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4 years ago

What is the solution to the system of linear equations graphed below?

Mathematics

1 answer:

OlgaM077 [116]4 years ago

5 0

Option C: There are no solutions

Explanation:

The linear equations is graphed.

We need to determine the solution of the system of equations.

The solution of the equations can be determined by finding the point of intersection of the two equations.

From the figure, it is obvious that the two equations are parallel to each other.

Also, the parallel lines have the same slope and the parallel lines never intersect.

Hence, the system consisting of parallel lines have no solution.

Therefore, the solution to the system of linear equations graphed is no solution.

Thus, Option C is the correct answer.

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Factor x^3+2x^2-9x-18, given that -2 is a zero.

Kisachek [45]

x³ + 2x² - 9x - 18 = 0

x²(x+2) - 9(x+2) = 0

(x + 2)(x² - 9) = 0

(x + 2)(x - 3)(x + 3) = 0

the answer is d.

4 0

4 years ago

SOMEONE PLEASE HELP ME! I promise i will mark brainlest, but i need the right answer for this. help is much appreciated <3 wh

dem82 [27]

Answer:

The middle one because there is only 1 value of y for every value of x.

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3 years ago

A parabola is graphed below.

Len [333]

Answer:

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

Here the vertex = (- 2, - 3 ) , then

y = a(x - (- 2))² - 3 , that is

y = a(x + 2)² - 3

To find a, substitute any point on the graph into the equation

Using the coordinates of the y- intercept (0, 5 )

5 = a(0 + 2)² - 3 ( add 3 to both sides )

8 = a(2)² = 4a ( divide both sides by 4 )

2 = a

y = 2(x + 2)² - 3 → B

3 0

3 years ago

Find the cdf F(x) associated with each of the following probability density functions. Sketch the graphs of f(x) and F(x).

wolverine [178]

Answer:

See steps below

Step-by-step explanation:

$\bf f(x)=3(1-x)^2\;(0$

$\bf F(x)=\int_{0}^{x}f(t)dt=\int_{0}^{x}3(1-t)^2dt=3\int_{0}^{x}(1-t)^2=1-(1-x)^3$

The cdf associated with f is

$\bf \boxed{F(x)=1-(1-x)^3}$ for 0<x<1

<h3>See picture 1 </h3>

The median is a point x such that

F(x) = ½

so, the median is

$\bf 1-(1-x)^3=1/2\rightarrow (1-x)^3=1/2\rightarrow \boxed{x=1-\sqrt[3]{2}}$

The 25th percentile equals the 1st quartile and is a point x such

F(x) = ¼

and the 25th percentile is

$\bf 1-(1-x)^3=1/4\rightarrow (1-x)^3=3/4\rightarrow \boxed{x=1-\sqrt[3]{3/4}}$

$\bf f(x)=\frac{1}{x^2}\;(1$

$\bf F(x)=\int_{1}^{x}f(t)dt=\int_{1}^{x}\frac{dt}{t^2}=1-\frac{1}{x}$

The cdf associated with f is

$\bf \boxed{F(x)=1-\frac{1}{x}}$ for x>1

<h3>See picture 2 </h3>

The median is

$\bf 1-\frac{1}{x}=\frac{1}{2}\rightarrow \boxed{x=2}$

The 25th percentile is

$\bf 1-\frac{1}{x}=\frac{1}{4}\rightarrow \boxed{x=4/3}$

f(x) = 1/3 for 0<x<1 or 2<x<4

$\bf F(x)=\int_{0}^{x}\frac{dt}{3}=\frac{x}{3}\;(0$

$\bf F(x)=\frac{1}{3}+\int_{2}^{x}\frac{dt}{3}=\frac{1}{3}+\frac{x-2}{3}=\frac{x-1}{3}\;(2$

The cdf associated with f is

$\bf F(x)=\frac{x}{3}$ for 0<x<1

$\bf F(x)=\frac{x-1}{3}$ for 2<x<4

<h3>See picture 3 </h3>

The median is

$\bf \frac{x-1}{3}=1/2\rightarrow x=1+3/2\rightarrow \boxed{x=5/2}$

The 25th percentile is

$\bf \frac{x}{3}=1/4\rightarrow \boxed{x=3/4}$

4 0

4 years ago

Round 725,390 to da nearest thousand

Komok [63]

Answer:

725,000

Step-by-step explanation:

The number to the right of 5 is 3, it's lower than five so your thousand is going to stay the same number. If your hundreds place is greater than 5, then it would be rounded to 726,000.

4 0

4 years ago