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3 years ago

Which is a solution to the equation? (х-2)(х + 5) = 18?

1 answer:

4 0

$\bf (x-2)(x+5)=18\implies \stackrel{\mathbb{F~O~I~L}}{x^2+3x-10}=18\implies x^2+3x-28=0 \\\\\\ (x-4)(x+7)=0\implies x= \begin{cases} 4\\ -7 \end{cases}$

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Find the area of rectangle PLUM

Cerrena [4.2K]

PL= sqrt(4^2+12^2)=4sqrt(1+9)=4sqrt(10)

Using similarity gives that AM=4/3.

ML=40/3

[PLMU]=40/3 * 4 = 160/3 (53.33)

PM= sqrt(4^2+16/9)=sqrt(160/9)=4sqrt(10)/3

4sqrt(10)*4sqrt(10)/3=160/3 or approximately 53.33

7 0

3 years ago

CDE E=90 c=58 CD=40 feet

Anastasy [175]

Answer:

D=-12 CDE=130

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Suppose that the functions r and s are defined for all real numbers x as follows.

mixas84 [53]

ANSWER

Given

$r(x) = x - 1$

and

$s(x) = 3 {x}^{2}$
ANSWER TO QUESTION 1

$(r + s)(x) = r(x) + s(x)$

$(r + s)(x) = x - 1 + 3 {x}^{2}$

$(r + s)(x) = 3 {x}^{2} + x - 1$

ANSWER TO QUESTION 2

$(r \times s)(x) = r(x) \times s(x)$

$(r \times s)(x) = (x - 1) \times 3 {x}^{2}$

$(r \times s)(x) = 3 {x}^{3} - 3 {x}^{2}$

ANSWER TO QUESTION 3

$( r - s)(x) = r(x) - s(x)$

$( r - s)(x) = x - 1 - 3 {x}^{2}$

$( r - s)( - 3) = (- 3) - 1 - 3 ({ - 3}^{2} )$

$( r - s)( - 3) = - 4 - 3 \times 9$

$( r - s)( - 3) = - 4 - 27 = - 31$

4 0

3 years ago

5(x-7) = 6(x+2)<br>please help ... I got 11x = -23??

eimsori [14]

Answer:

$x = -47$

Step-by-step explanation:

1. Use the distributive property

5 ( x - 7 ) = 6 ( x + 2 ) → 5x - 35 = 6x + 12

2. Subtract 5x from both sides of the equation

5x - 5x -35 = 6x - 5x + 12 → -35 = x + 12

3. Subtract 12 from both sides

-35 - 12 = x + 12 - 12 → -47 = x

4. So, the answer is

$x = -47$

4 0

3 years ago

Can someone Answer this?

Misha Larkins [42]

Answer:

The limit of the function as x approaches 1 is ∞

Vertical asymptote; x = 1

Step-by-step explanation:

The graphical approach to evaluating limits involves graphing the function, the graph of the function $\frac{1}{(x-1)^{2} }$ is contained in the attachment below.

The red vertical line is the graph of x = 1. As we approach the line x = 1 from the left, the value of the function, that is y becomes large and large indefinitely. That is the value of the function approaches infinity. The same case applies when we approach the vertical line x = 1 from the right.

Also noticeable is the fact that the function approaches the line x = 1 asymptotically. The function gets closer and closer to this line but actually never touches it. Therefore, the line x = 1 is our vertical asymptote of the function given

7 0

4 years ago