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allochka39001 [22]
3 years ago
11

Pablo graphs a system of equations. One equation is quadratic and the other equation is linear. What is the greatest number of p

ossible solutions to this system?
a.0
b.1
c.2
d.4
Mathematics
2 answers:
german3 years ago
8 0
The answer would be option c. A quadratic equation and a linear equation only involves two variables, an independent variable and a dependent variable. Therefore, having these two equation, the greatest number of possible solutions to this <span>system is two. </span>
Nady [450]3 years ago
7 0
For this case we have the following type of equations:
 Quadratic equation:
 y = ax ^ 2 + bx + c&#10;
 Linear equation:
 y = mx + b&#10;
 We observe that when equating the equations we have:
 ax ^ 2 + bx + c = mx + b&#10;
 Rewriting we have:
 ax ^ 2 + (b-m) x + (c-b) = 0&#10;
 We obtain a polynomial of second degree, therefore, the maximum number of solutions that we can obtain is 2.
 Answer:
 The greatest number of possible solutions to this system is:
 c.2
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Nastasia [14]

Answer:

0.08x = $5400-0.1x Add 0.1x to both sides of the equation.

0.18x = $5400 Divide both sides by 0.18

x = $3000 and $54000-$3000 = $24000

$3,000 is invested at 8%

$2,400 is invested at 10%

Step-by-step explanation:

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3 years ago
The function f(x)=(x-5)^2 +2 is not one-to-one. Identify a restricted domain that makes the function one-to-one, and find the in
Anon25 [30]

We have been given a quadratic function f(x)=(x-5)^{2} +2 and we need to restrict the domain such that it becomes a one to one function.

We know that vertex of this quadratic function occurs at (5,2).

Further, we know that range of this function is [2,\infty).

If we restrict the domain of this function to either (-\infty,5] or [5,\infty), it will become one to one function.

Let us know find its inverse.

y=(x-5)^{2}+2

Upon interchanging x and y, we get:

x=(y-5)^{2}+2

Let us now solve this function for y.

(y-5)^{2}=x-2\\&#10;y-5=\pm \sqrt{x-2}\\&#10;y=5\pm \sqrt{x-2}\\

Hence, the inverse function would be f^{-1}(x)=5+\sqrt{x-2} if we restrict the domain of original function to [5,\infty) and the inverse function would be f^{-1}(x)=5-\sqrt{x-2} if we restrict the domain to (-\infty,5].

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3 years ago
Absolute Value equations always have 2 answers.<br> True<br> False
Setler [38]
The answer is, "False". 
3 0
3 years ago
Find the area of quadrilateral ABCD
andreev551 [17]

Answer:

<em>A ≈ 28.5</em>

Step-by-step explanation:

a, b, c

P = a + b + c

Semiperimeter s = \frac{a+b+c}{2}

A = \sqrt{s(s-a)(s-b)(s-c)}

~~~~~~~~~~~~~~~

P_{ABC} = 4.3 + 2.89 + 6.81 = 14

s = 14 ÷ 2 = 7

A_{ABC} = \sqrt{7(7-4.3)(7-2.89)(7-6.81)} = √14.75901 ≈ 3.84

P_{BCD} = 8.59 + 7.58 + 6.81 = 22.98

s = 22.98 ÷ 2 = 11.49

A_{BCD} = \sqrt{11.49(11.49-8.59)(11.49-7.58)(11.49-6.81)} = √609.7343148 ≈ 24.6928

A_{ABCD} = 3.84 + 24.6928 ≈ <em>28.5</em>

3 0
3 years ago
Find the solutions to the equation below. check all that apply
BartSMP [9]

Answer:

A and E

Step-by-step explanation:

Given

20x² - 26x + 8 = 0 ( divide through by 2 )

10x² - 13x + 4 = 0

Consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term.

product = 10 × 4 = 40 and sum = - 13

The factors are - 5 and - 8

Use these factors to split the x- term

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5x(2x - 1) - 4(2x - 1) = 0 ← factor out (2x - 1) from each term

(2x - 1)(5x - 4) = 0

Equate each factor to zero and solve for x

2x - 1 = 0 ⇒ 2x = 1 ⇒ x = \frac{1}{2} → E

5x - 4 = 0 ⇒ 5x = 4 ⇒ x = \frac{4}{5} → A

6 0
3 years ago
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