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Misha Larkins [42]
3 years ago
10

PLEASE HELP QUICK!!!! What is the solution to the equation Two-thirds x + 1 = one-sixth x minus 7? x = negative 16 x = negative

4 x = 4 x = 16
Mathematics
2 answers:
spayn [35]3 years ago
8 0

Answer:

-16

Step-by-step explanation:

I solved it out.

Lunna [17]3 years ago
5 0

Answer:

it 16 boi

Step-by-step explanation:

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Which statement shows the Transitive Property of Equality? a. If AB = CD and AB =EF, then CD is not Equal to EF. b. If AB + BC =
steposvetlana [31]
The answer is c
AB = CD, CD = DE the AB = DE
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Write the equation in standard form 3x + 1 = 2y
yulyashka [42]
The standard form of an equation is
ax + by = c

In the question, we have the following formula
3x + 1 = 2y

To get the formula in standard form, we first have to subtract 2y from both sides (since we need to get 2y on the left side)
3x - 2y + 1 = 0

Finally we subtract 1 from both sides (since we need the number on the right side)
3x - 2y = - 1

Hence, 4. is the correct answer.
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3 years ago
What is the coefficient of the term of degree 8 in the polynomial below?
Nesterboy [21]
The correct answer is:  [D]:  "5 (five)" .
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<u>Note</u>:  The term of "degree 8" is the term with a variable with 
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3 years ago
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strojnjashka [21]

Answer:

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Step-by-step explanation:

7 0
2 years ago
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Write the equation in standard form for the circle that has a diameter with endpoints (22,0) and (2,0)
vovikov84 [41]

Answer:

(x - 12)² + y² = 100

Step-by-step explanation:

The standard form of the equation of a circle is;

(x - a)² + (y - b)² = r²

where:

a and b are the coordinates of the centre of the circle

r is the radius

We are given the coordinates of the endpoints of the diameter as; (22,0) and (2,0)

Thus, the centre of the circle would be at the mid point of the endpoints of the diameter.

Coordinates of the centre is;

((22 + 2)/2), (0 +0)/2))

This is;

(12, 0)

So, a = 12 and b = 0

Now,to get the radius r, we will use the formula;

r = √[(x2 - x1)² + (y2 - y1)²]

Where;

(x1, y1) and (x2, y2) are 2 points namely (12,0) and (22, 0)

r = √[(12 - 22)² + (0 - 0)²]

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Thus,equation of the circle is;

(x - 12)² + (y - 0)² = 10²

(x - 12)² + y² = 100

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3 years ago
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