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3 years ago

Whats the negation of this statement x+y=10

Mathematics

1 answer:

attashe74 [19]3 years ago

7 0

X = 10 - y

I hope this helps.:)

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What is the decimal representation of 2/ 10

kondor19780726 [428]

To represent 2/10 it can be 0.20 or 0.2

3 0

3 years ago

HSMath easy 10 points yk the vibes

kramer

The answer is four because 2 x 0 is 0 and then 12 divided by 3 is for

THE ANSWER IS 4

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3 years ago

Help me with math please, quickly

irga5000 [103]

Answer:

$3\neq -3$

False

Step-by-step explanation:

$y=3x-9\\\\x-\frac{1}{3}y=-3$

You substitute $y$ for the second equation.

$x-\frac{1}{3} y=-3\\\\x-\frac{1}{3}(3x-9)=-3$

Then you have to distribute $-\frac{1}{3}$ .

$x-x+3=-3$

Combine like terms.

$3\neq -3$

7 0

2 years ago

Read 2 more answers

15 points, no messing around, please do it right

cupoosta [38]

Answer:

416 is your answer!!!

Step-by-step explanation:

mak me brainliest and add me as a fiend

8 0

3 years ago

List all possible rational roots. Then use synthetic division to confirm which rational roots exist:

Kisachek [45]

Answer:

$\boxed{(1) \, x = \, \pm \dfrac{1}{2}, \pm 1, \pm2, \pm \dfrac{5}{2}, \pm 5, \pm 10; (2) \, x = -2}$

Step-by-step explanation:

2x³+ 6x² - x - 10 = 0

(1) Possible roots

The Rational Roots Theorem states that, if a polynomial has any rational roots, they will have the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.

$\text{Possible rational root} = \dfrac{ p }{ q } = \dfrac{\text{factor of constant term}}{\text{factor of leading coefficient}}$

In your function, the constant term is -10 and the leading coefficient is 2, so

$\text{Possible root} = \dfrac{\text{factor of 10}}{\text{factor of 2}}$

Factors of 10 = ±1, ±2, ±5, ±10

Factors of 2 = ±1, ±2

$\text{Possible roots are } \large \boxed{\mathbf{x = \pm \dfrac{1}{2}, \pm 1, \pm2, \pm \dfrac{5}{2}, \pm 5, \pm 10}}$

(2) Synthetic division

Rather than work through all 12 possibilities, I will do one that works.

$\begin{array}{r|rrrr}-2 & 2 & 6 & -1 & -10\\& & -4& -4 & 10\\& 2 & 2& -5 & 0\\\end{array}$

So, x = -2 is a root, and the quotient is 2x² + 2x - 5.

(3) Check for other rational roots

2x² + 2x - 5 = 0

D = b² - 4ac =2²- 4(2)(-5) = 4 + 40 = 44

√44 = 2√11, which is irrational.

Since irrational roots come in pairs, the cubic equation has two real, irrational roots and one rational root at x = -2.

6 0

3 years ago