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3 years ago

Given the 5th term is 103 and the come differences is -6 2/3 what is the 6th term?

Mathematics

1 answer:

LekaFEV [45]3 years ago

6 0

Answer:

102 1/3

Step-by-step explanation:

To find the 6th term, add the common difference to the 5th term:

103 + (-6 2/3) = 102 1/3

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Choose the best answer. The diagonals of a rectangle:

erastovalidia [21]

C.) Are the same length

No matter how long the rectangle is the diagonals always will measure the same length. Think about two sides of a square, they have to equal the same length because if they were not the same then the shape wouldn't be a square.

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3 years ago

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Consider the rational number, −0.4. Is the number greater than −2 and 1/3 but less than 4/5 ?

8_murik_8 [283]

Answer:

Step-by-step explanation:

Since the rational number -0.4 is a negative, it cannot have a greater value than 1/3 since it is a positive.

Though, the rest of the sentence is true.

-0.4 > -2

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3 years ago

a grocery store display has 2 boxes of cereal on the top row,3 boxes in the next row,4 boxes in the next row, and so on. How man

sveticcg [70]

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3 years ago

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Let alpha and beta be conjugate complex numbers such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}

miskamm [114]

Answer:

$-3+i\sqrt{3} , 1+\sqrt{3}$

Step-by-step explanation:

Given that alpha and beta be conjugate complex numbers

such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}.

Let

$\alpha = x+iy\\\beta = x-iy$

since they are conjugates

$\alpha-\beta = x+iy-(x-iy)\\= 2iy= 2i\sqrt{3} \\y =\sqrt{3}$

$\frac{\alpha}{\beta^2} }\\=\frac{x+i\sqrt{3} }{(x-i\sqrt{3})^2} \\=\frac{x+i\sqrt{3}}{x^2-3-2i\sqrt{3}} \\=\frac{x+i\sqrt{3}((x^2-3+2i\sqrt{3}) }{(x^2-3-2i\sqrt{3)}(x^2-3-2i\sqrt{3})}$

Imaginary part of the above =0

i.e. $\sqrt{3} (x^2-3)+2x\sqrt{3} =0\\x^2+2x-3=0\\(x+3)(x-1) =0\\x=-3,1$

So the value of alpha = $-3+i\sqrt{3} , 1+\sqrt{3}$

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3 years ago

FIND EF I WILL GIVE BRAINLIEST

Liono4ka [1.6K]

Can you explain ? i don't understood

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