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3 years ago

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Use the system of two linear equations to answer the question. 2x+5y=4 y=2x+8 Which of the following statements is correct about

the solution of the system? A The x-coordinate is 2. B The x-coordinate is -3. C The y-coordinate is -3. D The y-coordinate is 5.

1 answer:

Blababa [14]3 years ago

5 0

2x + 5(2x+8) = 4
2x + 10x + 40 = 4
12x = -36
x= -3
So it’s B

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Complete the square for 3x^2-12x=9

larisa [96]

Answer:

The solutions are -0.65 and 4.65.

Step-by-step explanation:

3x^2 - 12x = 9

Divide both sides by 3

x^2 - 4x = 3

To make a square with "x^2 - 4x", you need to have a "+ 4" after it so it can be simplified as (x-2)^2. So add 4 to both sides.

x^2 - 4x + 4 = 7

(x-2)^2 = 7

x - 2 = (square root) of 7

x = 2 (+/-) (square root) of 7.

x = {-0.65, 4.65}

The solutions are -0.65 and 4.65.

hope this helps:)sorry if it doesnt

8 0

3 years ago

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If a current of 12 amps produces 480 volts across a resistor then how many volts does a 15 amp produce?

LekaFEV [45]

Answer:

600 volts

Step-by-step explanation:

Given =

A = 12 ampere

V = 480 volts

A2 = 15 ampere

Solution =

R = V/I

= 480 / 12

= 40 ohms

R = V / I

V = R x I

= 40 x 15

= 600 volts

7 0

3 years ago

1. (i) Express 2704 as a product of its prime factors.

lidiya [134]

Answer:

all answers below

Step-by-step explanation:

(i) 2704 = 2^4 x 13^2

(ii) as both indices (the powers) are divisible by 2, it is a perfect square.

(iii) as both indices are not divisible by 3, it is not a perfect square.

8 0

3 years ago

Write out the form of the partial fraction decomposition of the function. Do not determine the numerical values of the coefficie

Dvinal [7]

For part (a), you have

$\dfrac x{x^2+x-6}=\dfrac x{(x+3)(x-2)}=\dfrac a{x+3}+\dfrac b{x-2}$
$x=a(x-2)+b(x+3)$

If $x=2$ , then $2=b(2-3)\implies b=-2$ .

If $x=-3$ , then $-3=a(-3-2)\implies a=\dfrac35$ .

So,

$\dfrac x{x^2+x-6}=\dfrac 3{5(x+3)}-\dfrac 2{x-2}$

For part (b), since the degrees of the numerator and denominator are the same, you first need to find the quotient and remainder upon division.

$\dfrac{x^2}{x^2+x+2}=\dfrac{x^2+x+2-x-2}{x^2+x+2}=1-\dfrac{x+2}{x^2+x+2}$

In the remainder term, the denominator $x^2+x+2$ can't be factorized into linear components with real coefficients, since the discriminant is negative $(1-4\times1\times2=-7)$ . However, you can still factorized over the complex numbers, so a partial fraction decomposition in terms of complexes does exist.

$x^2+x+2=0\implies x=-\dfrac12\pm\dfrac{\sqrt7}2i$
$\implies x^2+x+2=\left(x-\left(-\dfrac12+\dfrac{\sqrt7}2i\right)\right)\left(x-\left(-\dfrac12-\dfrac{\sqrt7}2i\right)\right)$
$\implies x^2+x+2=\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)$

Then you have

$\dfrac{x+2}{x^2+x+2}=\dfrac a{x+\dfrac12-\dfrac{\sqrt7}2i}+\dfrac b{x+\dfrac12+\dfrac{\sqrt7}2i}$
$x+2=a\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)+b\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)$

When $x=-\dfrac12-\dfrac{\sqrt7}2i$ , you have

$-\dfrac12-\dfrac{\sqrt7}2i+2=b\left(-\dfrac12-\dfrac{\sqrt7}2i+\dfrac12-\dfrac{\sqrt7}2i\right)$
$\dfrac32-\dfrac{\sqrt7}2i=-\sqrt7ib$
$b=\dfrac12+\dfrac3{2\sqrt7}i=\dfrac1{14}(7+3\sqrt7i)$

When $x=-\dfrac12+\dfrac{\sqrt7}2i$ , you have

$-\dfrac12+\dfrac{\sqrt7}2i+2=a\left(-\dfrac12+\dfrac{\sqrt7}2i+\dfrac12+\dfrac{\sqrt7}2i\right)$
$\dfrac32+\dfrac{\sqrt7}2i=\sqrt7ia$
$a=\dfrac12-\dfrac3{2\sqrt7}i=\dfrac1{14}(7-3\sqrt7i)$

So, you could write

$\dfrac{x^2}{x^2+x+2}=1-\dfrac{x+2}{x^2+x+2}=1-\dfrac {7-3\sqrt7i}{14\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)}-\dfrac {7+3\sqrt7i}{14\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)}$

but that may or may not be considered acceptable by that webpage.

5 0

3 years ago

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I Hope yall have an Amazing day!!

DIA [1.3K]

Yu too man :) :):) :)

6 0

3 years ago

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