Solve 8x-10y=12 and 4x-5y=5 by addition method with 0=2

Consider the equation below.

Eduardwww [97]

Answer:

The equation has no solutions .

Step-by-step explanation:

Given as :

<u>The linear equation is </u>

$\dfrac{3}{4}$ (24 -16 x) = $\dfrac{-2}{3}$ (18 x -27)

Or, $\dfrac{3}{4}$ ×24 - $\dfrac{3}{4}$ × 16 x = $\dfrac{-2}{3}$ × 18 x - $\dfrac{-2}{3}$ × 27

Or, $\dfrac{3}{4}$ ×24 - $\dfrac{3}{4}$ × 16 x = $\dfrac{-2}{3}$ × 18 x + $\dfrac{2}{3}$ × 27

Or, $\frac{3\times 24}{4}$ - $\frac{3\times 16 x}{4}$ = $\frac{-2\times 18 x}{3}$ + $\frac{2\times 27 x}{3}$

Or, 3 × 6 - 3 × 4 x = - 2 × 6 x + 2 × 9

Or, 18 - 12 x = - 12 x + 18

Or, 18 - 18 = - 12 x + 12 x

or, 0 = 0

∵ The linear equation has no variables , so the equation has no solution

Hence,The equation has no solutions . Answer

8 0

3 years ago

Please solve 108 = 2X-4(2X-9)

Fynjy0 [20]

108=2x-4(2x-9)
108=2x-8x+36
108=-6x+36
-36 -36
72=-6x
72/-6=-6x/-6
-12=x

3 0

3 years ago

The value 5 is an upper bound for the zeros of the function shown below.

Mice21 [21]

Answer:

The given statement that value 5 is an upper bound for the zeros of the function f(x) = x⁴ + x³ - 11x² - 9x + 18 will be true.

Step-by-step explanation:

Given

$f\left(x\right)\:=\:x^2\:+\:x^3\:-\:11x^2\:-\:9x\:+\:18$

We know the rational zeros theorem such as:

if $x=c$ is a zero of the function $f(x)$ ,

then $f(c) = 0$ .

As the $f\left(x\right)\:=\:x^2\:+\:x^3\:-\:11x^2\:-\:9x\:+\:18$ is a polynomial of degree $4$ , hence it can not have more than $4$ real zeros.

Let us put certain values in the function,

$f(5) = 448$ , $f(4) = 126$ , $f(3) = 0$ , $f(2) = -20$ ,

$f(1) = 0$ , $f(0) = 18$ , $f(-1) = 16$ , $f(-2) = 0$ , $f(-3) = 0$

From the above calculation results, we determined that $4$ zeros as

$x = -3, -2, 1,$ and $3$ .

Hence, we can check that

$f(x) = (x+3)(x+2)(x-1)(x-3)$

Observe that,

$for x > 3$ , $f(x)$ increases rapidly, so there will be no zeros for $x>3$ .

Therefore, the given statement that value 5 is an upper bound for the zeros of the function f(x) = x⁴ + x³ - 11x² - 9x + 18 will be true.

5 0

3 years ago

Graph the circle x2 + y2 + 10x + 6y + 18 = 0.

Minchanka [31]

Answer:

see below for a graph

Step-by-step explanation:

Completing the square for x and y terms can help you put this in the standard form.

(x^2 +10x) +(y^2 +6y) = -18

(x^2 +10x +25) +(y^2 +6x +9) = -18 +25 +9 . . . . add square of half linear coefficient

(x +5)^2 +(y +3)^2 = 16 . . . . . . . circle centered at (-5, -3) with radius 4

7 0

3 years ago

Read 2 more answers

Determine if Ginny factored correctly. If not, explain where she made an error. Ginny made a mistake in step 1 when she identifi

il63 [147K]

Quadratic equations are second-order equations. Ginney made a mistake when she identified b = 31. It should be b = –31.

<h3>What is a Quadratic Equation?</h3>

A quadratic equation is an equation that can be written in the form of

ax²+bx+c.

Where a is the leading coefficient, and

c is the constant.

In order to find the mistake that Ginny made, we will first find the factors ourselves, therefore, we will factorize the trinomial,

$6x^2 - 31x -30 = 0$

As we can see that the value of constants are,

a = 6

b = -31

c = 30

$6x^2 - 31x -30 = 0\\\\6x^2 - 36x+5x -30 = 0\\\\6x(x-6)+5(x-6)=0\\\\(6x+5)(x-6)=0$

Therefore, the factors of the trinomial $6x^2 - 31x -30 = 0$ are (6x+5) and (x-6).

If we compare our process with Ginny's process, we will find that she has taken the value of the constant b as 31 which is wrong.

Hence, Ginney made a mistake when she identified b = 31. It should be b = –31.

Learn more about Quadratic Equation:

brainly.com/question/17177510

3 0

3 years ago

Read 2 more answers