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1 month ago

Which represents the solution(s) of the graphed system of equations, y = x2 + 2x – 3 and y = x – 1?

Mathematics

2 answers:

babunello [11.8K]1 month ago

7 0

For solutions to exist, the equations need to be equal to one another. Thus, we can represent it as y=y, which implies:

x^2+2x-3=x-1 we subtract x from both sides

x^2+x-3=-1 then, we add 1 to both sides

x^2+x-2=0 proceed by factoring

(x+2)(x-1)=0, leading to two solutions:

x=-2 and 1; to obtain the corresponding y values, we can apply y=x-1

y(-2)=-3 and y(1)=0, so the solutions are represented as the points:

(-2,-3) and (1,0)

lawyer [12.5K]1 month ago

6 0

The intersection points of the given system of equations, y = x² + 2x – 3 and y = x – 1, are B. (–2, –3) and (1, 0). The

Discriminant of a quadratic equation ( ax² + bx + c = 0 ) is determined using the formula:

D = b² - 4 a c. The value of the Discriminant indicates the number of solutions as follows:

D < 0 → No Real Roots

D = 0 → One Real Root D > 0 → Two Real Roots. Now, let's solve the problem!

To find solutions for the above equations, we can apply the substitution method:

$y = y$ Setting each equation to zero gives us (x + 2) = 0 or (x - 1) = 0

. Thus, we find: x = -2 or x = 1. For x = -2:

the result is ( -2, -3 )

. When x = 1:

the outcome is ( 1, 0 )

Learn more about Solving Quadratic Equations by Factoring and Determining the Discriminant.

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Suppose we are interested in bidding on a piece of land and we know one other bidder is interested. The seller announced that th

Zina [12379]

Answer:

Step-by-step reasoning:

(a)

To have an accepted bid, it needs to exceed $10,000. Let bid x be a continuous random variable uniformly distributed between

$10,000 and $15,000

The range for accepted bidding is $[ {\rm{\$ 10,000, \$ 15,000}]$ , with b = $15000 and a = $10000.

The provided bidding range is [$10,000,$12,000]. The probability is determined as follows,

$\begin{array}{c}\\P\left( {X{\rm{ < 12,000}}} \right){\rm{ = }}1 - P\left( {X > 12000} \right)\\\\ = 1 - \int\limits_{12000}^{15000} {\frac{1}{{15000 - 10000}}} dx\\\\ = 1 - \int\limits_{12000}^{15000} {\frac{1}{{5000}}} dx\\\\ = 1 - \frac{1}{{5000}}\left[ x \right]_{12000}^{15000}\\\end{array}$

$=1- \frac{[15000-12000]}{5000}\\\\=1-0.6\\\\=0.4$

(b) The accepted bidding range is [$10,000,$15,000], where b = $15,000 and a =$10,000. The given bidding range is [$10,000,$14,000].

$\begin{array}{c}\\P\left( {X{\rm{ < 14,000}}} \right){\rm{ = }}1 - P\left( {X > 14000} \right)\\\\ = 1 - \int\limits_{14000}^{15000} {\frac{1}{{15000 - 10000}}} dx\\\\ = 1 - \int\limits_{14000}^{15000} {\frac{1}{{5000}}} dx\\\\ = 1 - \frac{1}{{5000}}\left[ x \right]_{14000}^{15000}\\\end{array} P(X14000)$

$=1- \frac{[15000-14000]}{5000}\\\\=1-0.2\\\\=0.8$

(c)

The optimal amount to bid for maximizing the probability of acquiring the property is calculated as,

The accepted bidding range is [$10,000,$15,000],

where b = $15,000 and a = $10,000. The provided bidding range is [$10,000,$15,000].

$\begin{array}{c}\\f\left( {X = {\rm{15,000}}} \right){\rm{ = }}\frac{{{\rm{15000}} - {\rm{10000}}}}{{{\rm{15000}} - {\rm{10000}}}}\\\\{\rm{ = }}\frac{{{\rm{5000}}}}{{{\rm{5000}}}}\\\\{\rm{ = 1}}\\\end{array}$

(d) If you know someone willing to pay you $16,000 for the property, would you still consider bidding less than the amount mentioned in part (c)? Why or why not?

5 0

2 months ago

What is the sum of 73.04 × 1.2

tester [12383]

87.648Step-by-step explanation:Multiplying decimals can be challenging. However, by breaking the decimal multiplication into two distinct equations, it becomes simpler. Calculating 73.04 x 1.2 can be expressed as (73.04 x 1) + (73.04 x 0.2). You already know that 73.04 x 1 equals 73.04. Next, compute 73.04 x 0.2, equivalent to 73.04 divided by 5, resulting in 14.608. Finally, add the two results (73.04 + 14.608) to arrive at the final answer of 87.648.

4 0

1 month ago