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Ganezh [65]
3 years ago
14

Find the value of b. Express the answer as a simplified radical.

Mathematics
2 answers:
Nimfa-mama [501]3 years ago
8 0
Py Pythagoras theorem ,we can find value of b
(AC)^2=(AB)^2+(BC)^2
(14)^2=(b)^2+(8)^2
=>196=b^2+64
=>196-64=b^2
=>132=b^2
=>2root3 =b
Hope this helps u...!!

Mama L [17]3 years ago
6 0

Here three sides of a right angled triangle given.

The three sides are b, 8 and 14.

The side with length 14 is the hypotenuse here. We have to find the side of length b.

To get b, we will use Pythagoras theorem here.

By using Pythagoras theorem we can write,

c^2=a^2+b^2, where c is the hypotenuse.

So here we can write,

14^2=8^2+b^2

196 = 8^2+b^2

196=64+b^2

To solve it for b, now we have to move 64 to the left side by subtracting it from both sides.

196-64 = 64+b^2-64

196-64 = b^2

132=b^2

Now to find b, we have to take square root to both sides. We will get,

\sqrt{132}= \sqrt{b^2}

\sqrt{132} =b

b= \sqrt{(4)(33)}

b= (\sqrt{4})(\sqrt{33})

b= 2\sqrt{33}

We have got the required answer for the side b.

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3 years ago
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HELP ASAP!! Identify the matrix transformation of ΔPQR, which has coordinates P(−5, −2),Q(−6, −3), and R(−2, −3), for reflection
podryga [215]

Answer:

Matrix transformation = \left[\begin{array}{ccc}-1&0\\0&1\end{array}\right]

Vertices of the new image: P'= (5,-2), Q'= (6,-3), R'= (2,-3)

Step-by-step explanation:

Transformation by reflection will produce a new congruent object in different coordinate. Reflection to y-axis made by multiplying the x coordinate with -1 and keep the y coordinate unchanged. The matrix transformation for reflection across y-axis should be: \left[\begin{array}{ccc}-1&0\\0&1\end{array}\right].

To find the coordinate of the vertices after transformation, you have to multiply the vertices with the matrix. The calculation of the each vertice will be:

P'= \left[\begin{array}{ccc}-1&0\\0&1\end{array}\right] \left[\begin{array}{ccc}-5\\-2\end{array}\right]= (5,-2)

Q'= \left[\begin{array}{ccc}-1&0\\0&1\end{array}\right] \left[\begin{array}{ccc}-6\\-3\end{array}\right]= (6,-3)

R'= \left[\begin{array}{ccc}-1&0\\0&1\end{array}\right] \left[\begin{array}{ccc}-2\\-3\end{array}\right]= (2,-3)

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Which figure is a counterexample to the following statement?
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The correct option is D.

Option A. isn't even about quadrilater, so we can immediately discard it.

Option B. statement is true, but has nothing to do with the point of the question. In fact, it is true that every square is in particular a rectangle, but in turn every rectangle is a parallelogram. So, there's no counterexample here

Option C. is false, because a dart is a parallelogram: both of its opposite sides are parallel.

Option D. finally presents a counterexample. In fact, The two bases of a trapezoid are parallel, but the two other sides are not. So, a trapezoid is not a parallelogram, even though it has a pair of parallel sides. This is way, in order to be a parallelogram, it is necessary for the quadrilateral to have two pairs of parallel sides.

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