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3 years ago

What is the sum of all the integers between square root of 12 and 52? I need to know now pls!

Mathematics

1 answer:

Alexxx [7]3 years ago

5 0

Answer: 7&8

Step-by-step explanation:

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HELP please and thanks.. :)

fomenos

Answer:

false: the event will never occur; the event will definitely occur; the event is likely to occur

true: there is a small chance that the event will occur

Step-by-step explanation:

7 0

2 years ago

What is an equation of the line that passes through the points (-4,-2)<br> and (8, 1)?

stiv31 [10]

Answer:

y=1/4x-1

Step-by-step explanation:

7 0

2 years ago

Karen has 1/2 cup of sugar. She is baking cookies and needs 1 1/3 cups for a recipe. How much more sugar does she need If she wa

Evgen [1.6K]

Answer:

3 1/2 cups

Step-by-step explanation:

Let's start by figuring out how much sugar Karen needs in total.

She needs (1+1/3) cups for her original recipe. To triple that, we need (1+1/3) 3 times. We can calculate this by multiplying it by 3.

(1+1/3) * 3 = 3 + 1 = 4

Karen needs 4 cups in total. She has 1/2 cup. The difference is how much more she needs. Therefore, she needs 4 - 1/2 = 3 1/2 cups

3 0

2 years ago

At the bank, Derek made 7 withdrawals, each in the same amount. His brother, John, made 5 withdrawls, each in the same amount. E

spayn [35]

At the bank, Derek made 7 withdrawals, each in the same amount. His brother, John, made 5 withdrawls, each in the same amount.

Let x be the amount of one of Derek's withdrawals

Each of John's withdrawals was $5 more than each withdrawal that Derek made.

x + 5 is t the amount of one of John's withdrawals

Derek made 7 withdrawals

So amount withdraw 7 times = 7x

John made 5 withdrawals

So amount withdraw 5 times = 5(x+5)

Both Derek and John withdrew the same amount of money in the end

(A) 7x = 5(x+5)

(B) Solve for x

7x = 5x + 25

Subtract 5x from both sides

2x = 25

Divide by 2

x = 12.5

we plug in 12.5 for x in 7x= 5x + 25

7(12.5) = 5(12.5) + 25

87.5 = 62.5+ 25

87.5 = 87.5

(D) Each brother withdrawal 87.5 dollars

6 0

2 years ago

A circle has the order pairs (-1, 2) (0, 1) (-2, -1) what is the equation . Show your work.

olga55 [171]

We know that:

$(x-a)^2+(y-b)^2=r^2$

is an equation of a circle.

When we substitute x and y (from the pairs we have), we'll get a system of equations:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}$

and all we have to do is solve it for a, b and r.

There will be:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}\\\\\\
\begin{cases}1+2a+a^2+4-4b+b^2=r^2\\a^2+1-2b+b^2=r^2\\4+4a+a^2+1+2b+b^2=r^2\end{cases}\\\\\\
\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\\\\\
$

From equations (II) and (III) we have:

$\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}$

and from (I) and (II):

$\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}$

Now we can easly calculate a and b:

$\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}$

Finally we calculate $r^2$ :

$a^2+b^2-2b+1=r^2\\\\\\\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}+1=r^2\\\\\\\dfrac{9}{4}+\dfrac{1}{4}-1+1=r^2\\\\\\\dfrac{10}{4}=r^2\\\\\\\boxed{r^2=\frac{5}{2}}$

And the equation of the circle is:

$(x-a)^2+(y-b)^2=r^2\\\\\\\left(x-\left(-\dfrac{3}{2}\right)\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}\\\\\\\boxed{\left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}}$

7 0

3 years ago