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Anastasy [175]
3 years ago
11

What transformation maps quadrilateral EFGH to quadrilateral QRSP?

Mathematics
2 answers:
irga5000 [103]3 years ago
7 0

Hello there

the answer to number 4 is.

90° degrees counterclockwise rotation

hope this helps

thank you

Best Regards Queen Z

sashaice [31]3 years ago
6 0

The markings indicate the figures are congruent. The orientation of both figures is clockwise, so no reflection is involved. The direction of QR appears to be 90° counterclockwise from the direction of EF, It appears as though the center of rotation is somewhere in the triangle EGS.

The transformation appears to be a 90° CCW rotation.

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Some one help me on this one
aleksandrvk [35]
The answer is III only, or D. 


We can start to solve this by knowing what the HL theorem means. The HL theorem, like its name implies, shows says that if a hypotenuse and leg of a triangle are congruent to the hypotenuse and leg of a different triangle, then the triangles are congruent. The only triangle that we see a hypotenuse congruent in is in figure III. In figure II, those congruent sides are both legs while in figure I we just see 2 congruent angles. Now in figure III, we can also see that two legs are congruent because of the reflexive property. That means that the answer is III, or D. 
5 0
3 years ago
Pls help...................
bearhunter [10]
First one b ≤ 9 good luck
5 0
3 years ago
Petes boat can travel 48 miles upstream in 4 hours. The return trip takes 3 hours. Find the speed of the boat in still water and
tankabanditka [31]
So... hmm bear in mind, when the boat goes upstream, it goes against the stream, so, if the boat has speed rate of say "b", and the stream has a rate of "r", then the speed going up is b - r, the boat's rate minus the streams, because the stream is subtracting speed as it goes up

going downstream is a bit different, the stream speed is "added" to boat's
so the boat is really going faster, is going b + r

notice, the distance is the same, upstream as well as downstream
thus   \bf \begin{cases}
b=\textit{rate of the boat}\\
r=\textit{rate of the river}
\end{cases}\qquad thus
\\\\\\

\begin{array}{lccclll}
&distance&rate&time(hrs)\\
&----&----&----\\
upstream&48&b-r&4\\
downstream&48&b+4&3
\end{array}
\\\\\\

\begin{cases}
48=(b-r)(4)\to 48=4b-4r\\\\
\frac{48-4b}{-4}=r\\
--------------\\
48=(b+r)(3)\\
-----------------------------\\\\
thus\\\\
48=\left[ b+\left(\boxed{\frac{48-4b}{-4}}\right) \right] (3)
\end{cases}

solve for "r", to see what the stream's rate is

what about the boat's? well, just plug the value for "r" on either equation and solve for "b"
5 0
3 years ago
Given that
VLD [36.1K]

-17/2

Step-by-step explanation:

30-8(8)+4y

30=64+4y

30-64=4y

-34=4y

y= -34/4

y= -17/2

hope it helps!

6 0
3 years ago
Solve the following quadratic equation.
Nonamiya [84]

Given:

Consider the given equation is

(x-18)^2=1

To find:

The values of x.

Solution:

We have,

(x-18)^2=1

Taking square root on both sides, we get

x-18=\pm\sqrt{1}

x-18=\pm 1

Adding 18 on both sides, we get

x-18+18=\pm 1+18

x=\pm 1+18

Now,

x=1+18 and x=-1+18

x=19 and x=17

Therefore, the correct option is A.

5 0
3 years ago
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