All categories

3 years ago

Please help the game hates me

Mathematics

1 answer:

kiruha [24]3 years ago

8 0

Answer:

negative

Step-by-step explanation:

a positive divided by a negative is a negative

You might be interested in

Two sets of quadratic expressions are shown below in various forms: Line 1: x2 + 3x + 2 (x + 1)(x + 2) (x + 1.5)2 − 0.25 Line 2:

Serga [27]

Its line 1

if we expand the second and third expressions they come back to the first.

The last one is called the vertex form of a parabola

5 0

3 years ago

How do I graph (5, -3), (0, -4)

Lorico [155]

Answer:

5 right and down 3 and (0 means the middle) so middle down 4

Step-by-step explanation:

lit : )

4 0

3 years ago

Solve 4y-1=7 plz step by step

miv72 [106K]

Hi,

Here we are going to be working on isolating the variable y, and seeing what its value equates to.

To do this, we must try and get the variable y on one side of the equation by itself.

Let's look at step one -

4y - 1 = 7

We want to get rid of the 1 since we need to isolate x. We do this by doing the inverse of its operation. Since 1 is negative, if we add positive 1 to it - we will get 0, thereby being closer to isolating y.

However, when we do something on one side of the equation we must do it on the other. This means we will add 1 on both sides.

4y - 1 + 1 = 7 + 1

4y = 8

Remember how I mentioned we do the inverse of the operation? In this case, 4 is multiplying y. The inverse operation of multiplication is division. So, to get rid of the 4 - we must divide 4y by 4, on both sides.

4y / 4 = 8 / 4

y = 2

We now know the variable y is equal to 2.

Hopefully, this helps.

3 0

3 years ago

Read 2 more answers

Determine which of the sets of vectors is linearly independent. A: The set where p1(t) = 1, p2(t) = t2, p3(t) = 3 + 3t B: The se

defon

Answer:

The set of vectors A and C are linearly independent.

Step-by-step explanation:

A set of vector is linearly independent if and only if the linear combination of these vector can only be equalised to zero only if all coefficients are zeroes. Let is evaluate each set algraically:

$p_{1}(t) = 1$ , $p_{2}(t)= t^{2}$ and $p_{3}(t) = 3 + 3\cdot t$ :

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (3 +3\cdot t) = 0$

$(\alpha_{1}+3\cdot \alpha_{3})\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot t = 0$

The following system of linear equations is obtained:

$\alpha_{1} + 3\cdot \alpha_{3} = 0$

$\alpha_{2} = 0$

$\alpha_{3} = 0$

Whose solution is $\alpha_{1} = \alpha_{2} = \alpha_{3} = 0$ , which means that the set of vectors is linearly independent.

$p_{1}(t) = t$ , $p_{2}(t) = t^{2}$ and $p_{3}(t) = 2\cdot t + 3\cdot t^{2}$

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot t + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (2\cdot t + 3\cdot t^{2})=0$

$(\alpha_{1}+2\cdot \alpha_{3})\cdot t + (\alpha_{2}+3\cdot \alpha_{3})\cdot t^{2} = 0$

The following system of linear equations is obtained:

$\alpha_{1}+2\cdot \alpha_{3} = 0$

$\alpha_{2}+3\cdot \alpha_{3} = 0$

Since the number of variables is greater than the number of equations, let suppose that $\alpha_{3} = k$ , where $k\in\mathbb{R}$ . Then, the following relationships are consequently found: