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ycow [4]
3 years ago
10

Find the distance from G to S.

Mathematics
1 answer:
OLEGan [10]3 years ago
4 0
The distance formula is the square root of (x1-x2)^2+ (y1-y2)^2

So it would be square root [ (-2-3)^2 + (6-(-2))^2]
So it would be square root(25+64)
Square root(89)= 9.433
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Which shows the equation below.....
sladkih [1.3K]
Option A is your answer. 

Subtract 3 from both sides to get zero on the right side of the equal sign. 

8 - 3 = 5. 

So that becomes your constant term in the equation. 
8 0
3 years ago
Read 2 more answers
Open-Ended Question<br>*Picture*
My name is Ann [436]
We have to calculate the volume of the right rectangular prism.
lenght=4 1/2 in=(4+1/2) in=9/2 in
width=5 in
height=3 3/4 in=(3+3/4) in=15/4 in

Volume (right rectangular prism = lenght x width x height.
volume=9/2 in * 5 in * 15/4 in=675/8 in³

we calculate the volume of this little cube.
volume=side³
volume=(1/4 in )³=1/64 in³

Now, we calculate the number of small cubes are needed to fit the right rectangular pris by the rule of three.

1 small cube----------------1/64 in³
x---------------------------------675/8 in³

x=(1 small cube * 675/8 in³) / 1/64 in³=5400 small cubes.

Answer: we need 5400 small cubes to fit the right rectangular prism.
8 0
3 years ago
Complete the table to represent how much each individual monthly deposit of $50 will be worth at the end of 12 months. Enter you
e-lub [12.9K]

The first month has the equation raised to 11 which is the number of months left in the period.

So each month the equation would be the same except it would be raised to 1 less than the previous month.

Month 2 = 50(1.003)^10

Month 3 = 50(1.003)^9

Month 4 = 50(1.003)^8

Month 5 = 50(1.003)^7

Month 6 = 50(1.003)^6

Month 7 = 50(1.003)^5

Month 8 = 50(1.003)^4

Month 9 = 50(1.003)^3

Month 10 = 50(1.003)^2

Month 11 = 50(1.003)^1

Month 12 = 50(1.003) ( the last month would not be raised to anything)

6 0
3 years ago
Read 2 more answers
How do I solve part A
seraphim [82]

Answer:

The solution of this expression is x_{1} = -1 and x_{2} = -\frac{1}{2}.

Step-by-step explanation:

The procedure for solution of exercise A is described below:

1) We expand the expression.

2) The resulting expression is rearranged into the form of a second order polynomial.

3) Roots are found by Quadratic Formula.

Step 1:

2\cdot x \cdot (x+1.5) = -1

2\cdot (x^{2}+1.5\cdot x) = -1

2\cdot x^{2} + 3\cdot x = -1

Step 2:

2\cdot x^{2}+3\cdot x +1 = 0

Step 3:

x_{1, 2} = \frac{-3\pm \sqrt{3^{2}-4\cdot (2)\cdot (1)}}{2\cdot (2)}

x_{1,2} = -\frac{3}{4}\pm \frac{1}{4}

x_{1,2} = \frac{-3\pm 1}{4}

The solution of this expression is x_{1} = -1 and x_{2} = -\frac{1}{2}.

4 0
3 years ago
Which number can each term of the equation be multiplied by to eliminate the fractions before solving? 6 – x + = 6 minus StartFr
zheka24 [161]

Answer:

We need to multiply 12 to each term to eliminate fractions.

Step-by-step explanation:

7 0
3 years ago
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