$

Compute the remaining integral.

$\displaystyle \int_0^1 \sin(x) \, dx = -\cos(x) \bigg|_{x=0}^{x=1} = -\cos(1) + \cos(0) = \boxed{1 - \cos(1)}$

We could also swap the order of integration variables by writing

$D = \left\{(x,y) \mid -1 \le y \le 0 \text{ and } -y \le x \le 1\right\}$

and

$\displaystyle \iint_D \cos(y+x) \, dA = \int_{-1}^0 \int_{-y}^1 \cos(y+x) \, dx\, dy$

and this would have led to the same result.

$\displaystyle \int_{-y}^1 \cos(y+x) \, dx = \sin(y+x)\bigg|_{x=-y}^{x=1} = \sin(y+1) - \sin(y-y) = \sin(y+1)$

$\displaystyle \int_{-1}^0 \sin(y+1) \, dy = -\cos(y+1)\bigg|_{y=-1}^{y=0} = -\cos(0+1) + \cos(-1+1) = 1 - \cos(1)$

7 0

8 months ago