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guajiro [1.7K]
2 years ago
12

Find YZ. please help!!!

Mathematics
1 answer:
Ad libitum [116K]2 years ago
6 0
Your picture is a bit blurry. What is the equation on WX and on ZY?
You might be interested in
How do you graph a circle x^2 + y^2=25 and the line is given y=2
Kazeer [188]
\bf \textit{equation of a circle}\\\\ 
(x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2
\qquad 
\begin{array}{lllll}
center\ (&{{ h}},&{{ k}})\qquad 
radius=&{{ r}}
\end{array}\\\\
-------------------------------\\\\
x^2+y^2=25\implies (x-0)^2+(y-0)^2=5^2

so notice above, the circle is centered at h,k or 0,0, the origin, and has a ratio of 5.

and y = 2, is just a horizontal line, check the picture below.

7 0
3 years ago
A 40-degree angle is translated 5 inches along a vector. What is the angle measurement, in degrees, of the Image?
vagabundo [1.1K]

Answer:

40°

Step-by-step explanation:

As clarified in an online document, a translation along a vector (which is a line in a plane) of a figure, is equivalent to a translation along a coordinate grid, and therefore, given that a translation is a form of rigid transformation, the the dimensions and inclinations of the rays forming the preimage are the same as those in the image and the angles measurement in the preimage and the image are equal.

Therefore, given that the angle measurement of the image is 40-degrees, the angle measurement of the image is also 40-degree (40°) angle.

8 0
3 years ago
Is sin theta=5/6, what are the values of cos theta and tan theta?
mina [271]

let's bear in mind that sin(θ) in this case is positive, that happens only in the I and II Quadrants, where the cosine/adjacent are positive and negative respectively.

\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{6}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-5^2}=a\implies \pm\sqrt{36-25}\implies \pm \sqrt{11}=a \\\\[-0.35em] ~\dotfill

\bf cos(\theta )=\cfrac{\stackrel{adjacent}{\pm\sqrt{11}}}{\stackrel{hypotenuse}{6}} \\\\\\ tan(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{adjacent}{\pm\sqrt{11}}}\implies \stackrel{\textit{and rationalizing the denominator}~\hfill }{tan(\theta )=\pm\cfrac{5}{\sqrt{11}}\cdot \cfrac{\sqrt{11}}{\sqrt{11}}\implies tan(\theta )=\pm\cfrac{5\sqrt{11}}{11}}

5 0
3 years ago
When looking at the graph of a 5th degree function, how can you determine if all of the zeros of the function are real?
Finger [1]

Answer:

If it cuts x-axis 5 times.

Step-by-step explanation:

When we look at the graph of a function we can see its real roots by looking at its graph

The intersecting points that is the number of times a line cutting x-axis will be the real root of the function

So, by looking at the 5th degree function the number of time that function cuts x-axis will be the number of real roots.

So, if we need to say all the zeroes or roots of the function are real means it will cut the x-axis 5 times.

Because a function will have the root equal to its degree.

4 0
3 years ago
It’s 2019! To celebrate the New Year, students are going to count to 2,019. It takes 1 second to count each number. If they begi
lys-0071 [83]

Answer:

Yes they can

Step-by-step explanation:

1 hour is 3600 seconds

6 0
3 years ago
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