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3 years ago

5

Can you guys help me in 2 3 4 5 pls

2 answers:

ddd [48]3 years ago

3 0

Help her with the dishes
Ask her whats wrong and how you can help
Pick it up and put it in the trash
Tell him its not nice and to treat other how you want to be treated.along with telling the teacher.

Helen [10]3 years ago

3 0

2. do the dishes first then do your hw
3.go up to her and ask her if everything is alright then come up with a solution to fix whats making her cry
4. Clean it up for them( if they want to be lazy let them just be the better person
5. Go get a teacher quick before things start to get rough

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Can someone please help me:)

Naddika [18.5K]

For a translation 4 units down you would subtract 4 from each y value and 5 units right you would add 5 to each x value.
Your resulting points would be.
A'(-1,1), B'(-1,-2), C'(3,-2), D'(3,2)

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3 years ago

Evaluate x3 − 2y2 − 3x3 + z4 if x = 3, y = 5, and z = −3.

White raven [17]

Answer:

Step-by-step explanation:

x^3-2y^2-3x^3+z^4

(3)^3-2(5)^3+(-3)^4

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3 years ago

Help plz i need help PLZ I HAVE A MATH TEST :(((((((((((

Naddik [55]

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3 0

3 years ago

Read 2 more answers

Need help!!!!! with this test pls IM SORRY FOR ASKING SO MANY TIMES BUT PLS HELP

Anika [276]

Answer:

I hope its 15/10

Step-by-step explanation:

please let me know if its wrong

6 0

2 years ago

Read 2 more answers

What does the fundamental theorem of algebra illustrate?

UNO [17]

Answer:

The fundamental theorem of algebra guarantees that a polynomial equation has the same number of complex roots as its degree.

Step-by-step explanation:

The fundamental theorem of algebra guarantees that a polynomial equation has the same number of complex roots as its degree.

We have to find the roots of this given equation.

If a quadratic equation is of the form $ax^{2}+bx +c=0$

Its roots are $\frac{-b+\sqrt{b^{2}-4ac } }{2a}$ and $\frac{-b-\sqrt{b^{2}-4ac } }{2a}$

Here the given equation is $2x^{2}-4x-1$ = 0

a = 2

b = -4

c = -1

If the roots are $x_{1} and x_{2}$ , then

$x_{1}$ = $\frac{-2+\sqrt{(-4)^{2}-4\times 2\times (-1)}}{2\times 2}$

= $\frac{4 +\sqrt{24}}{4}$

= $\frac{2+\sqrt{6} }{2}$

$x_{2}$ = $\frac{-2-\sqrt{(-4)^{2}-4\times 2\times (-1)}}{2\times 2}$

= $\frac{4 +\sqrt{8}}{4}$

= $\frac{2-\sqrt{6} }{2}$

These are the two roots of the equation.

6 0

3 years ago