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3 years ago

45 POINTS!! Please help me solve this Table I’m having a hard time this is my last chance to pass!

Mathematics

1 answer:

Anettt [7]3 years ago

8 0

Solution:

Given function:

f(x) = x - 1

Solution (x = -2):

f(x) = x - 1
=> -2 - 1
=> -3

Solution (x = 0):

f(x) = x - 1
=> 0 - 1
=> -1

Solution (x = 2):

f(x) = x - 1
=> 2 - 1
=> 1

Solution (x = 3):

f(x) = x - 1
=> f(x) = 3 - 1
=> 2

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Solve the following linear quadratic system of equations algebraically. y=^2+3x-2 y+3=5x

USPshnik [31]

Answer:

$x=1$

$y=2$

Step-by-step explanation:

$y=x^2+3x-2$ , $y+3=5x$

Replace all occurrences of $y$ in $y+3=5x$ with $x^2+3x-2.$

$(x^2+3x-2)+3=5x$

$y=x^2+3x-2$

Add $-2$ and 3.

$x^2+3x+1=5x$

$y=x^2+3x-2$

Subtract 5x from both sides of the equation.

$x^2+3x+1-5x=0$

$y=x^2+3x-2$

Subtract 5x from 3x.

$x^2-2x+1=0$

$y=x^2+3x-2$

Rewrite 1 as $1^2$ .

$x^2-2x+1^2=0$

$y=x^2+3x-2$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2x=2$ · $x$ · $1$

$y=x^2+3x-2$

Rewrite the polynomial.

$x^2-2$ · $x$ · $1$ $+$ $1^2=0$

$y=x^2+3x-2$

Factor using the perfect square

trinomial rule $a^2-2ab+b^2=(a-b)^2,$

where a = x and b = 1.

$(x-1)^2=0$

$y=x^2+3x-2$

Set the $x-1$ equal to 0.

$x-1=0$

$y=x^2+3x-2$

Add 1 to both sides of the equation.

$x=1$

$y=x^2+3x-2$

Replace all occurrences of $x$ in

$y=x^2+3x-2$ with 1.

$y=(1)^2+3(1)-2$

$x=1$

$y=2$

3 0

3 years ago

Rag the operation signs to make the equation true. An operation may be used once. More than once, or not at all × returned to ch

-BARSIC- [3]

Answer:

12/(7-4) + 5 * 3 = 19

Step-by-step explanation:

Given

Untrue Expression: 12 ( 7 − 4 ) + 5 3 = 19

Signs: +−×÷

Required

Use the given signs to make the equation true.

To solve this, the trial and error method will be adopted.

The trial and error method involves trying series of sign combinations in the given expression; whichever works is the right answer.

So, after few attempts the following makes the equation true

12/(7-4) + 5 * 3 = 19

To check if it's true or not; we simply expand the expression on the left hand side. This gives

12/3 + 5 * 3 = 19

4 + 5 * 3 = 19

4 + 15 = 19

19 = 19.

Hence, the sign combinations used in 12/(7-4) + 5 * 3 = 19 is correct.

3 0

3 years ago

Find two consecutive even integers such that the square of the second, decreased by twice the first is 52

Fiesta28 [93]

Let the first integer be $x$ . So, the next even number is $x+2$

Let's translate the formula: the square of the second is $(x+2)^2$ , twice the first is $2x$ . So, the equation is

$(x+2)^2 - 2x = 52$

Expand the square to get

$x^2+4x+4-2x = 52 \iff x^2+2x+4 = 52 \iff x^2+2x - 48 = 0$

The solutions to this equations are

$x = -8,\quad x = 6$

So, the possible consecutive even numbers are

$-8\text{ and } -6\quad\text{or}\quad 6\text{ and } 8$

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3 years ago

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Xelga [282]

(0,3) and (10, 15)
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so equation
y = 6/5x + 3

answer

A. Y=6/5x+3

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4 years ago

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Answer:

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