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Taya2010 [7]
3 years ago
7

Can someone help me pls

Mathematics
1 answer:
lesantik [10]3 years ago
7 0

A line segment bisector passes through the midpoint of the segment.  So the segment bisector of JK is M.

Also, the length of JM is 7x + 5, which we know is equal to 8x, so we just have to find x and then figure out the length.

7x + 5 = 8x   subtract 7x from both sides

5 = x

Does it work? Let's see. 7(5) + 5 = 40  AND  8(5) = 40

The length of JM is 40.

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Find either the maximum or the minimum value of the following quadratic equation. Be sure to show all of your work and identify
podryga [215]

Answer:

the minimum is (1,-9)

Step-by-step explanation:

y = 5x^2 - 10x - 4

since the parabola opens upward  5>0, this will have a minimum

it will occur along the axis of symmetry h=-b/2a

y =ax^2 +bx+c

h = -(-10)/2*5

h  = 10/10 =1

the minimum occurs at x =1

the y value for the minimum is calculated by substituting x =1 back into the equation

y = 5 * 1^2 - 10*1 -4

y = 5*1^2 -10 -4

y = 5-10-4

y = -9

the minimum is (1,-9)

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3 years ago
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70.2 divided by 4 = $17.55

or

$23.40 x .75 = $17.55

4 0
3 years ago
How many sqft is in a rectangle of 360 by 190 feet
monitta

Answer: 68400

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4 0
3 years ago
This one might not look familiar yet, but it's very similar to the distance problems you've already seen.
Debora [2.8K]

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Step-by-step explanation:

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3 years ago
Which sequence of transformations produces R'S'T' from RST? On a coordinate plane, triangle R S T has points (0, 0), (negative 2
arsen [322]

Answer:

Step-by-step explanation:

Given is a triangle RST and another triangle R'S'T' tranformed from RST

Vertices of RST are (0, 0), (negative 2, 3), (negative 3, 1).

Vertices of R'S'T' are  (2, 0), (0, negative 3), (negative 1, negative 1).

Comparing the corresponding vertices we find that x coordinate increased by 2 while y coordinate got the different sign.

This indicates that there is both reflection and transformation horizontally to the right by 2 units

So first shifted right by 2 units so that vertices became

(2,0) (0,3) (-1,1)

Now reflected on the line y=0 i.e. x axis

New vertices are

(2,0) (0,-3) (-1,-1)

8 0
4 years ago
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