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lara [203]
3 years ago
14

Jonathan places a star on a coordinate plane at (-2, -7). He wants to place another star across the y-axis, 5 points away. Where

will Jonathan place the other star?
Mathematics
1 answer:
VikaD [51]3 years ago
6 0

5 Points away could mean it is either (-2, -2) or (-2, -12)

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I’m doing a review but I forgot how to do this can someone help?
Daniel [21]

Answer:

-3x+10=5x-8

so basically its like grouping.

so answer is

8x=18?

Step-by-step explanation:

sorry if incorrect

6 0
3 years ago
Read 2 more answers
A rectangle's length is twice its width. if the perimeter is 120 cm, find the width of the rectangle.
dangina [55]
Let's use the formula to find the perimeter of a rectangle.
2l+2w=120
We know that the length is twice its width; the equation would be:L=2w
We can plug that in to the formula.
2(2w)+2w=120
4w+2w=120
6w=120
120÷6=20=w
Since the length is twice its width, 20×2=40. The length is 40 cm.
The width is 20 cm.
Tell me if this helps!!
4 0
3 years ago
A student claims that 2i is the only imaginary root of a polynomial equation that has real coefficients. Explain the student's m
____ [38]

Answer:

The Fundamental Theorem of Algebra assures that any polynomial  f(x)=0 whose degree is n ≥1 has at least one Real or Imaginary root. So by the Theorem we have infinitely solutions, including imaginary roots ≠ 2i

Step-by-step explanation:

1) This claim is mistaken.

2) The Fundamental Theorem of Algebra assures that any polynomial  f(x)=0 whose degree is n ≥1 has at least one Real or Imaginary root. So by the Theorem we have infinitely solutions, including imaginary roots ≠ 2i with real coefficients.

a_{0}x^{n}+a_{1}x^{2}+....a_{1}x+a_{0}

For example:

3) Every time a polynomial equation, like a quadratic equation which is an univariate polynomial one, has its discriminant following this rule:

\Delta < 0\\b^{2}-4*a*c

We'll have <em>n </em>different complex roots, not necessarily 2i.

For example:

Taking 3 polynomial equations with real coefficients, with

\Delta < 0

-4x^2-x-2=0 \Rightarrow S=\left \{ x'=-\frac{1}{8}-i\frac{\sqrt{31}}{8},\:x''=-\frac{1}{8}+i\frac{\sqrt{31}}{8} \right \}\\-x^2-x-8=0 \Rightarrow S=\left\{\quad x'=-\frac{1}{2}-i\frac{\sqrt{31}}{2},\:x''=-\frac{1}{2}+i\frac{\sqrt{31}}{2} \right \}\\x^2-x+30=0\Rightarrow S=\left \{ x'=\frac{1}{2}+i\frac{\sqrt{119}}{2},\:x''=\frac{1}{2}-i\frac{\sqrt{119}}{2} \right \}\\(...)

2.2) For other Polynomial equations with real coefficients we can see other complex roots ≠ 2i. In this one we have also -2i

x^5\:-\:x^4\:+\:x^3\:-\:x^2\:-\:12x\:+\:12=0 \Rightarrow S=\left \{ x_{1}=1,\:x_{2}=-\sqrt{3},\:x_{3}=\sqrt{3},\:x_{4}=2i,\:x_{5}=-2i \right \}\\

4 0
3 years ago
Simplify by combining like terms:<br> 3x - 10 + 2 + 6x
olya-2409 [2.1K]

Answer:

9x-8 is the simplified form

7 0
3 years ago
Someone help me please.
sattari [20]

The correct answer is C

4 0
3 years ago
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