1492 finding problem only using this numbers

Mathematics

1 answer:

Novay_Z [31]3 years ago

6 0

1,429 . 294= ? ÷ 14 = 30,009

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Tonya wrote the equation below which solves for x.

vivado [14]

Given parameters:

x = 150 minus 6 y

Interpreting this gives;

minus = subtraction = -

x = 150 - 6y

Now bring the variables together;

x + 6y = 150

The option that matches with this equivalent.

The other choices are wrong;

a, x + 150 = 6 y

x - 6y = 150 wrong

c. 2x + 3y = 75 mulitply through by 2;

2(2x + 3y) = 2(75)

4x + 6y = 150 wrong

d. 2x - 3y = 75

Multiply by 2;

2(2x -3y) = 2(75)

4x - 6y = 150

7 0

3 years ago

He first term of a sequence is 2, and each subsequent term is the reciprocal of the square of the preceding term. what is the po

polet [3.4K]

The first term is a₁ = 2.
Each subsequent term is the reciprocal of the square of the preceding term.

2nd term:
a₂ = 1/2² = 1/4

3rd term:
a₃ = 1/(1/4)² = 1/(1/16) = 16

4th term:
a₄ = 1/16² = 1/256

5th term:
a₅ = 1/(1/256)² = 1/(1/65536) = 65536
The positive square root of the fifth term is
√(65536) = 256

Answer: 256

4 0

3 years ago

Which expression is equivalent to (a^-3 b/a^-5 b3)^-3? Assume a=0, b=0.

prohojiy [21]

$Use:\\\\\dfrac{a^n}{a^m}=a^{n-m}\\\\(a^n)^m=a^{nm}\\\\a^{-n}=\dfrac{1}{a^n}\\\\(ab)^n=a^nb^n$

$\left(\dfrac{a^{-3}b}{a^{-5}b^3}\right)^{-3}=\left(a^{-3-(-5)}b^{1-3}\right)^{-3}=\left(a^{-3+5}b^{-2}\right)^{-3}=\left(a^2b^{-2}\right)^{-3}\\\\=(a^2)^{-3}(b^{-2})^{-3}=a^{2(-3)}b^{-2(-3)}=a^{-6}b^6=\dfrac{b^6}{a^6}$