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2 years ago

Why do we reverse the direction of the inequality symbol when multiplying or dividing by a negative number?

1 answer:

7 0

When you multiplied both sides by a NEGATIVE number, what WAS negative became positive, and what WAS positive became negative! Everything that was larger became small, and what was smaller became large. Everything REVERSED

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12 friends are at a party.8 Friends leave the party early. they have 3 1/2 pizzas. if they share the pizza evenly, how much pizz

Jobisdone [24]

There is not enough information did they share the pizza before the ones left or after that means there could be 2 probable outcomes

6 0

2 years ago

Can you check my Geometry answer? The perimeter of triangle ABC is 30 feet. AB=5x-7; BC=3x+1; AC=4x. List the angles of triangle

Alchen [17]

Yes, you're ok that's correct! :-)

3 0

2 years ago

Read 2 more answers

Determine whether the ordered pair is a solution to the inequality. (-1, -1); y Greaterthanorequalto - x + 7 (5, 3); y > 3/4

Art [367]

Answer:

No and Yes.

Step-by-step explanation:

The inequation is y ≥-x+7 and the pair is (x,y)=(-1,-1). Then:

-1≥ -(-1)+7

-1 ≥ 1+7

-1 ≥ 8, but this is not true, so (-1,-1) is not a solution of y ≥-x+7.

On the other hand, the inequation is y>(3/4)x-5 and the pair is (5,3). Then:

3>(3/4)(5)-5

3>(15/4)-5

3> -5/4 and this is true, so (5,3) is a solution of y>(3/4)x-5.

3 0

2 years ago

Graph and Label the Points on the coordinate plane below

Lapatulllka [165]

In order to graph these, you need to understand the X and Y axis.

The X-Axis contains the first number in your ordered pair, and goes left, right (horizontally) for example the first ordered pair you have.

(2,5 1/2). To do this, move 2 spaces to the right on your X-axis. now go 5 and a half spaces up, and plot the point.

(-2 1/2, 0.5) For this one, you move 2 and a half spaces to the left on the X-axis, and half a space up. put a point there.

(3.5, -4 1/2) move 3 and a half spaces to the right on your X-axis, and go 4 and a half spaces down. put a point there.

let me know if you have questions! :)

8 0

2 years ago

Write the equation of a polynomial of degree 3, with zeros 1, 2 and -1 where f(0)=2

drek231 [11]

<u>Answer:</u>

The equation of a polynomial of degree 3, with zeros 1, 2 and -1 is $x^{3}-2 x^{2}-x+2=0$

<u>Solution:</u>

Given, the polynomial has degree 3 and roots as 1, 2, and -1. And f(0) = 2.

We have to find the equation of the above polynomial.

We know that, general equation of 3rd degree polynomial is

$F(x)=x^{3}-(a+b+c) x^{2}+(a b+b c+a c) x-a b c=0$

where a, b, c are roots of the polynomial.

Here in our problem, a = 1, b = 2, c = -1.

Substitute the above values in f(x)

$F(x)=x^{3}-(1+2+(-1)) x^{2}+(1 \times 2+2(-1)+1(-1)) x-1 \times 2 \times(-1)=0$

$\begin{array}{l}{\rightarrow x^{3}-(3-1) x^{2}+(2-2-1) x-(-2)=0} \\ {\rightarrow x^{3}-(2) x^{2}+(-1) x-(-2)=0} \\ {\rightarrow x^{3}-2 x^{2}-x+2=0}\end{array}$

So, the equation is $x^{3}-2 x^{2}-x+2=0$

Let us put x = 0 in f(x) to check whether our answer is correct or not.

$\mathrm{F}(0) \rightarrow 0^{3}-2(0)^{2}-0+2=2$

Hence, the equation of the polynomial is $x^{3}-2 x^{2}-x+2=0$

3 0

2 years ago