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3 years ago

Will give brainliest for BEST answer!

Mathematics

1 answer:

Nataliya [291]3 years ago

8 0

Use substitution and substitute y into the other equation. One of the equations already gives you y in terms of x, so use that and substitute it into the other equation. y = 3x - 4

Plug into the other equation: -3y = -9x + 12
-3(3x-4) = -9x + 12
-9x + 12 = -9x + 12
This is an identity. So that means that any value of x makes this equation true. So B.

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Multiply the binomial and the trinomial. <br> (x-2)(x^2+2x+4)

malfutka [58]

Answer:

Step-by-step explanation:

To solve this problem, we need to multiple $x$ and -2 by $x^{2} + 2x + 4$ and add the results together:

$x(x^{2} + 2x + 4)$

$x^{3} + 2x^{2} + 4x$

$-2(x^{2} + 2x + 4)$

$-2x^{2} - 4x - 8$

Adding the two together give the following:

$(x^{3} + 2x^{2} + 4x) + (-2x^{2} - 4x - 8)$

$x^{3} - 8$

7 0

3 years ago

Help me PLZ with the exact answer

olchik [2.2K]

Y = 4/-3 - 1

The slope can either be -4/3 or 4/-3 and the y-intercept is (0,1)

4 0

3 years ago

Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai

marta [7]

$f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1$

Lets check with every option

(a) [-4,-3]

We plug in -4 for x and -3 for x

$f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55$

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3 for x and -2 for x

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.

(c) [-2,-1]

We plug in -2 for x and -1 for x

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1 for x and 0 for x

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0 for x and 1 for x

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.

(f) [1,2]

We plug in 1 for x and 2 for x

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

$f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1], (d) [-1,0], (f) [1,2]

8 0

3 years ago

Read 2 more answers

Use DeMoivre's Theorem to find (3cis*pi/6)^3

Sonbull [250]

The correct value of (3cis(pi/6))³ is 27i.

<h3>What is Complex Number?</h3>

Complex numbers are numbers that consist of two parts — a real number and an imaginary number. Complex numbers are the building blocks of more intricate math, such as algebra.

Given the complex number in polar coordinate expressed as

z = r(cos∅+isin∅)

zⁿ = {r(cos∅+isin∅)}ⁿ

According to DeMoivre’s Theorem;

zⁿ = rⁿ(cosn∅+isinn∅)

Given the complex number;

(3cis(pi/6))³

= {3(cosπ/6 + isinπ/6)}³

Using DeMoivre’s Theorem;

= 3³(cos3π/6 + isin3π/6)

= 3³(cosπ/2 + isinπ/2)

= 3³(0 + i(1))

= 27i

Thus, the correct value of (3cis(pi/6))³ is 27i.

Learn more about Complex number from:

brainly.com/question/10251853

#SPJ1

7 0

2 years ago

Has anybody gotten a 100% on their FLVS Math segment exam?

Flura [38]

Answer:

i have

Step-by-step explanation:

3 0

2 years ago

Will give brainliest for *BEST* answer!

Will give brainliest for BEST answer!