Answer: Choice A
![\sqrt[4]{3^3}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B3%5E3%7D)
The fourth root of 3^3
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Explanation:
The exponent of 1/4 means "fourth root", so,
![3^{1/4} = \sqrt[4]{3}](https://tex.z-dn.net/?f=3%5E%7B1%2F4%7D%20%3D%20%5Csqrt%5B4%5D%7B3%7D)
The outer exponent 3 can be moved inward like this
Then we can use the rule
![x^{m/n} = \sqrt[n]{x^m}](https://tex.z-dn.net/?f=x%5E%7Bm%2Fn%7D%20%3D%20%5Csqrt%5Bn%5D%7Bx%5Em%7D)
This means
![3^{3/4} = \sqrt[4]{3^3}](https://tex.z-dn.net/?f=3%5E%7B3%2F4%7D%20%3D%20%5Csqrt%5B4%5D%7B3%5E3%7D)
Answer:
a+2b-d=1, 3, 5, 7
Step-by-step explanation:
(ax^2+bx+3)(x+d)
ax^3+bx^2+3x+adx^2+bdx+3d
ax^3+bx^2+adx^2+3x+bdx+3d=x^3+6x^2+11x+12
ax^3=x^3, a=1
bx^2+adx^2=6x^2
x^2(b+ad)=6x^2
b+ad=6
b+(1)d=6
b+d=6
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3x+bdx=11x
x(3+bd)=11x
3+bd=11
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b=6-d
3+(6-d)d=11
3+6d-d^2=11
3-11+6d-d^2=0
-8+6d-d^2=0
d^2-6d+8=0
factor out,
(d-4)(d-2)=0
zero property,
d-4=0, d-2=0
d=0+4=4,
d=0+2=2
b=6-4=2,
b=6-2=4.
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a+2b-d=1+2(2)-2=1+4-2=5-2=3
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a+2(4)-4=1+8-4=9-4=5
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a+2(2)-4=1+4-4=5-4=1
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a+2(4)-2=1+8-2=9-2=7
Answer: B
Step-by-step explanation:
2/9 = 2(5)/9(5) = 10/45
You can multiply the numerator and the denominator by the same and the fraction will always be equivalent.
2 to 9 is equivalent to 10 to 45
