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Natali [406]
3 years ago
7

Find the equation of the line through point (3,2) and perpendicular to 3x + 5y = 15

Mathematics
1 answer:
Vesnalui [34]3 years ago
6 0

Answer:

y=5/3x-3

Step-by-step explanation:

first, let's put 3x+5y=15 into y=mx+b form

move 3x to the other side

5y=-3x+15

divide by 5

y=-3/5x+3

now, perendicular lines have slopes that are negative and reciprocal.

With our new equation we can automatically find the slope, which would be 5/3.

now, we need to find b

plug the points (3,2) into y=5/3x+b

2=5/3(3)+b

2=5+b

-3=b

so that means that our equation will now be

y=5/3x-3.

Hope this helps!

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Solve for y Y/3 + 25 = 31
ruslelena [56]

Answer:

y = 18

Step-by-step explanation:

8 0
2 years ago
. Osteen has 32 marbles in his collection. 18 of his marbles are red, four are blue, and the rest are
nalin [4]

Answer: B

Explanation:  

32 marbles in total.

18 red marbles

4 blue marbles

10 green marbles.

You are asked what part of Osteen's marbles are red?

Probability = number of possible outcomes/total number of out outcomes.

P(Red) = 18/32 = 9/16

Both, 18 and 32 are divisible by 2 which gives 9/16. Since 9/16 cannot be reduced any further, we live it as it is. If you wanted to plug this into a calculator that would give 0.5625 which is 56.25% of Osteen's marbles are red.

P(Blue) = 4/32 = 1/8. 1/8 = 0.125 which 12.5% of his marbles are blue.

P(Green) = 10/32 = 5/16. 5/16= 0.3125 which 31.25 of his marbles are green. When you add 0.5625 + 0.125 + 0.3125 = 1 which is 100%. All probabilities add up to 1.

Hope this elaborate explanation helps.

6 0
2 years ago
One-fifth of a number x is less than 8. Find the greatest possible prime number x.
GuDViN [60]
Hello there, I am unable to use the “less than” symbol. Sorry :( .

1/5x "less than" 8
x “Less than” 40

Prime number is 37,41.
Therefore x is 37.
8 0
2 years ago
Use Cramer's rule to find the solution to the following system of linear equations.
slavikrds [6]

Answer:

x = 53

y = 30

Step-by-step explanation:

Step(I):-

Given equations are

x -2y =-7 ...(I)

5x-9y =-5 ..(ii)

The matrix form  AX = B

                    \left[\begin{array}{ccc}1&-2\\ 5  & -9\\\end{array}\right] \left[\begin{array}{ccc}x\\y\\\end{array}\right] = \left[\begin{array}{ccc}-7\\-5\\\end{array}\right]

The determinant

= \left|\begin{array}{ccc}1&-2\\5&-9\\\end{array}\right| = -9+10 =1

By using Cramer's Rule

Δ₁ =      \left[\begin{array}{ccc}-7&-2\\\\-5&-9\end{array}\right]



The determinant is     Δ₁ = -9 X -7 - (10 ) = 53

x = Δ₁ / Δ

x = 53

The determinant

Δ₂ =



Δ₂ = -5 +35

     

 

y = Δ₂/Δ =  30

     

5 0
3 years ago
Read 2 more answers
The distance of a golf ball from the hole can be represented by the right side of a parabola with vertex (-1,8). The ball reache
ddd [48]

Answer:

Let's suppose that the hole is at y = 0m, where x is the time variable.

we know that:

The vertex is (-1s, 8m).

(i suppose x in seconds and y in meters)

At x = 1s, the ball reaches the hole, so we also have the point:

(1s, 0m).

Remember that the vertex of a quadratic equation y = a*x^2 + b*x + c is at:

x = -b/2a,

then we have:

-1 = -b/2a.

Then we have tree equations:

8m = a*(-1s)^2 + b*-1s + c

0m = a*(1s)^2 + b*1s + c

-1s = -b/2a.

First we should isolate one variable in the third equation, and then replace it in one of the other two:

1s*2a = b.

So we can replace b in the first two equations bi 1s*2a.

8m = a*1s^2 - 1s*2a*1s + c

0m = a*1s^2 + 1s*2a*1s + c

We can simplify both equations and get:

8m = a*( 1s^2 - 2s^2) + c = -a*1s^2 + c.

0m = a*(1s^2 + 2s^2) + c = a*3s^2 + c.

Easily we can isolate c in the second equation and then replace it into the first equation:

c = -a*3s^2

The first equation becomes:

8m = -a*1s^2 - a*3s^2 = -a*4s^2

a = 8m/-4s^2 = -2m/s^2.

Now with a, we can find the values of c and b.

c = -a*3s^2 = -(-2m/s^2)*3s^2 = 6m.

b = 1s*2a = 1s*(-2m/s^2) = -2m/s.

Then the equation is:

y = (-2m/s^2)*t^2 + (-2m/s)*t + 6m

5 0
3 years ago
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