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3 years ago

Which expression has a value of 15 when n = 7

Mathematics

2 answers:

expeople1 [14]3 years ago

6 0

The answer to this would be c

ahrayia [7]3 years ago

4 0

The answer to this question would be D

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Why do you multiply by a power of 10 when writing a repeating decimal as a rational number

stira [4]

1.28 = 128 * 10^-2

128 * (1/10^2)

128 / 10^2

128 / 100

1.28

Because, it's literally the same thing mathematically if you worked it out. Though, as I'm sure you're aware, it's just a much shorter and cleaner way of writing out large or small numbers.

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4 years ago

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lisabon 2012 [21]

Answer:

In technical terms, the slope of the line is the change in y over the change in x. But I just like to think of it as rise over run. To find the slope of the line, pick two points on the line.

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3 years ago

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A number patter has 75 as its first term. The rule for the pattern is subtract 6. What is the sixth term?

denpristay [2]

39 because you have to subtract six 6 times so 6*6=36 so 75-36=39

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3 years ago

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Can someone please help for brainlest

Klio2033 [76]

Answer:125√3:6

Step-by-step explanation:

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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

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

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3 years ago