How do you add integers with different signs?

2 answers:

7 0

A Because whenever you’re adding integers that have the same thing you keep the same signs in at the absolute value of each number

hjlf3 years ago

5 0

B because if you have something like 2+-3 it would equal -1 because 3-2 is 1 and -3 is the larger number

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ololo11 [35]

Answer:

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Step-by-step explanation:

7 0

3 years ago

Which of the following equations do not represent a proportional relationship?

Dmitry [639]

The correct answer for the question is y=x

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3 years ago

Find the x intercept of the parabola with vertex 3,-2 and y intercept 0,7 round to the nearest hundredth

12345 [234]

An x-intercept is namely a "solution" or "zero" or "root" often called, and when that happens, y = 0, just like with any other x-intercept.

$\bf ~~~~~~\textit{parabola vertex form}
\\\\
\begin{array}{llll}
\boxed{y=a(x- h)^2+ k}\\\\
x=a(y- k)^2+ h
\end{array}
\qquad\qquad
vertex~~(\stackrel{}{ h},\stackrel{}{ k})\\\\
-------------------------------\\\\
\begin{cases}
h=3\\
k=-2
\end{cases}\implies y=a(x-3)^2-2
\\\\\\
\textit{we also know that }
\begin{cases}
x=0\\
y=7
\end{cases}\implies 7=a(0-3)^2-2
\\\\\\
9=9a\implies \cfrac{9}{9}=a\implies 1=a\qquad therefore\qquad \boxed{y=(x-3)^2-2}$

so what is its x-intercept anyway?

$\bf \stackrel{y}{0}=(x-3)^2-2\implies 2=(x-3)^2\implies \pm\sqrt{2}=x-3
\\\\\\
\pm\sqrt{2}+3=x\qquad therefore\qquad (\pm\sqrt{2}+3~~~~,~~~~0)$

5 0

3 years ago

If two polynomial equations have real solutions, then will the equation that is the result of adding, subtracting, or multiplyin

kirill115 [55]

No. A polynomial equation in one variablel ooks like P(x) = Q(x), where P and Q are polynomials.

Consider polynomial equations x^2 = 3 and x^2 = 1.

Obviously they have real solutions.

Subtract the two polynomial equations:

(x^2 - x^2) = (3 - 1)

0 = 2...

We get the polynomial equation 0 = 2. We call this a polynomial equation because single constants are also by definition polynomials.

Obviously 0 = 2 has no real solution.