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3 years ago

12

2a + 6 = 6 2a = Subtract 6 from both sides Divide both sides by 2

1 answer:

Alla [95]3 years ago

8 0

Answer:

0

Step-by-step explanation:

2a+6=6

-6=-6

2a=0

2 2

a=0

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What is the relationship between the following ordered pairs 0,6 2,5 4,4 and 6,3 keep in mind it will be a two step process

aliya0001 [1]

Find the first rate of change...

(5-6)/(2-0)=(4-5)/(4-2)=(3-4)/6-4)=-1/2

So this equation has a constant velocity which means that it is linear and can be expressed as y=mx+b where m=slope=velocity and b=y-intercept, value of y when x=0...we saw earlier that the velocity or slope is -1/2 so:

y=-x/2+b, and we can use any point to solve for b, I'll use (6,3)

3=-6/3+b

3=-2+b

b=5

So the relationship if the function:

y=-x/2+5 or more neatly...

y=(10-x)/2

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3 years ago

Pick correct graph from multiple choice options. A.B.C.D

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Answer:

The answer would be C.

Step-by-step explanation:

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3 years ago

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Write an equation for the line that passes through (-1,4) with a slope of -10

Maurinko [17]

Answer:

y = - 10x - 6

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Here slope m = - 10, hence

y = - 10x + c ← is the partial equation

To find c substitute (- 1, 4) into the partial equation

4 = 10 + c ⇒ c = 4 - 10 = - 6

y = - 10x - 6 ← equation of line

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3 years ago

A veterinarian treated 7 dogs this morning. The list below gives the weights (in pounds) of each dog. , 6, 46, 21, 9, 21, 5143 F

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Step-by-step explanation:

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3 years ago

Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$

And the probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{6}$

$P(1) = \frac{1}{3}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{6}$

$P(2) = \frac{1}{2}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{6}$

P(Odd Number) is then calculated as:

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{7}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{7}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

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3 years ago