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2 years ago

Please help Linear equations

Mathematics

1 answer:

svlad2 [7]2 years ago

8 0

Answer:

2 and 4

Step-by-step explanation:

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Hey can you please help me

anygoal [31]

Answer:

Step-by-step explanation:

General Form of an equation of a line is given by: $y=mx+b$

Where,

m is the slope, and

b is the y-intercept (point where line cuts the y-axis)

Also, parallel lines have equal slopes.

Equation of first plane is given as $d=3.67t+4.81$

Comparing this with general form of a line, we see that 3.67 is the slope and 4.81 is the y-intercept.

The second plane is going parallel to this. So slope should be same. Hence second plane has slope of 3.67. We can write:

$d=3.67t+c$

To find the equation of second plane, we need to solve for c. Also, a point given for second plane is $(t=0, d=2.14)$ , substituting these values into the equation we got, we get the value of c:

$2.14=3.67(0)+c\\2.14=0+c\\c=2.14$

We already know m and now we know c, so we can write final equation as:

$d=3.67t+2.14$

Answer choice C is right.

8 0

3 years ago

Y= -2x+3 fill in the table using this function rule.

sp2606 [1]

Answer:

there is no table!

Step-by-step explanation:

7 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

Can you exchange the coordinates of points ( x 1, y 1) and ( x 2, y 2) in the distance formula and still find the correct distan

notka56 [123]

Answer:

yes

Step-by-step explanation:

As an example

(x₁, y₁ ) = (1,2) and (x₂, y₂ ) = (5,3), then

d = $\sqrt{(5-1)^2+(3-2)^2}$ = $\sqrt{4^2+1^2}$ = $\sqrt{16+1}$ = $\sqrt{17}$