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2 years ago

Which angles are alternate

Mathematics

2 answers:

Artyom0805 [142]2 years ago

7 0

Answer:

KLO and POL

Step-by-step explanation:

fomenos2 years ago

6 0

Answer:

POL and KLO

Step-by-step explanation:

Alternate interior angles are between the parallel lines and on opposite sides of the transversal

NOL and MLO

POL and KLO

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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

Factor x^4 + x y^3 + x^3 + y^3 completely. Show your work.

andrey2020 [161]

$\bf x^4+xy^3+x^3+y^3\impliedby \textit{now let's do some grouping}
\\\\\\
(x^4+xy^3)~~+~~(x^3+y^3)\implies x(x^3+y^3)~~+~~(x^3+y^3)
\\\\\\
\stackrel{\stackrel{common}{factor}}{(x^3+y^3)}(x+1)\\\\
-------------------------------\\\\
\textit{now recall that }\qquad \qquad \textit{difference of cubes}
\\ \quad \\
a^3+b^3 = (a+b)(a^2-ab+b^2)\qquad
a^3-b^3 = (a-b)(a^2+ab+b^2)\\\\
-------------------------------\\\\
(x+y)(x^2-xy+y^2)(x+1)$

5 0

3 years ago

Question 1 (3 points) What value of p makes this equation true? 3p - 8 = -23 . O p = -6 Op = -5 Op = 5 p = 6

olchik [2.2K]

The value of p that makes the given equation true is equal to: B. p = -5

<u>Given the following equation:</u>

$3p - 8 = -23$

To find a value of p that makes the given equation true:

In this exercise, you're required to determine a value of p that satisfies the given equation true such that when substituted into the equation, it has a true result or outcome.

Rearranging the equation by collecting like terms, we have:

$3p - 8 = -23\\\\3p=-23+8\\\\3p=-15\\\\p=\frac{-15}{3}$

p = -5

Find more information: brainly.com/question/3600420

7 0

3 years ago

Question 4<br> Find the slope and y-intercept in the table. Simplify as needed.

Agata [3.3K]

Answer:

Y - intercept is 1 , slope is 3

4 0

3 years ago

Please answer thanks a lot!

Evgen [1.6K]

Vol (pyr) = 1/3 b × h, where h = 15 and b = base = area of triangular base = 1/2 b×h, where h = 12 and b = 13
V = 1/3 (1/2×12×13)×15
V = (1/3×1/2×12)×13×15
V = 2×13×15 = 30×13 = 390 in^3

6 0

4 years ago

Which angles are alternate ​

Which angles are alternate