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Minchanka [31]
3 years ago
15

How do i solve this equation please help !!

Mathematics
1 answer:
Serggg [28]3 years ago
6 0

Answer:

a = 4.8

Step-by-step explanation:

9 - 4a + 12 = 6a - 27

21 - 4a = 6a - 27

48 - 4a = 6a

48 = 10a

a = 4.8

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Does this graph represent a function ? Please help thanks
Ne4ueva [31]

Answer:

Yes, it is a constant function with no slope.

3 0
4 years ago
The sides of a triangle are in the ratio 3:4:5 what is the length of each side if the perimeter of the triangle is 90 cm?
polet [3.4K]

Answer:

The answer to your question is:

side 1 = 22.5 cm

side 2 = 30 cm

side 3 = 37.5 cm

Step-by-step explanation:

 ratio = 3:4:5

Perimeter = 90 cm

First we get the sum of the numbers of the proportions

                          3 + 4 + 5 = 12

now divide the perimeter by the number of parts (12)

             

                           90 / 12 = 7.5

Side 1 = 3(7.5) = 22.5 cm

Side 2 = 4(7.5) = 30 cm

Side 3 = 5(7.5) = 37.5 cm

     Perimeter = 22.5 + 30 + 37.5 = 90 cm

7 0
3 years ago
Hi . i need help with this question on my geometry test !
Simora [160]

Answer:

x = 2

y = 7

Step-by-step explanation:

ON = LM

therefore

12x - 5 = 4x + 11

      +5         +5

12x = 4x + 16

-4x    -4x

8x = 16

/8     /8

x = 2

NM = OL

therefore

x + 7 = 2y - 5

(2) + 7 = 2y - 5

9 = 2y - 5

+5        +5

14 = 2y

/2     /2

7 = y

3 0
3 years ago
I need to know a equivalent fraction for number 7
AlekseyPX
9/10 is equivalent to 90/100 and the answer under mine or above or the other answer should be reported. They got points for a question they did not answer correctly.
7 0
3 years ago
A rancher has 140 feet of fence with which to enclose three sides of a rectangular garden (the fourth side is a cliff wall and w
Scilla [17]
A rancher wants to enclose a rectangular field with 220 ft of fencing.
One side is a river and will not require a fence.
What is the maximum area that can be enclosed?
:
The field requires only 3 sides of fence, therefore
L + 2W = 220
L = (220-2W); we can use this for substitution in the Area equation
:
Area
A = L*W
replace L with (220-2W)
A = W(220-2W)
A = -2W^2 + 220W
A quadratic equation, max A occurs at the axis of symmetry, x = -b/(2a)
In this equation
W = %28-220%29%2F%282%2A-2%29
W = 55 ft is the width for max area
:
Find the max area
A = -2(55^2) + 220(55)
A = -2(3025) + 12100
A = 6050 sq/ft is the max area
:
:
Confirm this with the dimensions calculated; 110 * 55 = 6050
3 0
3 years ago
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