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stepan [7]
3 years ago
14

Okay I got a question

Mathematics
1 answer:
Maslowich3 years ago
7 0
What’s the question?
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Somebody please help with this problem
Mrrafil [7]

Step-by-step explanation:

We have been given that AE=BE and \angle1\cong \angle2.

We can see that angle CEA is vertical angle of angle DEB, therefore, m\angle CEA=m\angle DEB as vertical angles are congruent.

We can see in triangles CEA and DEF that two angles and included sides are congruent.

\angle 1\cong \angle 2

AE=BE

\angle CEA\cong\angle DEB or \angle 3\cong \angle 4  

Therefore, \Delta CEA\cong \Delta DEB by ASA postulate.

Since corresponding parts of congruent triangles are congruent, therefore CE must be congruent to DE.

5 0
3 years ago
A vector with magnitude 9 points in a direction 190 degrees counterclockwise from the positive x axis. Write in component form
Over [174]

Answer:

\vec{v}= \text{ or } \approx

Step-by-step explanation:

Component form of a vector is given by \vec{v}=, where i represents change in x-value and j represents change in y-value. The magnitude of a vector is correlated the Pythagorean Theorem. For vector \vec{v}=, the magnitude is ||v||=\sqrt{i^2+j^2.

190 degrees counterclockwise from the positive x-axis is 10 degrees below the negative x-axis. We can then draw a right triangle 10 degrees below the horizontal with one leg being i, one leg being j, and the hypotenuse of the triangle being the magnitude of the vector, which is given as 9.

In any right triangle, the sine/sin of an angle is equal to its opposite side divided by the hypotenuse, or longest side, of the triangle.

Therefore, we have:

\sin 10^{\circ}=\frac{j}{9},\\j=9\sin 10^{\circ}

To find the other leg, i, we can also use basic trigonometry for a right triangle. In right triangles only, the cosine/cos of an angle is equal to its adjacent side divided by the hypotenuse of the triangle. We get:

\cos 10^{\circ}=\frac{i}{9},\\i=9\cos 10^{\circ}

Verify that (9\sin 10^{\circ})^2+(9\cos 10^{\circ})^2=9^2\:\checkmark

Therefore, the component form of this vector is \vec{v}=\boxed{}\approx \boxed{}

6 0
3 years ago
Please need help asap!!
worty [1.4K]

A property of rectangles is that opposite sides are congruent. In this case, since QT and RS are opposite sides, they are equal to each other. With this knowledge, you can form the equation 50=4x+8 to solve for x.

Firstly, subtract both sides by 8: 42=4x

Next, divide both sides by 4 and your answer will be 10.5=x, or the third option.

6 0
3 years ago
[1x-6]>1 what are the solutions
a_sh-v [17]

Answer:

x > 7

(I think this is Right but not sure yet )

3 0
3 years ago
HELP ME!! WILL MARK BRAINLIEST DUE TODAY!!<br> there are two parts to it !
Varvara68 [4.7K]

Answer:

Part A: A      

Step-by-step explanation:

Part B: -6 - x - 0

4 0
3 years ago
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