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Anna35 [415]
3 years ago
6

Write an equation ​(a) in​ slope-intercept form and ​(b) in standard form for the line passing through ​(22​,99​) and perpendicu

lar to 33xplus+55yequals=11.
Mathematics
1 answer:
liberstina [14]3 years ago
5 0
Y-99=-1/33(x-22)

a). y = 1/33x + 98 1/3

b) 33y = x + 109 1/3
x - 33y = -109 1/3
You might be interested in
Evaluate the expression for the given values.<br><br> 2x + (3y + z2)<br> X = 2, y = 3, z = 4
bazaltina [42]

Answer:

29

Step-by-step explanation:

2(2)+(3(3)+(4²))

4+(9+16)

4+25

29

pls if this was h helpful pls tag brainiest

3 0
2 years ago
X - y = 6
Dominik [7]

the system has no solution.

Option C is correct.

Step-by-step explanation:

We need to solve the system of equations by substitution

x - y = 6\,\,\,eq(1)\\x = y + 2\,\,\,eq(2)

Putting value of x from eq(2) into eq(1)

(y+2)-y=6\\y+2-y=6\\0+2=6\\0=6-2\\0\neq4

As as 0≠4is not true, we cannot find the value of y so the system has no solution.

Option C is correct.

Keywords: System of equations

Learn more about system of equations at:

  • brainly.com/question/7490805
  • brainly.com/question/9045597
  • brainly.com/question/13168205

#learnwithBrainly

4 0
3 years ago
Please help me on this
AlekseyPX

9514 1404 393

Answer:

  10 ft

Step-by-step explanation:

The midsegment length is the average of the other two lengths.

  (QR +PS)/2 = TU

  QR +PS = 2×TU . . . . . . . . . . . . . . . multiply by 2

  QR = 2×TU -PS = 2(14 ft) -18 ft . . . subtract PS; substitute given values

  QR = 10 ft

8 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}\\\\\\&#10;\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}\\\\\\&#10;\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}\\\\\\&#10;

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
Somebody help.. im so lost!!
aleksley [76]
What type of math is this?
4 0
3 years ago
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