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Archy [21]
3 years ago
5

HELP HELP HURRY!!!!!

Mathematics
2 answers:
Anettt [7]3 years ago
5 0

Answer:

Point R located on ray PQ.

Step-by-step explanation:

Given : Diagram

To find : Which point is located on ray PQ.

Solution : We have given <--------M---------N---------O------P--------Q----------R--------S--------->.

Ray : A part of a line with a start point but no end point (it goes to infinity).

We can see from the diagram ray PQ start from P but it has no end point.

So , point R ans S located on ray PQ but we have option R

Therefore, D. point R located on ray PQ.

irinina [24]3 years ago
4 0
If the entire ray is PQ, then all of those points lie on the ray.
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Help me please....ASAP...................................
Marysya12 [62]
Area of square = 5 x 5 = 25 cm^2
area of rectangle = 25 x 10 = 250 cm^2
area of figure = 25 + 250 = 275 cm^2

hope it helps
6 0
3 years ago
Find the missing number in this proportion. 3/16 = 5/?
Ludmilka [50]
3/16=5/x
times both sides by 16x (aka cross multiply
3x=5*16
3x=80
divide both sides by 3
x=80/3
x=26 and 2/3
the missing number is 80/3 or 26 and 2/3
8 0
3 years ago
Determine whether the equation represents a direct variation 2y=5x+1
olga nikolaevna [1]
<span>2y/2=5x+1/2
y=5x/2+1/2
no it's not a direct variation because of 1/2</span>
8 0
3 years ago
Y=-x^2+6x-4<br> Find the Axis of Symmetry and the Vertex, Also solve the whole equation.
Inessa [10]

Answer:

1. <em>Axis of symmetry</em>: x = 3

2. <em>Vertex:</em> (3,5)

3. <em>Solution of the equation</em>:

<u />

  • x-intercepts:

               (3-\sqrt{5},0) \\\\(3+\sqrt{5},0)

  • y-intercept: (0, -4)

Explanation:

<u>1. Equation:</u>

    y=-x^2+6x-4

<u>2. </u><em><u>Axis of symmetry:</u></em>

That is the equation of a parabola, whose standard form is:

                 y=ax^2+bx+c

Where:    

                 a=-1;b=6;c=-4

The axis of symmetry is the vertical line with equation:

               x=-b/2a

Substitute  a=-1,\text{ and }b=6

          x=-6/[(2)(-1)]=-6/(-2)=3

Thus, the axis of symmetry is:

            x=3

<em><u>3. Vertex</u></em>

<em><u /></em>

The x-coordinate of the vertex is equal to the axys of symmetry, i.e x = 3.

To find the y-xoordinate, substitute this value of x into the equation for y:

               y=-x^2+6x-4\\\\y=-(3)^2+6(3)-4=-9+18-4=5

Therefore, the vertex is (3, 5)

<u>4. Find the x-intercepts</u>

The x-intercepts are the roots of the equation, which are the points wher y = 0.

        y=-x^2+6x-4=0

Use the quadratic equation:

       x=\frac{-b\pm \sqrt{b^2-4ac} }{2a} \\\\x=\frac{-6\pm\sqrt{(-6)^2-4(-1)(-4)} }{2(-1)}\\\\x_1=3-\sqrt{5} \\\\x_2=3+\sqrt{5}

<u>5. Find the y-intercept</u>

<u />

The y-intercet is the value of y when x=0:

         y=-x^2+6x-4\\\\y=-(0)^2+6(0)-4=-4

7 0
3 years ago
Which value of w makes 14 = 11+<br> 6a true statement?<br> 8<br> Choose 1 answer:<br> PLEASEEEE HELP
Lorico [155]

Answer:

I think D

Step-by-step explanation:

7 0
2 years ago
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