All categories

3 years ago

PLS HELP ........................

Mathematics

1 answer:

lisov135 [29]3 years ago

3 0

Answer:

Step-by-step explanation:

First, check to see which graph has a line going through the point (2,3). B and D are the only ones that have lines going through (2,3) (A comes close but it is not quite).

Next, you need to see which line would be parallel to the equation 3x-y=2. When two lines are parallel, they have the same slope. You have to turn that equation into point-slope formula (y=mx+b with m being the slope and b being the y-intercept). First, subtract 3x from both sides. You will then get -y=-3x+2. Then y needs to be alone (right now it has a -1 attached). Divide both sides by -1 to get y=3x-2. The number in front of the x is the slope, in this case, 3 or 3/1 (we do not care about the y-intercept, it will not help us in this problem since we are looking for a line parallel to this equation and so our line will not have the same y-intercept as this other equation). Since the line in parallel to that equation, we know that this line also has a slope of 3/1. Find the line between B and D that has a slope of 3/1, you get B (the line goes up 3 over 1).

You might be interested in

Which of the following cannot be used to express the number 8? eight more than zero +8 eight below zero positive eight

MrRa [10]

Eight below zero would express the value -8, negative eight,

but not 8.

7 0

3 years ago

Read 2 more answers

Failure to adequately plan for retirement can result in

valkas [14]

<span>A) <span>financial dependence on children. :)</span></span>

8 0

3 years ago

John looked at his bank account and noticed that he had $2,029.90 in his account. All that he could remember is that he created

Over [174]

We will have that for the previous year, he had in the account:

$m_{y6}=2029.90-\frac{2029.90\cdot6}{100}\Rightarrow m_{y6}=1908.106$

For the year previous to that one, he had:

$m_{y5}=1908.106-\frac{1908.106\cdot6}{100}\Rightarrow m_{y5}=1793.61964$

For the previous year to that, he had:

$m_{y4}=1793.61964-\frac{1793.61964\cdot6}{100}\Rightarrow m_{y4}=1686.002462$

For the year previous to that one, he had:

$m_{y3}=1686.002462-\frac{1686.002462\cdot6}{100}\Rightarrow m_{y3}=1584.842314$

For the second year that the money was in the account, there was:

$m_{y2}=1584.842314-\frac{1584.842314\cdot6}{100}\Rightarrow m_{y2}=1489.751775$

And the orinial ammount of money that was in the account was:

$m_{y1}=1489.751775-\frac{1489.751775\cdot6}{100}\Rightarrow m_{y1}=1400.366669$

From this, we know that Jhon had originally $1400.37 in the bank account.

6 0

1 year ago

the point 3,2 lies on a circle.what is the exact length of the radius of this circle if the center is located at (-1,-4)

Crazy boy [7]

You can use the distance formula to solve this.

Distance Formula:
$d = \sqrt{(X_2 -X_1)^2 + (Y_2 - Y_1)^2}$

I am going to use point (3,2) as point 1 and (-1,-4) as point 2
Insert the points into the formula
$d = \sqrt{(X_2 -X_1)^2 + (Y_2 - Y_1)^2}$
$d = \sqrt{((-1) - 3)^2 + ((-4) - 2)^2}$
Solve:
$d = \sqrt{((-1) - 3)^2 + ((-4) - 2)^2}$
$d = \sqrt{(-4)^2 + (-6)^2}$
$d = \sqrt{16 + 36}$
$d = \sqrt{52}$
$d = \sqrt{52} = \sqrt{4}\sqrt{13}$
$d = \sqrt{52} = 2 \sqrt{13}$
$d = 2 \sqrt{13}$
$radius = 2 \sqrt{13} = 7.2111$

5 0

3 years ago

A LOOOOOOOOT OF POINTS AND THANK YOU AND BRAINLIEST!!!!!!!!

Anarel [89]

Answer:

Step-by-step explanation:

vertex is 1 units left.

y is always positive.

y=|x+1|

5 0

3 years ago

Read 2 more answers