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2 years ago

Which linear inequality is represented by the graph?

Mathematics

1 answer:

Dovator [93]2 years ago

3 0

Answer:

The answer to your question is the second choice (y ≥ 1/3x - 1)

Step-by-step explanation:

Process

1.- Find two points of the graph

A (0, -1)

B (3, 0)

2.- Find the slope

m = (0 + 1)/(3 - 0)

m = 1/3

3.- Find the equation of the line

y - y1 = m(x - x1)

y + 1 = 1/3(x - 0)

y + 1 = 1/3x

y = 1/3x - 1

4.- Find the equation of the inequality

We need the upper part of the line so the inequality must be

y ≥ 1/3x - 1

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In the figure below, EH GJ and FI intersects EH at G as shown below.

Pachacha [2.7K]

Answer:

D. 15°

Step-by-step explanation:

EGF and HGF are equal and 75°

since JGH is 90° we can say JGI is 90 - 75 = 15°

5 0

2 years ago

Find the x-intercepts of the parabola with

12345 [234]

Answer:

x-intercepts are (0, 0) and (-6, 0)

Step-by-step explanation:

equation of a parabola in vertex form: y = a(x - h)² + k

where (h, k) is the vertex

Substituting the given vertex (-3, -18) into the equation:

y = a(x + 3)² - 18

If the y-intercept is (0, 0) then substitute x=0 and y=0 into the equation and solve for a:

0 = a(0 + 3)² - 18

⇒ 0 = a(3)² - 18

⇒ 0 = 9a - 18

⇒ 9a = 18

⇒ a = 2

Therefore, y = 2(x + 3)² - 18

To find the x-intercepts, set the equation to 0 and solve for x:

2(x + 3)² - 18 = 0

Add 18 to both sides: 2(x + 3)² = 18

Divide both sides by 2: (x + 3)² = 9

Square root both sides: x + 3 = ±3

Subtract 3 from both sides: x = ±3 - 3

so x = 3 - 3 = 0

and x = -3 - 3 = -6

So x-intercepts are (0, 0) and (-6, 0)

8 0

2 years ago

Read 2 more answers

Is 6^-2 positive or negative? Explain.

Gnoma [55]

Answer:

It's positive when you have 6^-2 it's the same as having $(\frac{1}{6})^{2} \\$ . So 6^-2 it's 1/36.

Step-by-step explanation:

7 0

3 years ago

Find the general solution of x'1 = 3x1 - x2 + et, x'2 = x1.

Ann [662]

Note that if ${x_2}'=x_1$ , then ${x_2}''={x_1}'$ , and so we can collapse the system of ODEs into a linear ODE:

${x_2}''=3{x_2}'-x_2+e^t$
${x_2}''-3{x_2}'+x_2=e^t$

which is a pretty standard linear ODE with constant coefficients. We have characteristic equation

$r^2-3r+1=\left(r-\dfrac{3+\sqrt5}2\right)\left(r+\dfrac{3+\sqrt5}2\right)=0$

so that the characteristic solution is

${x_2}_C=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}$

Now let's suppose the particular solution is ${x_2}_p=ae^t$ . Then

${x_2}_p={{x_2}_p}'={{x_2}_p}''=ae^t$

and so

$ae^t-3ae^t+ae^t=-ae^t=e^t\implies a=-1$

Thus the general solution for $x_2$ is

$x_2=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}-e^t$

and you can find the solution $x_1$ by simply differentiating $x_2$ .

7 0

3 years ago

F(x) = -x + 5, what is the value of f(-4)?<br> I

liberstina [14]

Answer:

Step-by-step explanation:

F(x)=-x+5

f(-4)= -(-4)+5

f(-4)= 4+5

f(-4)= 8

Only we have to substitute value of x then after simplification we will get answer..Hope this will help

7 0

2 years ago

Read 2 more answers