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3 years ago

Answer question 11 and 12

Mathematics

1 answer:

Darya [45]3 years ago

5 0

Answer:

11. y = -2x - 4.

12. y = x - 5

Step-by-step explanation:

11. the given points on the line are ((-2, 0) and (0, -4)

The slope of the line

= difference in y values / corresponding difference in x values

= ( - 4 - 0) / 0 - (-2))

= -4 / 2

= -2.

The general form of a straight line is y = mx + c where m = the slope and c = y value of the intercept on the y-axis.

So the equation is:

y = -2x - 4.

12.

The points are (1, -4) and (3, -2).

The slope is (-2 - (-4)) / (3 - 1)

= 2/2

= 1.

So the equation is

y = 1x + c

We find c by plugging in the point (1, -4):

-4 = 1 + c

c = -4 - 1

c = -5.

The equation is y = x - 5.

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Please answer 10 point + brainlyest = 1 big thank u

solmaris [256]

<em><u>Step-by-step explanation:</u></em>

FIRST, we want to understand every property:

Associative Property: The associative property states that you can add or multiply regardless of how the numbers are grouped so (5 + 4) + 3 = 5 + (4 + 3)

Commutative Property: Commutative is the one that refers to moving stuff around so 3 + 2 = 2 + 3

Additive Inverse Property: This is the number that when added to the original number, equals 0 or it's the opposite of the number so 5 <- is the original number and -5 <- is the additive inverse.

Simplify: Is just to add like terms, and make the equation the simplest.

1. ORIGINAL EXPRESSION

2. Additive Inverse Property

3. Commutative property

4. Associative Property

5. Simplify

6. Simplify

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3 years ago

Find the area of the rhombus 6cm, 8cm

Igoryamba

Answer:

A=24cm²

Step-by-step explanation:

Area=1/2(product of the diagonals)

A=1/2(6×8)

A=1/2(48)

divide by 2

A=24cm²

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3 years ago

Sam entered three functions into a graphing calculator, y1 = x + 2, y2 = x2 + 2, and y3 = 2x. A portion of the table created by

V125BC [204]

For this case we have the following functions:
$y1 = x + 2

y2 = x ^ 2 + 2

y3 = 2 ^ x$

When x = 0 we have:
For y1:
$y1 = 0 + 2

y1 = 2$
For y2:
$y2 = 0 ^ 2 + 2

y2 = 0 + 2

y2 = 2$
Therefore, we have to:
$y1 = y2
$

When x = 5 we have:
For y2:
$y2 = 5 ^ 2 + 2

y2 = 25 + 2

y2 = 27$
For y3:
$y3 = 2 ^ 5

y3 = 32$
Therefore, we have to:
$y2 \ \textless \ y3
$

When x = -1 we have:
For y1:
$y1 = -1 + 2

y1 = 1$
For y2:
$y2 = (-1) ^ 2 + 2

y2 = 1 + 2

y2 = 3
$
For y3:
$y3 = 2 ^ {-1}

y3 = 1/2

y3 = 0.5$
Therefore, we have to:
$y3 \ \textless \ y1 \ \textless \ y2
$

Answer:
When x = 0, y1 = y2
When x = 5, y2 <y3
When x = -1, y3 <y1 <y2

4 0

3 years ago