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Len [333]
3 years ago
13

How many 6-letter words can be made from the letters WASTEFUL if repetition of letters is allowed?

Mathematics
1 answer:
Tanzania [10]3 years ago
4 0
8 letters to pick from and you can repeat letters soo
8x8x8x8x8x8
= 8^6
= 262144
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Help on this please asap
Juli2301 [7.4K]

Answer:

87.75

Step-by-step explanation:

You are correct

3 0
2 years ago
The area of the shaded section is 63 square units. What is the value of x ? Help
Juli2301 [7.4K]
I found an image that corresponds to the above question.
The shape was a rectangle. Its width is 7 units, its length is x.
The shaded area of the rectangle forms a trapezoid.

Area of a trapezoid = [(a+b)/2] * h

63 sq. units = (7+x)/2 * 7
63/7 = (7+x)/2
9 = (7+x)/2
9*2 = 7 + x
18 = 7 + x
18 - 7 = x
11 = x

The value of x is 11 units. It is the length of the rectangle.

Area of the trapezoid = (a+b)/2 * h
63 = (7+11)/2 * 7
63 = 18/2 * 7
63 = 9 * 7
63 = 63 

7 0
3 years ago
Read 2 more answers
Which is a reasonable frist step that can be used to solve the equation 4x+3(x+2)=5(2x-3)
vagabundo [1.1K]

Answer:

they are not equal if that was what you were looking for

Step -by-step explanation:

8 0
3 years ago
Read 2 more answers
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
To factor 9x^2-16 you can first rewrite the expression as:
natka813 [3]
I hope this helps you

4 0
3 years ago
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