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3241004551 [841]
3 years ago
8

How do i solve this?

Mathematics
1 answer:
vodka [1.7K]3 years ago
5 0

Step-by-step explanation:

all I see is a piece of paper

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What is the best approximation of the solution to the system of the nearest integer values
Natali5045456 [20]

Answer:

  • (2, 3.5) or (2, 3)

Step-by-step explanation:

<em>See attached.</em>

The intersection of the lines is the solution.

<u>Approximate values of the coordinates is:</u>

  • (2, 3.5) or (2, 3) if integer values are required

7 0
3 years ago
Read 2 more answers
What is the square root of 6 - the square root of 12?<br><br> Please answer fast!
dmitriy555 [2]

Answer:

1.01 or 1.0

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
PLEASE EXPLAIN!
stepladder [879]

Answer:

15

Step-by-step explanation:

5(2)(3)-2(2)+13+7(2)-6(2)(3)-4(2)+(2)(3)

30-4+13+14-36-8+6

15

6 0
3 years ago
if a quadratic equation with real coefficients has a discriminant of -36 then what type of roots does it have
Aleks04 [339]

Step-by-step explanation:

We have,

If a quadratic equation with real coefficients has a discriminant of -36.

The general form of quadratic equation is :

ax^2+bx+c=0

The discriminant of this equation is : D=b^2-4ac

If D=0, it will have 1 real roots

If D>0, it will have 2 real roots

If D<0, it will have no real roots

We have,

D = -36 < 0, so, the quadratic equation will have no real roots.

5 0
3 years ago
Hey guys<br>im new here<br>please solve this for me with steps!<br>ill mark as the best answer​
Vinil7 [7]

Answer:

The factors of  2(x+y)^2-9(x+y)-5 is ((x+y)-5)(2x+2y+1)

Step-by-step explanation:

Given polynomial

=>2(x+y)^2-9(x+y)-5

To Find:

The factors of the polynomial =?

Solution:

Lets assume  k = (x+y)

Then 2(x+y)^2-9(x+y)-5 can be written as 2k^2-9k-5

Now by using quadratic formula

k =\frac{-b\pm\sqrt{(b^2-4ac}}{2a}

where

a= 2

b= -9

c= -5

Substituting the values, we get

k =\frac{-b\pm\sqrt{(b^2-4ac)}}{2a}

k =\frac{-(-9) \pm \sqrt{((-9)^2-4(2)(-5)}}{2(2))}

k =\frac{-(-9) \pm \sqrt{(81+40)}}{4}

k =\frac{-(-9) \pm \sqrt{(121)}}{4}

k =\frac{-(-9) \pm 11}}{4}

k= \frac{ 9 \pm 11}{4}

k =  \frac{20}{4}                         k =  \frac{-2}{4}    

k_1 =5                                      k_2 = -\frac{1}{2}

2k^2-9k-5= 2(k-5)(k+\frac{1}{2})

Solving the RHS we get

\frac{2}{2}(k-5)(2k+1)

(k-5)(2k+1)

Substituting k = x+y

((x+y)-5)(2(x+y+1)

((x+y)-5)(2x+2y+1)

5 0
3 years ago
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