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4 years ago

9

What is the word form of each decimal 0.2

2 answers:

tia_tia [17]4 years ago

6 0

What you think it is

8090 [49]4 years ago

4 0

Answer:

zero and two. . .

Step-by-step explanation:

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Find the inverse. Show your work. <br><br> It should probably be a fraction.

djyliett [7]

2/1 is the answer. Thank ya

5 0

3 years ago

Which polynomials are in standard form?

Kruka [31]

Answer:

Regular type is called the form in which polynomials are mostly used. The parameters are calibrated to the lowest degree. The polynomial x4 + 2x3+x+11 is standard, for instance, because four are most strong and three and one are the strongest.

Step-by-step explanation:

3 0

3 years ago

Can you simplify -4X - X =

vlabodo [156]

Sure ... exactly the way you simplify "2 + 2" when it comes up in conversation ... perform the indicated operation wherever possible. In simplified form, -4x-x is written as. " -5x ".

3 0

3 years ago

Save me the headache

maxonik [38]

$(9\sin2x+9\cos2x)^2=81$

Taking the square root of both sides gives two possible cases,

$9\sin2x+9\cos2x=9\implies\sin2x+\cos2x=1$

or

$9\sin2x+9\cos2x=-9\implies\sin2x+\cos2x=-1$

Recall that

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

If $\alpha=2x$ and $\beta=\dfrac\pi4$ , we have

$\sin\left(2x+\dfrac\pi4\right)=\dfrac{\sin2x+\cos2x}{\sqrt2}$

so in the equations above, we can write

$\sin2x+\cos2x=\sqrt2\sin\left(2x+\dfrac\pi4\right)=\pm1$

Then in the first case,

$\sqrt2\sin\left(2x+\dfrac\pi4\right)=1\implies\sin\left(2x+\dfrac\pi4\right)=\dfrac1{\sqrt2}$

$\implies2x+\dfrac\pi4=\dfrac\pi4+2n\pi\text{ or }\dfrac{3\pi}4+2n\pi$

(where $n$ is any integer)

$\implies2x=2n\pi\text{ or }\dfrac\pi2+2n\pi$

$\implies x=n\pi\text{ or }\dfrac\pi4+n\pi$

and in the second,

$\sqrt2\sin\left(2x+\dfrac\pi4\right)=-1\implies\sin\left(2x+\dfrac\pi4\right)=-\dfrac1{\sqrt2}$

$\implies2x+\dfrac\pi4=-\dfrac\pi4+2n\pi\text{ or }-\dfrac{3\pi}4+2n\pi$

$\implies2x=-\dfrac\pi2+2n\pi\text{ or }-\pi+2n\pi$

$\implies x=-\dfrac\pi4+n\pi\text{ or }-\dfrac\pi2+n\pi$

Then the solutions that fall in the interval $[0,2\pi)$ are

$x=0,\dfrac\pi4,\dfrac\pi2,\dfrac{3\pi}4,\pi,\dfrac{5\pi}4,\dfrac{3\pi}2,\dfrac{7\pi}4$

5 0

3 years ago

Read 2 more answers

Can anyone think of Think of a quadratic equation that has two (2) real number solutions? If you can, please write the equation

serious [3.7K]

$1( 1) {}^{2} + 1(1) - 2 = 0$

1(2)^2+3(2)-10 = 0

4 0

3 years ago