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2 years ago

What are all the triangles in geometry? And the angles too

Mathematics

1 answer:

Vikki [24]2 years ago

4 0

Equilateral triangle: all sides are equal lengths, all angles are equal too

Scalene triangle: has three different sides, all angles are different

Isosceles triangle: 2 sides are equal, one isn’t, 2 angles are the same, one is different

Acute triangle: angles are less than 90 degrees

Obtuse triangle: angles are larger than 90 degrees

Right triangle: angle forms a 90 degree angle

Those are the triangles :)

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Draw two lines that are perpendicular. What type of angle is formed where the perpendicular lines intersect?

Vladimir79 [104]

a right angle (90 degrees)

4 0

2 years ago

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2/3 of a number is 28 what is that number

Sidana [21]

Answer:

28×3÷2=14×2×3÷2=14×3=42

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2 years ago

The difference between the two roots of the equation 3x^2+10x+c=0 is 4 2/3 . Find the solutions for the equation.

andrezito [222]

Answer:

Given the equation: $3x^2+10x+c =0$

A quadratic equation is in the form: $ax^2+bx+c = 0$ where a, b ,c are the coefficient and a≠0 then the solution is given by :

$x_{1,2} = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ......[1]

On comparing with given equation we get;

a =3 , b = 10

then, substitute these in equation [1] to solve for c;

$x_{1,2} = \frac{-10\pm \sqrt{10^2-4\cdot 3 \cdot c}}{2 \cdot 3}$

Simplify:

$x_{1,2} = \frac{-10\pm \sqrt{100- 12c}}{6}$

Also, it is given that the difference of two roots of the given equation is $4\frac{2}{3} = \frac{14}{3}$

i.e,

$x_1 -x_2 = \frac{14}{3}$

Here,

$x_1 = \frac{-10 + \sqrt{100- 12c}}{6}$ , ......[2]

$x_2= \frac{-10 - \sqrt{100- 12c}}{6}$ .....[3]

then;

$\frac{-10 + \sqrt{100- 12c}}{6} - (\frac{-10 + \sqrt{100- 12c}}{6}) = \frac{14}{3}$

simplify:

$\frac{2 \sqrt{100- 12c} }{6} = \frac{14}{3}$

$\sqrt{100- 12c} = 14$

Squaring both sides we get;

$100-12c = 196$

Subtract 100 from both sides, we get

$100-12c -100= 196-100$

Simplify:

-12c = -96

Divide both sides by -12 we get;

c = 8

Substitute the value of c in equation [2] and [3]; to solve $x_1 , x_2$

$x_1 = \frac{-10 + \sqrt{100- 12\cdot 8}}{6}$

$x_1 = \frac{-10 + \sqrt{100- 96}}{6}$ or

$x_1 = \frac{-10 + \sqrt{4}}{6}$

Simplify:

$x_1 = \frac{-4}{3}$

Now, to solve for $x_2$ ;

$x_2 = \frac{-10 - \sqrt{100- 12\cdot 8}}{6}$

$x_2 = \frac{-10 - \sqrt{100- 96}}{6}$ or

$x_2 = \frac{-10 - \sqrt{4}}{6}$

Simplify:

$x_2 = -2$

therefore, the solution for the given equation is: $-\frac{4}{3}$ and -2.

3 0

2 years ago

PLEASE HELP!!!!!<br> Find AB and M

DENIUS [597]

Answer:

Step-by-step explanation:

triangle ABC is an equilateral so all angles and sides are equal to each other

AB = 6

triangle DEF is an isosceles so two of its angles are equal to each other

a triangles angles add up to 180°

to find the measure of these angles we’ll write this equation:

x + x + 40 = 180

and solve for x

2x + 40 = 180

2x = 140

x = 70°

so mF is 70°

7 0

2 years ago

15. Find to the nearest degree the measure of the acute angle that the graph

Vikentia [17]

Check the picture below.

make sure your calculator is in Degree mode.

7 0

2 years ago

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