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3 years ago

Can you please help me. I think it is x=0 and x=3

Mathematics

1 answer:

stiks02 [169]3 years ago

6 0

When the graph is constant, you have to look for when it draws a straight line.
Look at the x axis, since it asks for the x value.

Your answer is correct! When x = 0, to x = 3, the function is constant.

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If the cost of plastering 4 walls of a room at Rs 45 per sq. m is Rs 8100, find the area of the 4 walls of the room

olga55 [171]

Step-by-step explanation:

Rs8,100 / (Rs45 per square meter)

= 180 square meters.

The area of the 4 walls of the room is 180 sq.m.

3 0

2 years ago

What is the last number in the number sequence given below? 10 : 10 : 20 : 45 : 110 : 300 : ?

MrRissso [65]

To get from the first number to the second number the equation is 10x0.5+5=10
Thereafter the equation to get to the next number is 10x1.0+10=20
Thereafter the equation to get the next number is 20x1.5+15=45
Thereafter the equation to get the next number is 45x2.0+20=110
You can notice the pattern, the first number is what the last number was, you increase the second number by 0.5 and you increase the third number by 5.
Following this pattern to find the last number you do 110x2.5+25. This equation results in the number 300. Now time for the last number. You input 300x3.0+30. The answer results in 930. Therefore, the missing number is 930.

3 0

3 years ago

Read 2 more answers

Please help me I need it right now

Nana76 [90]

Answer:

X+y=180, Y=Z, X+Z=180

Step-by-step explanation:

Hope this helps. i remember doing this.

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

In an inequality between two numbers, -0.8 is located below the other number on the vertical number line. Which condition would

BigorU [14]

The answer is B. If -0.8 was below the other number, the other number would be above it.

5 0

2 years ago

Read 2 more answers