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Firlakuza [10]
3 years ago
5

Compare the quantities in Column A and Column B.

Mathematics
1 answer:
GaryK [48]3 years ago
6 0

The y-intercept of 2y = 3x - 4 and the y-intercept of 4x - 2y=4 are equal

The equations are given as:

2y = 3x - 4

4x - 2y=4

Make y the subject in both equations

<u>First equation</u>

2y = 3x - 4

y = \frac 32x - 2

<u>Second equation</u>

4x - 2y=4

y = 2x - 2

A linear equation is represented as:

y = mx + b

Where b represents the y-intercept

So: By comparison,

b_1 = 2 --- the y-intercept of the first equation

b_2 = 2 --- the y-intercept of the second equation

2 = 2.

Hence, the y-intercept of 2y = 3x - 4 and the y-intercept of 4x - 2y=4 are equal

Read more about y-intercepts at:

brainly.com/question/4015585

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Each week, heather's company has $5000 in fixed costs plus an additional $250 for each system produced. The company is able to p
Pepsi [2]

Answer:

Step-by-step explanation:

Step one:

Given data for Heather's company

Fixed cost = $5000

Additional cost = $250

in a week there are 168 hours

hence they will produce a total of 168/5= 33.6

=34 systems

the expression for the total cost can be modeled as

c=mn+f

where c=total cost

 m= the Additional cost

n= number of systems produced per hour

  h= the number of  systems produced per hour

   f= fixed cost

Step two:

c=mn+f

given that n=34, we can find the total cost as

c=250(34)+5000

c=8500+5000

c=13500

the total cost for each week is $13,500

3 0
3 years ago
I need some serious help at the moment. can someone help me with this problem?
IRINA_888 [86]

Answer:

You get 0.95


Step-by-step explanation:

Change each percentage to decimals.

70% is equal to 0.70

25% is equal to .25

15% is equal to .15

Add the decimal of pizza to the decimal of salad.


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
Find z. Write your answer in simplest radical form.
Elenna [48]
Z=8 mm
hope this helped
4 0
3 years ago
This caked was sliced parallel to the base
Greeley [361]
Square (added characters)
3 0
4 years ago
Read 2 more answers
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