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3 years ago

ABCD is a quadrilateral.

Mathematics

2 answers:

wariber [46]3 years ago

4 0

6x+90=360

6x=270

x=45

No lines are not parallel

weqwewe [10]3 years ago

3 0

Answer:

A) 45=x

B) Yes, since both A and B are 90°

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How do we graph f(x)=x-7

babymother [125]

The graph of f(x) is a straight line passing through points (0, -7) and (7, 0)

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3 years ago

Find the answers please?

vodka [1.7K]

Answer:

<1 and <5 by the corresponding angles theorem

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3 years ago

Read 2 more answers

If you bent length 44 cm is bent into a circle find the radius of the circle?If the same wire is bent into the shape of a square

Tju [1.3M]

Answer:

r = 7 cm; l = 11cm

The circle encloses more area than the square

Step-by-step explanation:

(a) Radius of circle

The formula for the circumference of a circle is

A = 2πr

r = A/(2π)

Data:

C = 44 in

Calculation:

r = 44/(2 × 22/7)

= 44/2 × 7/22

= 2/2 × 7

= 7 in

(b) Side of square

P = 4l

l = P/4

= 44/4

= 11 in

(c) Areas

(i) Circle

A = πr²

= 22/7 × 7²

= 22/7 × 49

= 22 × 7

= 154 cm²

(ii) Square

A = l²

= 11²

= 121 cm²

The circle encloses more area than the square.

5 0

4 years ago

Explain the difference between the union and intersection of two sets.

Margaret [11]

Answer:

The union of two sets is a new set that contains all of the elements that are in at least one of the two sets.

Step-by-step explanation:

7 0

3 years ago

General solutions of sin(x-90)+cos(x+270)=-1<br> {both 90 and 270 are in degrees}

mixer [17]

Answer:

$\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

Step-by-step explanation:

Given:

$\sin (x-90^{\circ})+\cos(x+270^{\circ})=-1$

First, note that

$\sin (x-90^{\circ})=-\cos x\\ \\\cos(x+270^{\circ})=\sin x$

So, the equation is

$-\cos x+\sin x= -1$

Multiply this equation by $\frac{\sqrt{2}}{2}:$

$-\dfrac{\sqrt{2}}{2}\cos x+\dfrac{\sqrt{2}}{2}\sin x= -\dfrac{\sqrt{2}}{2}\\ \\\dfrac{\sqrt{2}}{2}\cos x-\dfrac{\sqrt{2}}{2}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos 45^{\circ}\cos x-\sin 45^{\circ}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos (x+45^{\circ})=\dfrac{\sqrt{2}}{2}$

The general solution is

$x+45^{\circ}=\pm \arccos \left(\dfrac{\sqrt{2}}{2}\right)+2\pi k,\ \ k\in Z\\ \\x+\dfrac{\pi }{4}=\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{4}\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

4 0

4 years ago