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3 years ago

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On a coordinate plane, a solid straight line has a positive slope and goes through (negative 4, 0) and (0, 2). Everything to the

right of the line is shaded. Which linear inequality is represented by the graph? y ≤ One-halfx + 2 y ≥ One-halfx + 2 y ≤ One-thirdx + 2 y ≥ One-thirdx + 2

2 answers:

Paha777 [63]3 years ago

9 0

Answer:

$y\le\dfrac{1}{2}x+2$

Step-by-step explanation:

On a coordinate plane, a solid straight line has a positive slope and goes through (negative 4, 0) and (0, 2). This line has the equation

$y-0=\dfrac{2-0}{0-(-4)}(x-(-4))\\ \\y=\dfrac{1}{2}(x+4)\\ \\y=\dfrac{1}{2}x+2$

The origin belongs to the shaded region, so its coordinates must satisfy the inequality. Since

$\dfrac{1}{2}\cdot 0+2=2\ge 0,$

then the correct inequality is

$y\le\dfrac{1}{2}x+2$

zzz [600]3 years ago

4 0

Answer:

A

y ≤ $\frac{1}{2}$ x + 2

Step-by-step explanation:

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What is the answer for3+8x=4

Schach [20]

Answer: x=1/8

Step-by-step explanation:

Subtract 3 from both sides

3+ 8x=4

8x=1

Divide by 8 on both sides

x=1/8

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"the life length x (in years) of a dvd player" is exponentially distributed with mean 5 years. what is the probability that a mo

loris [4]

The exponential distribution is:
$\lambda e^{- \lambda x}$
where $\lambda = \frac{1}{mean}=\frac{1}{5}$

The probability we want is how likely will a dvd player last more than 8 years, given it has already lasted 5 years
To find this, you use conditional probability.
$P(A|B) = \frac{P(A and B)}{P(B)}$
where A is P(x>8) and B is P(x>5)
To find these probabilities, integrate over the distribution:
$P(x \ \textgreater \ N) = \int_N^{\infty} \frac{1}{5} e^{-x/5} dx = e^{-N/5}$

Sub into conditional probability formula:

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Final Answer: Given a dvd player is more than 5 years old, the probability that it will last another 3 more years is about 54.9%

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3 years ago

What is the area of the parallelogram shown below in square units? 9 units 20 units

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Maths functions question!!

Marina86 [1]

Answer:

5) DE = 7 units and DF = 4 units

6) ST = 8 units

$\textsf{7)} \quad \sf OM=\dfrac{3}{2}\:units$

8) x ≤ -3 and x ≥ 3

Step-by-step explanation:

<u>Information from Parts 1-4:</u>

brainly.com/question/28193969

$f(x)=-x+3$

$g(x)=x^2-9$

A = (3, 0) and C = (-3, 0)

<h3><u>Part (5)</u></h3>

Points A and D are the <u>points of intersection</u> of the two functions.

To find the x-values of the points of intersection, equate the two functions and solve for x:

$\implies g(x)=f(x)$

$\implies x^2-9=-x+3$

$\implies x^2+x-12=0$

$\implies x^2+4x-3x-12=0$

$\implies x(x+4)-3(x+4)=0$

$\implies (x-3)(x+4)=0$

Apply the zero-product property:

$\implies (x-3)= \implies x=3$

$\implies (x+4)=0 \implies x=-4$

From inspection of the graph, we can see that the x-value of point D is <u>negative</u>, therefore the x-value of point D is x = -4.

To find the y-value of point D, substitute the found value of x into one of the functions:

$\implies f(-4)=-(-4)=7$

Therefore, D = (-4, 7).

The length of DE is the difference between the y-value of D and the x-axis:

⇒ DE = 7 units

The length of DF is the difference between the x-value of D and the x-axis:

⇒ DF = 4 units

<h3><u>Part (6)</u></h3>

To find point S, substitute the x-value of point T into function g(x):

$\implies g(4)=(4)^2-9=7$

Therefore, S = (4, 7).

The length ST is the difference between the y-values of points S and T:

$\implies ST=y_S-y_T=7-(-1)=8$

Therefore, ST = 8 units.

<h3><u>Part (7)</u></h3>

The given length of QR (⁴⁵/₄) is the difference between the functions at the same value of x. To find the x-value of points Q and R (and therefore the x-value of point M), subtract g(x) from f(x) and equate to QR, then solve for x:

$\implies f(x)-g(x)=QR$

$\implies -x+3-(x^2-9)=\dfrac{45}{4}$

$\implies -x+3-x^2+9=\dfrac{45}{4}$

$\implies -x^2-x+\dfrac{3}{4}=0$

$\implies -4\left(-x^2-x+\dfrac{3}{4}\right)=-4(0)$

$\implies 4x^2+4x-3=0$

$\implies 4x^2+6x-2x-3=0$

$\implies 2x(2x+3)-1(2x+3)=0$

$\implies (2x-1)(2x+3)=0$

Apply the zero-product property:

$\implies (2x-1)=0 \implies x=\dfrac{1}{2}$

$\implies (2x+3)=0 \implies x=-\dfrac{3}{2}$

As the x-value of points M, Q and P is negative, x = -³/₂.

Length OM is the difference between the x-values of points M and the origin O:

$\implies x_O-x_m=o-(-\frac{3}{2})=\dfrac{3}{2}$

Therefore, OM = ³/₂ units.

<h3><u>Part (8)</u></h3>

The values of x for which g(x) ≥ 0 are the values of x when the parabola is above the x-axis.

Therefore, g(x) ≥ 0 when x ≤ -3 and x ≥ 3.

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1 year ago

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<img src="https://tex.z-dn.net/?f=%28x%20-%201%29%28%20x%20-%20%28%20%5Cfrac%7B1%7D%7B2%7D%20%2B%203i%29%28x%20-%20%28%20%5Cfrac

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$(X-1)(\frac{37}{4} -3ix+\frac{x}{2} )$

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3 years ago