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MatroZZZ [7]
2 years ago
10

Help pls :))))))))))))

Mathematics
1 answer:
Aneli [31]2 years ago
6 0

Answer:

yes, the triangle are congruent by SAS

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Does the following scenario represent independent or dependent events?
ra1l [238]
Independent
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Can I have brainliest
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3 years ago
Find the missing endpoint if S is the midpoint of RT <br><br>R(-9,4) and S(2,-1) ; Find T​
liraira [26]

Answer:

T is at (13,-6)

Step-by-step explanation:

x-coordinate of T: from -9 to 2, there's 11 units, so you add 11 to 2 to find the x-coordinate of T, which is 13.

y-coordinate of T: from 4 to -1 there's -5 units, so you subtract 5 from -1 to find the y-coordinate of T, which is -6.

4 0
3 years ago
The question is in the picture, please help!!<br> a. (16/7 , 6/7)<br> b.(4 , 6/7)<br> c.( 2 , 6/7)
lianna [129]

Answer:

b.(4,6/7)

Step-by-step explanation:

i hope i was able to help with the question

4 0
3 years ago
(3+root3)(3-root 2) is equal t<br><br><br><br> Pls its urgent
olganol [36]

Answer:

9 - √6

Step-by-step explanation:

(3 + √3)(3 - √2) ➡ 9 - 3√3 + 3√3 - √6

9 - √6

7 0
3 years ago
What does the fundamental theorem of algebra illustrate?
UNO [17]

Answer:

The fundamental theorem of algebra guarantees that a polynomial equation has the same number of complex roots as its degree.

Step-by-step explanation:

The fundamental theorem of algebra guarantees that a polynomial equation has the same number of complex roots as its degree.

We have to find the roots of this given equation.

If a quadratic equation is of the form ax^{2}+bx +c=0

Its roots are \frac{-b+\sqrt{b^{2}-4ac } }{2a} and \frac{-b-\sqrt{b^{2}-4ac } }{2a}

Here the given equation is 2x^{2}-4x-1 = 0

a = 2

b = -4

c = -1

If the roots are x_{1} and x_{2}, then

x_{1} = \frac{-2+\sqrt{(-4)^{2}-4\times 2\times (-1)}}{2\times 2}

                       = \frac{4 +\sqrt{24}}{4}

                       = \frac{2+\sqrt{6} }{2}

x_{2} = \frac{-2-\sqrt{(-4)^{2}-4\times 2\times (-1)}}{2\times 2}

                        = \frac{4 +\sqrt{8}}{4}

                        = \frac{2-\sqrt{6} }{2}

These are the two roots of the equation.

6 0
3 years ago
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