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3 years ago

The sum of the interior angles of a triangle measures 180°. T or F?

1 answer:

7 0

This is true. The sum of interior angles always add up to 180<span>°.

Hopefully this helps! If you have any more questions or don't understand, feel free to DM me, and I'll get back to you ASAP! :)
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How do I solve Special quadratics?

mel-nik [20]

http://www.purplemath.com/modules/specfact.htm

7 0

2 years ago

Doug hits a baseball straight towards a 15 ft high fence that is 400 ft from home plate. The ball is hit 2.5ft above the ground

Scorpion4ik [409]

Answer:

$125.4\ \text{m/s}$

Step-by-step explanation:

u = Initial velocity of baseball

$\theta$ = Angle of hit = $30^{\circ}$

x = Displacement in x direction = 400 ft

y = Displacement in y direction = 15 ft

$y_0$ = Height of hit = 2.5 ft

$a_y$ = g = Acceleration due to gravity = $32.2\ \text{ft/s}^2$

t = Time taken

Displacement in x direction

$x=u_xt\\\Rightarrow x=u\cos\theta t\\\Rightarrow t=\dfrac{x}{u\cos\theta}\\\Rightarrow t=\dfrac{400}{u\cos30^{\circ}}\\\Rightarrow t=\dfrac{400}{u\dfrac{\sqrt{3}}{2}}\\\Rightarrow t=\dfrac{800}{u\sqrt{3}}$

Displacement in y direction

$y=y_0+u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow y=y_0+u\sin\theta t+\dfrac{1}{2}a_yt^2\\\Rightarrow 15=2.5+u\sin30^{\circ}(\dfrac{800}{u\sqrt{3}})+\dfrac{1}{2}\times -32.2\times (\dfrac{800}{u\sqrt{3}})^2\\\Rightarrow \dfrac{400}{\sqrt{3}}-\dfrac{10304000}{3u^2}-12.5=0\\\Rightarrow 218.44=\dfrac{10304000}{3u^2}\\\Rightarrow u=\sqrt{\dfrac{10304000}{3\times218.44}}\\\Rightarrow u=125.4\ \text{m/s}$

The minimum initial velocity needed for the ball to clear the fence is $125.4\ \text{m/s}$

5 0

3 years ago

ILL GIVE BRAINELST

murzikaleks [220]

Answer:

A = 12 ft

Step-by-step explanation:

Perimeter ( of SQUARE) B = 64

then each side = 64/4 = 16

AREA of SQUARE C = 400

side X side = 400

then side = sqrt 400

Now use Pyhtagorean Theorem to find leg of right triangle that is

side A Side B hypotenuse Side C

A^2 + 16^2 = (sqrt 400)^2

A^2 = 400 - 256

A^2 = 144 then A = 12 Ft

6 0

2 years ago

Simplify the expression

kow [346]

Answer:

Step-by-step explanation:

$\dfrac{5x^3-8x^2-4x}{6x^2-18x+12}\\\\6(x^2-3x+2)\ne0\ \iff\ x=\frac{3\pm\sqrt{9-8}}{2}\ne0\ \iff\ x\ne2\ \wedge\ x\ne1\\\\\\\dfrac{5x^3-8x^2-4x}{6x^2-18x+12}=\dfrac{x(5x^2-8x-4)}{6(x^2-3x+2)}=\dfrac{x(5x^2-10x+2x-4)}{6(x^2-2x-x+2)}=\\\\\\=\dfrac{x[5x(x-2)+2(x-2)]}{6[x(x-2)-(x-2)]} =\dfrac{x(x-2)(5x+2)}{6(x-2)(x-1)}=\dfrac{x(5x+2)}{6(x-1)}=\dfrac{5x^2+2x}{6x-6}\\\\\\f(x)=5x^2+2x\\\\g(x)=6x-6$

4 0

2 years ago

Can anyone help with this

vesna_86 [32]

I would be able to but i left my maths book at school and that has the explanation in, sorry :(

8 0

3 years ago

Read 2 more answers