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2 years ago

pls pls pls help i give brainest and i will give 37 points and i will give a thanks plssss sove 22 and 23

Mathematics

1 answer:

Alex787 [66]2 years ago

8 0

<h2>Answer:</h2><h2>what do u mean solve 22&23</h2>

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NO LINKS!! Use the method of substitution to solve the system. (If there's no solution, enter no solution). Part 11z

Roman55 [17]

Answer:

$(x,y)=\left(\; \boxed{-1,-8} \; \right)\quad \textsf{(smaller $x$-value)}$

$(x,y)=\left(\; \boxed{5,16} \; \right)\quad \textsf{(larger $x$-value)}$

Step-by-step explanation:

Given system of equations:

$\begin{cases}y=x^2-9\\y=4x-4\end{cases}$

To solve by the method of substitution, substitute the first equation into the second equation and rearrange so that the equation equals zero:

$\begin{aligned}x^2-9&=4x-4\\x^2-4x-9&=-4\\x^2-4x-5&=0\end{aligned}$

Factor the quadratic:

$\begin{aligned}x^2-4x-5&=0\\x^2-5x+x-5&=0\\x(x-5)+1(x-5)&=0\\(x+1)(x-5)&=0\end{aligned}$

Apply the <u>zero-product property</u> and solve for x:

$\implies x+1=0 \implies x=-1$

$\implies x-5=0 \implies x=5$

Substitute the found values of x into the <u>second equation</u> and solve for y:

$\begin{aligned}x=-1 \implies y&=4(-1)-4\\y&=-4-4\\y&=-8\end{aligned}$

$\begin{aligned}x=5 \implies y&=4(5)-4\\y&=20-4\\y&=16\end{aligned}$

Therefore, the solutions are:

$(x,y)=\left(\; \boxed{-1,-8} \; \right)\quad \textsf{(smaller $x$-value)}$

$(x,y)=\left(\; \boxed{5,16} \; \right)\quad \textsf{(larger $x$-value)}$

6 0

10 months ago

Read 2 more answers

The product of two consecutive positive integers is 132. What are the integers?

alina1380 [7]

Answer:

The integers are 11 and 12

Step-by-step explanation:

11 and 12 are consecutive positive integers, and 11 x 12 = 132.

5 0

2 years ago

What is the true solution to the equation below? In e^in x + in e^in x^2 = 2 in 8

mariarad [96]

To solve the given equation, w need to review some rules:
$(1) \ ln \ e = 1 \\ 
(2) \ ln \ b^{a} = a*ln \ b \\
(3) \ ln \ a + ln \ b = ln \ (ab) \\
(4) \ ln \ a - ln \ b = ln \ \frac{a}{b} \\$

The given equation is :
$ln \ e^{ln \ x} + ln \ e^{ln \ x^{2}} = 2 \ ln \ 8$
$ln \ x * ln \ e + ln \ x^{2} * ln \ e = 2 \ ln \ 8$ ⇒⇒⇒⇒ rule (2)
$ln \ x + ln \ x^{2} = ln \ 8^{2}$ ⇒⇒⇒⇒ rule (1) and rule (2)

$ln \ ( x* x^{2} ) = ln \ 64$ ⇒⇒⇒⇒ rule (3)
removing the nature logarithm from both sides

$x^{3} = 64 = 4^{3}$
∴ x = 4

So, the correct answer is option (2) ⇒⇒⇒⇒ x = 4

5 0

2 years ago

Read 2 more answers

Add or subtract. (3/x-2)+(4/x+5)

Alex787 [66]

The answer is 7x+7/x^2+3x-10

3 0

2 years ago

1250 / 16 show work please answer fast

OLga [1]

Answer:

78.125

Step-by-step explanation:

sorry i can show the work :(

5 0

2 years ago