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3 years ago

I need help please?!):

Mathematics

1 answer:

Serggg [28]3 years ago

3 0

Answer:

The answer is D.

Step-by-step explanation:

The first thing I did was substitute the variables in for x and y. This is the numbers in the answer choices: (x,y). I don't know if there is an easier way to do this, but you can replace the two numbers in for the variables since it's multiple choice.

For example,

A: -2+2*4=6 Isn't right

B: 1+2*-1= -1 Nope

C: (0,0) you can already tell it's not right because anything multiplied by 0 is 0.

D: -4+2*1=-2 This is the correct answer.

You might be interested in

Y=-3(x+4)² + 1<br> Vertex to standard

Bess [88]

Answer:

Correct

Step-by-step explanation:

7 0

3 years ago

You work 4 hours on Monday, 4.5 hours on Tuesday, 7.25 hours on Thursday, and 12 hours on Saturday. If you get paid $11.52 per h

ch4aika [34]

Answer:

$319.43

Step-by-step explanation:

Monday= 4 * $11.52= $46.08

Tuesday= 4.5 * $11.52= $51.84

Thursday= 7.25 * $11.52= $83.27

Saturday= 12 * $11.52= $138.24

When you add all of those together you get $319.43

hope this helped (:

(this took so long so this better be right...<3)

6 0

3 years ago

Is 12n2+12n+1 prime over the set of polynomials with rational coefficients? Need help ASAP

professor190 [17]

Answer:

No, the given polynomial is not prime.

Step-by-step explanation:

Given: a polynomial $12n^2+12n+1$

To check: whether the given polynomial is prime or not.

Solution:

A prime polynomial is a polynomial that cannot be factored into polynomials of lower degree and has integers as coefficients.

In the given polynomial, coefficients are integers.

Also, the polynomial $12n^2+12n+1$ cannot be factored further into polynomials of lower degree.

So, the given polynomial is not prime.

8 0

4 years ago

3(x+1)=5(x−2)+7<br> solve for x

bearhunter [10]

X=3

3(3+1)=12

5(3-2)+7=12

8 0

2 years ago

Reflection Over Parallel Lines Please complete the attached reflection. Thanks!

Allushta [10]

Answer: A(3, -5)

B(6, -2)

C(9, -2)

Step-by-step explanation:

If we have a point (x, y), and we do a reflection over the axis y = a, then the only thing that will change in our point is the value of x.

Now, the distance between x and a must remain constant before and after the reflection.

so if x - a = d

then the new position of the point will be:

(a - d, y) = (2a - x, y).

I will use that relationship for the 3 points

We start with the point (1, -5)

The reflection over y = -1 leaves.

The distance between 1 and -1 is = 1 - (-1) = 2.

Then the new point is (-1 - 2, -5) = (-3, -5)

Now we do a reflection over y = 1, so D = -3 - 1 = -2

Then the new point is:

A = (1 -(-2), -5) = (3, -5)

B) (2, -2)

Reflection over y = -1.

distance, d = 2 - (-1) = 3

the point is (-1 - 3, -2) = (-4, -2)

Now, a reflection over y = 1.

The distance is D = -4 - 1 = -5

The new point is (1 - (-5), 2) = (6, -2)

C) (5, -2)

reflection over y = -1

Distance: D = 5 - ( - 1) = 6

New point: (-1 - 6, -2) = (-7, -2)

Reflection over y = 1.

Distance D = -7 - 1 = -8

New point ( 1 - (-8), -2) = (9, -2)

4 0

3 years ago