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2 years ago

Please help! Giving brainlist!

Mathematics

2 answers:

Sauron [17]2 years ago

8 0

Answer:

∠AXB or ∠PXQ or ∠AXQ or ∠PXB or ...

Step-by-step explanation:

A vertical angle is formed from the rays that are opposite the rays that form the original angle.

<h3>opposite rays</h3>

The rays forming angle CXD are rays XC and XD. The ray opposite XC will have end point X, and can be named by either of the two points shown on the ray: A or P. That is, rays XA and XP are opposite ray XC.

Similarly, rays XB and XQ are opposite ray XD.

<h3>vertical angle</h3>

The vertical angle can be named using either of the first pair of opposite rays with either of the second pair:

∠AXB, ∠PXB, ∠AXQ, ∠PXQ

Of course, the ray names can be reversed in the angle names:

∠BXA, ∠BXP, ∠QXA, ∠QXP

Altogether, there are 8 ways to name the vertical angle to ∠CXD.

aalyn [17]2 years ago

3 0

The angle that is vertical (opposite) with ∠CXD = ∠AXB

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Write the equation of the parabola that has its x-intercepts at (1+<img src="https://tex.z-dn.net/?f=%5Csqrt%7B5%7D" id="TexForm

VladimirAG [237]

Answer:

$y=2x^2-4x-8$

Step-by-step explanation:

<u>Factored form of a parabola</u>

$y=a(x-p)(x-q)$

where:

p and q are the x-intercepts.
a is some constant.

Given x-intercepts:

(1+√5, 0)
(1-√5, 0)

Therefore:

$\implies y=a(x-(1+\sqrt{5}))(x-(1-\sqrt{5}))$

$\implies y=a(x-1-\sqrt{5})(x-1+\sqrt{5})$

To find a, substitute the given point (4, 8) into the equation and solve for a:

$\implies a(4-1-\sqrt{5})(4-1+\sqrt{5})=8$

$\implies a(3-\sqrt{5})(3+\sqrt{5})=8$

$\implies4a=8$

$\implies a=2$

Therefore, the equation of the parabola in factored form is:

$\implies y=2(x-1-\sqrt{5})(x-1+\sqrt{5})$

Expand so that the equation is in standard form:

$\implies y=2(x^2-x+\sqrt{5}x-x+1-\sqrt{5}-\sqrt{5}x+\sqrt{5}-5)$

$\implies y=2(x^2-x-x+\sqrt{5}x-\sqrt{5}x+\sqrt{5}-\sqrt{5}+1-5)$

$\implies y=2(x^2-2x-4)$

$\implies y=2x^2-4x-8$

6 0

1 year ago

Someone help me with this question please.

grandymaker [24]

It’s 27 because u multiple the first 2 number and it’s a negative then add the rest to it

5 0

3 years ago

Solve for XX. Assume XX is a 2×22×2 matrix and II denotes the 2×22×2 identity matrix. Do not use decimal numbers in your answer.

sveticcg [70]

The question is incomplete. The complete question is as follows:

Solve for X. Assume X is a 2x2 matrix and I denotes the 2x2 identity matrix. Do not use decimal numbers in your answer. If there are fractions, leave them unevaluated.

$\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ =<em>I</em>.

First, we have to identify the matrix <em>I. </em>As it was said, the matrix is the identiy matrix, which means

<em>I</em> = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

So, $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Isolating the X, we have

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ - $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Resolving:

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}2-1&8-0\\-6-0&-9-1\end{array}\right]$

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, we have a problem similar to A.X=B. To solve it and because we don't divide matrices, we do X=A⁻¹·B. In this case,

X= $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ ⁻¹· $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, a matrix with index -1 is called Inverse Matrix and is calculated as: A . A⁻¹ = I.

So,

$\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ · $\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

9a - 3b = 1

7a - 6b = 0

9c - 3d = 0

7c - 6d = 1

Resolving these equations, we have a= $\frac{2}{11}$ ; b= $\frac{7}{33}$ ; c= $\frac{-1}{11}$ and d= $\frac{-3}{11}$ . Substituting:

X= $\left[\begin{array}{ccc}\frac{2}{11} &\frac{-1}{11} \\\frac{7}{33}&\frac{-3}{11} \end{array}\right]$ · $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$