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IRINA_888 [86]
3 years ago
5

What does this expression represent?

Mathematics
2 answers:
liberstina [14]3 years ago
5 0

Answer:

Step-by-step explanation:

KengaRu [80]3 years ago
4 0

A: a number divided by ten.  **g/10**

B: a number to the third power times by ten.  **10g³**

C: a number to the third power divided by ten.   **g³/10**

D: a number to the second power divided by ten. **g²/10**

answer: C

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The amount of interest earned from a savings account was $580. What was the interest rate if $1432 was deposited 6 years before
SSSSS [86.1K]

Answer:


Step-by-step explanation:

Solving our equation

r = 580 / ( 1432 × 6 ) = 0.06750466

r = 0.06750466

converting r decimal to a percentage

R = 0.06750466 * 100 = 6.7505%/year

The interest rate required to

accumulate simple interest of $ 580.00

from a principal of $ 1,432.00

over 6 years is 6.7505% per year.

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3 years ago
Find the equation of a line parallel to the line 2x-y-9=0 and passing through the point of intersection of the lines 5x+y+4=0 an
Cerrena [4.2K]

Answer:

y=2x-3

Step-by-step explanation:

1. Using the point of intersections, we use the substitution method to find the coordinates of the line parallel to 2x-y-9=0

5x+y+4=0

y= -4-5x

substituting the value of y in 2x+3y=1:

2x +3(-4-5x)=1\\\\2x-12-15x-1=0\\-13x=13x\\x= \frac{13}{-13}= -1

substituting x=-1 in y= -4-5x:

y= 1 (upon solving, you should get this)

(x,y)= (-1,1)

2. Using y=mx+c and making y the subject of the formula 2x-y-9=0 and using the coordinate we found earlier, we will find the equation of the parallel line. (We make y the subject of the formula to find the gradient)

2x-y-9=0\\2x-9=y\\y= 2x-9

y= mx+c\\

-1= 2 x 1 +c

-3=c

  • y= 2x-3

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3 years ago
38/12 what is it in cents and dollars
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4 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

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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
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