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3 years ago

In the figure, j || k and m∠1 = 63° What is m∠6?

Mathematics

1 answer:

MakcuM [25]3 years ago

6 0

M∠6=117°. since j║k, that means that ∠1 and ∠5 are corresponding angles, so m∠1=m∠5 (m∠5=63°). ∠6 is the supplement of ∠5 so their combined angle measurements must equal to 180°. so m∠6+m∠5=180°, m∠6+63=180°. solve for m∠6. then you come up with m∠6=117°. Hope this helps!

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Give a reason why the number 0.393993999399993999993… is irrational

Zigmanuir [339]

It never ends. it is a never ending number and those aren’t rational

7 0

3 years ago

A rugby player passes the ball 7.75 m across the field, where it is caught at the same height as it left his hand. At what angle

prisoha [69]

Answer-

At 15.91° the ball was thrown.

Solution-

In the question it is given that the ball was released and caught from the same height.

We know that,

$R=\frac{u^{2}\sin 2\Theta}{g}$

Where,

R = Range of the projectile = 7.75 m,

u = Initial speed = 12 m/s

θ = Angle at which it was released

g = Acceleration due to gravity = 9.8 m/s²

Putting all the values,

$\Rightarrow 7.75 = \frac{12^{2}\sin 2\Theta}{9.8}$

$\Rightarrow \sin 2\Theta=\frac{7.75 \times 9.8}{144} =0.5274$

$\Rightarrow 2\theta=\sin^{-1} 0.5274=31.83 \ or \ 148.17$

$\Rightarrow \theta=\frac{31.83}{2} \ or \ \frac{148.17}{2} =15.91 \ or \ 74.08$

$\Rightarrow \theta=15.91^{\circ} \ (smaller \ angle)$

3 0

3 years ago

Read 2 more answers

CAN SOMEONE PLEASE JUST ANSWER THIS ASAP FOR BRAINLIEST!!!

BARSIC [14]

Answer:

F=7xy²

Step-by-step explanation:

Rewrite the question

4xy * F = 28x^2y^3 Divide by 4xy

$\Text{F = }\dfrac{28x^2y^3}{4xy}$

Divide the 4 into the 28.

Divide the x into the x^2

Divide the y into the y^3

$\Text{F=}\dfrac{7x * x * y * y * y}{4xy}$

$\Text{F=}7*x*y*y$

$\Text{F=}7*x*y^2$

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5 0

3 years ago

Using the quadratic formula to solve x2 + 20 = 2x, what are the values of x?

cluponka [151]

Step-by-step explanation:

x² + 20 = 2x

x² − 2x + 20 = 0

x = [ -b ± √(b² − 4ac) ] / 2a

x = [ -(-2) ± √((-2)² − 4(1)(20)) ] / 2(1)

x = [ 2 ± √(4 − 80) ] / 2

x = [ 2 ± √(-76) ] / 2

x = [ 2 ± 2i√19 ] / 2

x = 1 ± i√19

3 0

3 years ago

How to prove this???

swat32

$\cos^3 2A + 3 \cos 2A \\
\Rightarrow \cos 2A (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (1 - \sin^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - \sin^2 2A) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - (2\sin A \cos A)(2\sin A \cos A)) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - 4\sin^2 A \cos^2 A) \\ 
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$

go to right side now

$4( \cos^6 A - \sin^6 A)\\
\Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A)$

use $x^3 - y^3 = (x-y)(x^2 + xy + y^2)$ and $x^3 + y^3 = x^2 - xy + y^2$

$4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A) \\
\Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\
~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A)$

so $\sin^2 A + \cos^2 A = 1$

$4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\ ~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 + \cos A \sin A )(1- \cos A \sin A ) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \cos^2 A \sin^2 A )\\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \sin^2 A \cos^2 A ) \\
 \Rightarrow Left hand side$

4 0

3 years ago