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3 years ago

WILL MARK BRAINLIEST!!

Mathematics

2 answers:

Ket [755]3 years ago

7 0

Answer:

The correct answer is the third one

plz mark me brainsliest!!!!!

bija089 [108]3 years ago

3 0

Answer:

3 rd one

Step-by-step explanation:

Please mark me as brainliest, vote 5.0 and thank me to support

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Can anyone help with my algebra 1 homework??

Anni [7]

Answer:

I believe the final answer is x=3

Step-by-step explanation:

when trying to solve equations, always sort things out and prioritize things, that way it's easier to get to the final answer.

x+2x-5=4

3x-5=4 just move the -5 to the other side and make it a positive

3x=9

x+y=4

y=1

3+1=4

3 0

3 years ago

PLZ HURRY DIS IS TIMED !!!!

Bogdan [553]

Answer:

The 4-5

Step-by-step explanation:

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3 years ago

Fantom [35]

Answer:

what are the requirements??

5 0

3 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

Given A = {x | x < 1}, B = {x | x ≥ 5}, and C = {x | x = 5}, match the following items. 1. A B Ø 2. A C {x | x ≥ 5} 3. B C {x

ivann1987 [24]

Answer:

The matching is shown below.

Step-by-step explanation:

The given sets are defined as

A = {x | x < 1}

It means all the values of x which are less than 1.

B = {x | x ≥ 5}

It means all the values of x which are greater than or equal to 5.

C = {x | x = 5}

It means all the value of x is 5.

Union of two sets contains all the elements of both sets.

Intersection of two sets contains only common elements of both sets.

The matching is shown below.

Sets Correct value

1. A ∪ B {x | x < 1 or x ≥ 5}

2. A ∪ C {x | x < 1 or x = 5}

3. B ∪ C {x | x ≥ 5}

4. A ∩ B Ø

5. B ∩ C {x | x = 5}

6 0

3 years ago