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Nadusha1986 [10]
3 years ago
15

Plzzz help me with this ​

Mathematics
1 answer:
Illusion [34]3 years ago
5 0
Ok, to prove the two triangles, we need reasoning:

1) the side AC is common (as you can see, both triangles share the side AC)

2) angle DCA is equal to angle BAC (this is due to alternate angles)

3) angle BCA is equal to angle DA (again, this is due to alternate angles)
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4 0
3 years ago
The sum of two numbers is twenty-five. One number is five less then the other number. Find the larger number
ololo11 [35]

Answer:

15

Step-by-step explanation:

10 is 5 less than fifteen.

Sorry if its wrong

5 0
3 years ago
Represent this statement as an equation: The product of a number and 4 plus 2 is 14.
DiKsa [7]

Answer:

4x+2=14

Step-by-step explanation:

8 0
3 years ago
Rewrite the following without an exponent.<br> (-5)-2 (the -2 is the exponent)
yuradex [85]

Answer: -5 x -5

The exponent two means the other number multiplied 2 times, so the equation is -5 x -5. The answer to the equation is -25.

Hope this helped you out!

3 0
3 years ago
Find all real solutions to the equation (x² − 6x +3)(2x² − 4x − 7) = 0.
Jet001 [13]

Answer:

x = 3 + √6 ; x = 3 - √6 ; x = \frac{2+3\sqrt{2}}{2} ;  x = \frac{2-(3)\sqrt{2}}{2}

Step-by-step explanation:

Relation given in the question:

(x² − 6x +3)(2x² − 4x − 7) = 0

Now,

for the above relation to be true the  following condition must be followed:

Either  (x² − 6x +3) = 0 ............(1)

or

(2x² − 4x − 7) = 0 ..........(2)

now considering the equation (1)

(x² − 6x +3) = 0

the roots can be found out as:

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

for the equation ax² + bx + c = 0

thus,

the roots are

x = \frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times(3)}}{2\times(1)}

or

x = \frac{6\pm\sqrt{36-12}}{2}

or

x = \frac{6+\sqrt{24}}{2} and, x = x = \frac{6-\sqrt{24}}{2}

or

x = \frac{6+2\sqrt{6}}{2} and, x = x = \frac{6-2\sqrt{6}}{2}

or

x = 3 + √6 and x = 3 - √6

similarly for (2x² − 4x − 7) = 0.

we have

the roots are

x = \frac{-(-4)\pm\sqrt{(-4)^2-4\times2\times(-7)}}{2\times(2)}

or

x = \frac{4\pm\sqrt{16+56}}{4}

or

x = \frac{4+\sqrt{72}}{4} and, x = x = \frac{4-\sqrt{72}}{4}

or

x = \frac{4+\sqrt{2^2\times3^2\times2}}{2} and, x = x = \frac{4-\sqrt{2^2\times3^2\times2}}{4}

or

x = \frac{4+(2\times3)\sqrt{2}}{2} and, x = x = \frac{4-(2\times3)\sqrt{2}}{4}

or

x = \frac{2+3\sqrt{2}}{2} and, x = \frac{2-(3)\sqrt{2}}{2}

Hence, the possible roots are

x = 3 + √6 ; x = 3 - √6 ; x = \frac{2+3\sqrt{2}}{2} ; x = \frac{2-(3)\sqrt{2}}{2}

7 0
3 years ago
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