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3 years ago

What is the final reason???? I will give brainliest!

Mathematics

1 answer:

DaniilM [7]3 years ago

8 0

Well CD and CB are both equal to each other because they’re the same angles , it’s a reflection

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The y intercept of the line 5y + 2x -3=0

Alecsey [184]

5y+2x-3=0

5y=-2x+3

y=-2/5 x+3

Sope intercept of the line is y=-2/5 x+ 3/5.

The intercept is 3/5.

Check: x=0 when y=3/5

5y+2x-3=0

5(3/5)+2x-3

3+2x-3=0

2x=0

x=0

Correct

The answer is 3/5.

5 0

3 years ago

23j-21g+20e-13+52e-37j+45-49e<br> (Like terms)

eduard

The solution to the algebraic expression is: 23e - 21g - 14j + 32

What are algebraic expressions?

Algebraic expressions are mathematical expressions that contain variables, coefficients, and arithmetic operations such as addition, subtraction, division, and multiplication.

Solving algebraic expressions are an important part of mathematics as it helps to improve the aptitude and solving skills of the students.

From the given information, we have;

23j - 21g + 20e - 13 + 52e - 37j + 45 - 49e

let's rearrange by taking the like terms to the same sides;

= 23j -37j - 21g + 20e + 52e - 49e - 13 + 45

= -14j - 21g + 23e + 32

= 23e - 21g - 14j + 32

Therefore, we can conclude that the solution to the algebraic expression is: 23e - 21g - 14j + 32

Learn more about solving algebraic expressions here:

brainly.com/question/4344214

#SPJ1

3 0

2 years ago

Use the discriminant to determine what type of roots the equations will have, and categorize the equations according to their ro

topjm [15]

Step-by-step explanation:

The discriminant of the quadratic equation $ax^2+bx+c=0$ :

$\Delta=b^2-4ac$

If Δ < 0, then the equation has two complex roots $x=\dfrac{-b\pm\sqrt\Delta}{2a}$

If Δ = 0, then the equation has one repeated root $x=\dfrac{-b}{2a}[/tex If Δ > 0, then the equation has two discint roots [tex]x=\dfrac{-b\pm\sqrt\Delta}{2a}$

$1.\ x^2-4x+2=0\\\\a=1,\ b=-4,\ c=2\\\\\Delta=(-4)^2-4(1)(2)=16-8=8>0,\ \bold{two\ distinct\ roots}\\\sqrt\Delta=\sqrt8=\sqrt{4\cdot2}=2\sqrt2\\\\x=\dfrac{-(-4)\pm2\sqrt2}{2(1)}=\dfrac{4\pm2\sqrt2}{2}=2\pm\sqrt2\\\\==============================\\\\2.\ 5x^2-2x+3=0\\\\a=5,\ b=-2,\ c=3\\\\\Delta=(-2)^2-4(5)(3)=4-60=-56$

$3.\ 2x^2+x-6=0\\\\a=2,\ b=1,\ c=-6\\\\\Delta=1^2-4(2)(-6)=1+48=49>0,\ \bold{two\ distinct\ roots}\\\sqrt\Delta=\sqrt{49}=7\\\\x=\dfrac{-1\pm7}{(2)(2)}\\\\x_1=\dfrac{-8}{4}=-2,\ x_2=\dfrac{6}{4}=\dfrac{3}{2}\\\\==============================\\\\4.\ 13x^2-4=0\qquad\text{add 4 to both sides}\\\\13x^2=4\qquad\text{divide both sides by 13}\\\\x^2=\dfrac{4}{13}\to x=\pm\sqrt{\dfrac{4}{13}},\ \bold{two\ distinct\ roots}\\\\==============================$

$5.\ x^2-6x+16=0\\\\a=1,\ b=-6,\ c=16\\\\\Delta=(-6)^2-4(1)(16)=36-64=-28$

$7.\ 4x^2+11=0\qquad\text{subtract 11 from both sides}\\\\4x^2=-11\qquad\text{divide both sides by 4}\\\\x^2=-\dfrac{11}{4}\to x=\pm\sqrt{-\dfrac{11}{4}}\\\\x=\pm\dfrac{\sqrt{11}}{2}\ i,\ \bold{two\ complex\ roots}$