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AfilCa [17]
2 years ago
12

The perimeter of the triangle at right is 52 units. Write and solve an equation based on the information in the diagram. Use you

r
solution for to calculate the length of each side of the triangle. Be sure to confirm that your answer is correct. Homework Help
7x-4/
19
Equation:
Solve: z
10x + 3
Length of each side:
Red side:
7x - 4
19
Blue Side:
Yellow Side:
10r+3
(hp

Mathematics
1 answer:
marysya [2.9K]2 years ago
3 0

Answer:

  • equation: (7x-4) +19 +(10x+3) = 52
  • x = 2
  • red: 10
  • blue: 19
  • yellow: 23

Step-by-step explanation:

The equation is based on the relation that the perimeter is equal to the sum of the side lengths.

  P = red + blue + yellow

  52 = (7x -4) +(19) +(10x +3)

  52 = 17x +18 . . . . . . . . . . . . . . simplify

  34 = 17x . . . . . . . . . . . . . subtract 18

  2 = x . . . . . . . . . . . . divide by 17

__

  red = 7(2) -4 = 10

  blue = 19

  yellow = 10(2) +3 = 23

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iogann1982 [59]

Hey there!

Markup is the percentage that the cost of an item is increased.

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I hope this helps!

8 0
3 years ago
What is the height, in inches, of a parallelogram that has a base of 3 feet and an area of 324 square inches?
AlekseyPX
The area of a parallelogram is just:

A=bh, we are given b=3ft, so b=36in and A=324in^2 so

324=36h  dividing both sides by 36

h=9

So the height is 9 inches.
7 0
3 years ago
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The area of a parallelogram With a height of 6 meters is 126 square meters. What is the base length of the parallelogram?
suter [353]
Ok
A=bh
plug in
126 square meters=6b
126/6                        6/6
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6 0
3 years ago
Find the measures of the numbered angles in each rhombus
Svetlanka [38]

1)

1 = 38

2 = 38

3 = 38

4 = 38

2)

1 = 32

2 = 90

3 = 58

4 = 32

Let me know if I'm wrong.

4 0
3 years ago
Prove that the roots of x2+(1-k)x+k-3=0 are real for all real values of k​
masha68 [24]

Answer:

Roots are not real

Step-by-step explanation:

To prove : The roots of x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0 are real for all real values of k ?

Solution :

The roots are real when discriminant is greater than equal to zero.

i.e. b^2-4ac\geq 0b

2

−4ac≥0

The quadratic equation x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0

Here, a=1, b=1-k and c=k-3

Substitute the values,

We find the discriminant,

D=(1-k)^2-4(1)(k-3)D=(1−k)

2

−4(1)(k−3)

D=1+k^2-2k-4k+12D=1+k

2

−2k−4k+12

D=k^2-6k+13D=k

2

−6k+13

D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))

For roots to be real, D ≥ 0

But the roots are imaginary therefore the roots of the given equation are not real for any value of k.

6 0
3 years ago
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