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denis-greek [22]
2 years ago
11

Please help!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
Temka [501]2 years ago
6 0

Answer:

bro I have no idea I'm just screwing around

Step-by-step explanation:

you ask your step sister

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Which term describes alliance line segment that connects a vertex of a triangle to a point on the line containing the opposite s
uranmaximum [27]

Answer:

Altitude

Step-by-step explanation:

An altitude of a triangle always extends from a vertex to the opposite base (side) of the triangle, and it is always at a right angle. It is also known as the "height".

3 0
3 years ago
Show that AC and BD bisect each other if A(–3, –5), B(2, –3), C(3, 5), and D(–2, 3). Find AC and BD
Mazyrski [523]

Answer

Multipy all of them

Step-by-step explanation:

mulitiply all of them

7 0
3 years ago
Read 2 more answers
Brad worked 40 hours last week and earned $394. What was Brad's hourly pay in dollars and cents?
ZanzabumX [31]
Brad's pay is $9.85 per hour
5 0
3 years ago
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Which point gives the vertex of ƒ(x) = –x2 + 4x – 3?
Dmitriy789 [7]

Answer:

The vertex is (2,1)

Step-by-step explanation:

ƒ(x) = –x^2 + 4x – 3

Factor out the negative

   = -(x^2 -4x+3)

 Factor

What 2 numbers multiply to +3 and add to -4

-3*-1 = 3

-3+-1 = -4

f(x) =  -( x-3)(x-1)

Find the zeros

0 = -( x-3)(x-1)

0 = x-3   0 = x-1

x=3            x=1

The x value of the vertex is 1/2 way between the two zeros

(3+1)/2 = 4/2 =2

To find the y value, substitute x=2 in

f(2) =  -( 2-3)(2-1)

       =-(-1)(1) = 1

The vertex is (2,1)

4 0
3 years ago
Iv)<br>6x+3y=6xy<br>2x + 4y= 5xy​
Margaret [11]

Answer:

Ok, we have a system of equations:

6*x + 3*y = 6*x*y

2*x + 4*y = 5*x*y

First, we want to isolate one of the variables,

As we have almost the same expression (x*y) in the right side of both equations, we can see the quotient between the two equations:

(6*x + 3*y)/(2*x + 4*y) = 6/5

now we isolate one off the variables:

6*x + 3*y = (6/5)*(2*x + 4*y) =  (12/5)*x + (24/5)*y

x*(6 - 12/5) = y*(24/5  - 3)

x = y*(24/5 - 3)/(6 - 12/5) = 0.5*y

Now we can replace it in the first equation:

6*x + 3*y = 6*x*y

6*(0.5*y) + 3*y = 6*(0.5*y)*y

3*y + 3*y = 3*y^2

3*y^2 - 6*y = 0

Now we can find the solutions of that quadratic equation as:

y = \frac{6 +- \sqrt{(-6)^2 - 4*3*0} }{2*3} = \frac{6 +- 6}{6}

So we have two solutions

y = 0

y = 2.

Suppose that we select the solution y = 0

Then, using one of the equations we can find the value of x:

2*x + 4*0 = 5*x*0

2*x = 0

x = 0

(0, 0) is a solution

if we select the other solution, y = 2.

2*x + 4*2 = 5*x*2

2*x + 8 = 10*x

8 = (10 - 2)*x = 8x

x = 1.

(1, 2) is other solution

8 0
3 years ago
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