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3 years ago

Which graph best represents the equation y = -2x + 5?

Mathematics

1 answer:

Setler [38]3 years ago

7 0

Answer:

B) The top right choice.

Step-by-step explanation:

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this is a hard math question the volume of a prism is the product of its height and area of its base, v = bh. a rectangular pris

Lerok [7]

V = bh
16y^4 + 16y^3 + 48y^2 = 16y^2(y^2 + y + 3)
Therefore, the prism has a base area of 16y^2 square units and height of y^2 + y + 3 units

8 0

3 years ago

How many arangments of letters in PARALLEL

coldgirl [10]

There are 8! ways to arrange the 8 letters. Due to the repeated L (3×) and A (2×), only one out of (2!)(3!) = 12 of these is unique.

The number of unique arrangements is 8!/(2!*3!) = 3,360

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3 years ago

P(x)=Third-degree, with zeros of −3, −1, and 2, and passes through the point (1,12).

Mila [183]

Answer:

The polynomial is:

$p(x) = -x^3 - 2x^2 + 5x + 6$

Step-by-step explanation:

Zeros of a function:

Given a polynomial f(x), this polynomial has roots $x_{1}, x_{2}, x_{n}$ such that it can be written as: $a(x - x_{1})*(x - x_{2})*...*(x-x_n)$ , in which a is the leading coefficient.

Zeros of −3, −1, and 2

This means that $x_1 = -3, x_2 = -1, x_3 = 2$ . Thus

$p(x) = a(x - x_{1})*(x - x_{2})*(x-x_3)$

$p(x) = a(x - (-3))*(x - (-1))*(x-2)$

$p(x) = a(x+3)(x+1)(x-2)$

$p(x) = a(x^2+4x+3)(x-2)$

$p(x) = a(x^3+2x^2-5x-6)$

Passes through the point (1,12).

This means that when $x = 1, p(x) = 12$ . We use this to find a.

$12 = a(1 + 2 - 5 - 6)$

$-12a = 12$

$a = -\frac{12}{12}$

$a = -1$

Thus

$p(x) = -(x^3+2x^2-5x-6)$

$p(x) = -x^3 - 2x^2 + 5x + 6$

6 0

3 years ago

Hey Mahfia, please help! Find an equation in standard form for the hyperbola with vertices at (0, ±10) and asymptotes at

Dafna1 [17]

Answer:

$\huge\boxed{ \red{ \boxed{ \tt{ \frac{ {y}^{2} }{ {10}^{2} } - \frac{ {x}^{2} }{ {12}^{2} } = 1}}}}$

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>

conic sections
PEMDAS

<h3>tips and formulas:</h3>

$\sf hyperbola \:equation : \\ \sf \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1$
vertices of hyperbola:(±a,0) and (0,±b) if reversed
$\sf \: asymptotes : \\ y = \pm\frac{b}{a} x$