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tangare [24]
3 years ago
10

What is 12x-10y=-19 in slope intercept form?

Mathematics
1 answer:
KengaRu [80]3 years ago
8 0

Answer:

-10y=-12x-19

-y=  <u>-12x-19</u>

           10

y=  <u>12x + 19</u>

            10

Step-by-step explanation:

form slope intercept its y=mx+b

so you find y what equal .

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Which graph best represents the solution to the system of equations shown below? (5 points) y = −4x − 19 y = 2x − 1 A coordinate
Oksana_A [137]

Answer:

Wait wait wait this makes no sense?

Step-by-step explanation:

3 0
2 years ago
2a+3b=5<br> b=a-5 help please
diamong [38]

Answer:

2a + 3b = 5

b = a - 5

Step-by-step explanation:

2a + 3b = 5

b = a - 5

You can write this another way:

2a + 3(a-5) = 5 ( I added the second formula in the first one )

Now you gotta factor out:

2a + 3a - 15 = 5

Here im gonna do plus 15

2a + 3a = 20 ==> 5a = 20

Now if you divide by 5, you get: a = 4

Now you can fill in the second formula again

b = a - 5 (ill fill this in now)

b = 4 - 5 = -1

So, this makes:

a = 4 , b = -1

6 0
3 years ago
Read 2 more answers
Find a polynomial function of least degree having only real coefficients, a leading coefficient of 1, and roots of 2-√6 , 2+ √6
horsena [70]

Answer:

x^{4} -18x^{3}+104x^{2} -172x-100

Step-by-step explanation:

The 3 roots are given out of which 2 are real and 1 is imaginary. For a polynomial of least degree having real coefficients, it must have a complex conjugate root as the 4th root. Therefore, based on 4 roots, the least degree of polynomial will be 4. Finding the polynomial having leading coefficient=1 and solving it based on multiplication of 2 quadratic polynomials, we get:  

\\\\x_{1} = 2-\sqrt{6} \\x_{2} = 2+\sqrt{6} \\x_{3}=7-i \\x_{4}=7+i \\\\P(x)=1(x-x_{1})(x-x_{2} )(x-x_{3} )(x-x_{4} ) \\\\=(x-(2-\sqrt{6}))(  x-(2+\sqrt{6} )) (x-(7-i))( x-(7+i))\\=((x-2)+\sqrt{6})( ( x-2)-\sqrt{6} ) ((x-7)+i)( (x-7)-i)\\=((x-2)^{2} -(\sqrt{6} )^{2} )((x-7)^{2}-(i)^{2})\\=(x^{2} -4x-2)(x^{2} -14x+50)\\=x^{4} -18x^{3}+104x^{2} -172x-100\\

7 0
3 years ago
Pls pls pls help
Natasha_Volkova [10]

Answer:

Parallelograms I, II, and IV

Step-by-step:

Area of parallelograms:

I. A=3*5=15 units squared

II. A=5*3=15 units squared

III. A=4*4=16 units squared

IV. A=5*3=15 units squared

So, parallelograms I, II, and IV have the same area of 15 units squared.

5 0
3 years ago
Is the following statement true or false?
a_sh-v [17]
Yes it is true that when you write a check you must write the amount to be paid in both numbers and words for security purposes
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3 years ago
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