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3 years ago

Can someone please help me please!!!!!!!!!

Mathematics

1 answer:

Alla [95]3 years ago

7 0

What do they mean by describe the diagram

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A bucket of paint has spilled on a tile floor. The paint flow can be expressed with the function p(t) = 6t, where t represents t

alukav5142 [94]

P(t)=6t

A(p)=πp^2, since p(t)=6t

A(t)=π(p(t))^2

A(t)=π(6t)^2

A(t)=36πt^2, so when t=8 and approximating π≈3.14

A(8)≈36(3.14)(8^2)

A(8)≈36(3.14)64

A(8)≈7234.56 u^2

3 0

3 years ago

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Three consecutive integers have a sum of –21. Which equation can be used to find the value of the three numbers?

Ksju [112]

Answer:

The third one, x+(x+1)+(x+2)=-21 because x, x+1 and x+2 are three consecutive numbers.

6 0

3 years ago

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The point (–7, –3) is located in which quadrant?

andrezito [222]

Quadrant 3 is the answer

7 0

3 years ago

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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

3 years ago

Need help ASAP

madreJ [45]

Answer:

A 1/4

Step-by-step explanation:

2/3 x = 1/6

2x/3 = 1/6

2x/3 = 1/2x3

The answer is 1/4

8 0

3 years ago

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