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liubo4ka [24]
2 years ago
9

Which represents the equation 6x+3y = 12 when solved for y?

Mathematics
2 answers:
stich3 [128]2 years ago
6 0

Answer:

C: Y= -2x+4

Step-by-step explanation:

You can thank me later

ra1l [238]2 years ago
4 0

Answer:

Y = -2x + 4

Step-by-step explanation:

6x + 3y = 12

3y = -6x + 12

When dividing both sides of the addition problem get divided by 3, so

y = -2x + 4

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Connor bought a baseball card eight years ago that has depreciated (goes down in value) by $120 since he purchased it. Find the
Flura [38]

Answer:

-$15

Step-by-step explanation:

To find the average yearly change, divide -120 by 8, since it has decreased by $120 over 8 years.

-120/8

= -15

So, the average yearly change in the value of the card is -$15

6 0
3 years ago
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WHAT NUMBER OCCURS FREQUENTLY<br><br><br><br><br>ILL GIVE BRAINLIEST<br>​
xxMikexx [17]

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I think it's 2 because it has a larger number of dots than other numbers.

Step-by-step explanation:

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2 years ago
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Field book of an land is given in the figure. It is divided into 4 plots . Plot I is a right triangle , plot II is an equilatera
Kazeer [188]

Answer:

Total area = 237.09 cm²

Step-by-step explanation:

Given question is incomplete; here is the complete question.

Field book of an agricultural land is given in the figure. It is divided into 4 plots. Plot I is a right triangle, plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)

From the figure attached,

Area of the right triangle I = \frac{1}{2}(\text{Base})\times (\text{Height})

Area of ΔADC = \frac{1}{2}(\text{CD})(\text{AD})

                        = \frac{1}{2}(\sqrt{(AC)^2-(AD)^2})(\text{AD})

                        = \frac{1}{2}(\sqrt{(13)^2-(19-7)^2} )(19-7)

                        = \frac{1}{2}(\sqrt{169-144})(12)

                        = \frac{1}{2}(5)(12)

                        = 30 cm²

Area of equilateral triangle II = \frac{\sqrt{3} }{4}(\text{Side})^2

Area of equilateral triangle II = \frac{\sqrt{3}}{4}(13)^2

                                                = \frac{(1.73)(169)}{4}

                                                = 73.0925

                                                ≈ 73.09 cm²

Area of rectangle III = Length × width

                                 = CF × CD

                                 = 7 × 5

                                 = 35 cm²

Area of trapezium EFGH = \frac{1}{2}(\text{EF}+\text{GH})(\text{FJ})

Since, GH = GJ + JK + KH

17 = \sqrt{9^{2}-x^{2}}+5+\sqrt{(15)^2-x^{2}}

12 = \sqrt{81-x^2}+\sqrt{225-x^2}

144 = (81 - x²) + (225 - x²) + 2\sqrt{(81-x^2)(225-x^2)}

144 - 306 = -2x² + 2\sqrt{(81-x^2)(225-x^2)}

-81 = -x² + \sqrt{(81-x^2)(225-x^2)}

(x² - 81)² = (81 - x²)(225 - x²)

x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴

144x² - 11664 = 0

x² = 81

x = 9 cm

Now area of plot IV = \frac{1}{2}(5+17)(9)

                                = 99 cm²

Total Area of the land = 30 + 73.09 + 35 + 99

                                    = 237.09 cm²

7 0
3 years ago
A small business earns a profit of $4,500 in
olchik [2.2K]

Answer:

$2750/month

Step-by-step explanation:

7 0
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A man jogs to the shop and back in half an hour on the way he jogs at 4m/s and on the way back he jogs at 5m/s ignoring time spe
pochemuha
Ratio 4:5
There are 9 total.
30 mins = 1800 secs. 
1800 divided by 9 = 200
200 secs = 3mins 20secs
200 is one "part" of the ratio, so multiply it by 4, then 5
800:1000
                           This is the important one
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13mins, 20secs: 16mins 40secs
Distance is speed x time,
16mins x 4 = 64
40 secs = 2/3
 x 4 = 1.6666
Distance to shop - 65.6m

5 0
3 years ago
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