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Oksanka [162]
3 years ago
5

Y= x² + 5x – 3 y - x = 2

Mathematics
1 answer:
boyakko [2]3 years ago
7 0

Answer:

if you need graph

is as following

1) y=x2+5x-3

2) y-x=2

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What are the two categories of fees?
Nesterboy [21]

Step-by-step explanation:

Types of fees and charges

Monthly account keeping or service fee. The fee you pay for an organisation to manage your bank account. ...

Internet banking fee. ...

EFTPOS transaction fee. ...

ATM transaction fee. ...

Non-bank or foreign ATM fee. ...

Telephone banking transaction fee. ...

Branch withdrawal fee. ...

Cheque withdrawal fee.

3 0
2 years ago
Read 2 more answers
Help whit that 3 questions please and thanks
Serjik [45]

Answer:

80.

4s+5c=46\\3s+c=22

81. Each apple cost $0.50

82. One pound of chocolate = $3

Step-by-step explanation:

80.

Let sold of 1 senior citizen ticket be "s", and

cost of 1 child ticket be "c"

First Day, 4 senior citizen ticket and 5 child tickets sold, equaled $46, so the equation we can write is:

4s+5c=46

Second Day, 3 senior citizen ticket and 1 child ticket sold, equaled $22, so the equation we can write is:

3s+c=22

So, the system of equation (2 equations) written above can be solved simultaneously to find our answer.

4s+5c=46\\3s+c=22

81.

Let price of 1 apple be "a" and price of 1 banana be "b"

Since 3 apples and 2 banana cost $2, so we can write:

3a + 2b = 2

Then,

4 apples and 4 bananas cost $3, so we can write:

4a + 4b = 3

We can solve for the variable b in equation 1. Shown below:

3a + 2b = 2\\2b=2-3a\\b=\frac{2-3a}{2}

Now we put this value of b into equation 2 and solve for cost of 1 apple, a. Shown below:

4a+4b=3\\4a+4(\frac{2-3a}{2})=3\\4a+4-6a=3\\2a=1\\a=\frac{1}{2}\\a=0.5

So each apples cost $0.5

82.

Let 1 pound of chocolate be cost "c"

Let 1 pound of peanut be cost "p"

4 lb of chocolate and 2 lb of peanut = $16, thus we can write:

4c + 2p = 16

Next,

2 lb of chocolate and 3 lb of peanut costed $12, so we can write:

2c + 3p = 12

We solve first equation for p:

4c + 2p = 16\\2p=16-4c\\p=\frac{16-4c}{2}\\p=8-2c

Now, we put this expression for p into equation 2 and solve for c:

2c + 3p = 12\\2c + 3(8-2c) = 12\\2c+24-6c=12\\4c=12\\c=3

So, 1 pound of chocolates cost $3

5 0
4 years ago
Expanded Form and Area Model- Compare the two methods. How were those methods different? Explain which model you found most help
lisov135 [29]
On expanded form is for example 139 in expanded form would be (100)+(30)+(9)and area model is the model for multiplication
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
An estimate for -√93 is ________.<br> (a) -8.1<br> (b) -8.5<br> (c) -9.1<br> (d) -9.5
prohojiy [21]
D hope this helps I think
7 0
4 years ago
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