(40 points + Brainliest)

Mathematics

2 answers:

Sladkaya [172]3 years ago

7 0

The answer would be B because y=mx+b

Harman [31]3 years ago

3 0

Answer:

b) y = -5x -7

i had this question and this option was correct for me. i have a few friends from different schools but we took the same kind of test and they had varying answers so this might not be correct for some but i hope it helps someone in the future.

~hope i helped ouo have a nice rest of ur day~

lots of love,

lee

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five consecutive terms of an arithmetic sequence have a sum of 40. The product of the middle and the two end terms is 224. Find

ratelena [41]

Let the five terms be: a, a + d, a + 2d, a + 3d, a + 4d, then
a + a + d + a + 2d + a + 3d + a + 4d = 5a + 15d = 40
i.e. a + 3d = 8
Also, (a + 2d)(a + 3d)(a + 4d) = 224
(a + 3d - d)(a + 3d)(a + 3d + d) = 224
(8 - d)(8)(8 + d) = 224
(8 - d)(8 + d) = 224/8 = 28
64 - d^2 = 28
d^2 = 64 - 28 = 36
d = sqrt(36) = 6
But a + 3d = 8
a + 3(6) = 8
a = 8 - 18 = -10

Therefore, the term of the sequence is: -10, -10 + 6, -10 + 2(6), -10 + 3(6), -10 + 4(6)
= -10, -4, -10 + 12, -10 + 18, -10 + 24
= -10, -4, 2, 8, 14

4 0

3 years ago

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Pavlova-9 [17]

Answer: 15w

Step-by-step explanation:

6 0

3 years ago

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

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

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