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stepladder [879]
3 years ago
13

Which graph shows a rate of change of one-half between –4 and 0 on the x-axis? On a coordinate plane, a straight line with a pos

itive slope crosses the x-axis at (6, 0) and the y-axis at (0, 3). Solid circles appear on the line at (negative 4, negative 1), (0, 3). On a coordinate plane, a parabola opens up. It goes through (negative 5.25, 4), has a vertex of (0, negative 3), and goes through (5.25, 4). Solid circles appear on the parabola at (negative 4, 1), (0, negative 3). On a coordinate plane, a parabola opens down. It goes through (negative 5.5, negative 4), has a vertex of (0, 4), and goes through (5.5, negative 4). Solid circles appear on the parabola at (negative 4, 0), (0, 4). On a coordinate plane, a curved line opens up and left in quadrant 2. It is asymptotic to the negative x-axis and positive y-axis. Solid circles appear on the line at (negative 4, 0.25), (0, 4).
Mathematics
1 answer:
inysia [295]3 years ago
8 0

Answer:

First option

Step-by-step explanation:

There is a mistake in the description of the first function.  

On a coordinate plane, a straight line with a positive slope crosses the x-axis at (-6, 0) and the y-axis at (0, 3). Solid circles appear on the line at (-4, 1), (0, 3).

Given two points of a function, (x1, y1) and (x2, y2), its rate of change between them is:

Rate of change: (y2 - y1)/(x2 - x1)

Points of the 1st graph: (-4, 1), (0, 3).

Rate of change: (3- 1)/(0 - (-4)) = 1/2

Points of the 2nd graph: (-4, 1), (0, -3).

Rate of change: (-3 - 1)/(0 - (-4)) = -1

Points of the 3rd graph: (-4, 0), (0, 4).

Rate of change: (4 - 0)/(0 - (-4)) = 1

Points of the 4th graph: (-4, 0.25), (0, 4).

Rate of change: (4 - 0.25)/(0 - (-4))   = 0.9375

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Find the average rate of change of f(x)=2x^2-7x from x=2 to x=6
scoundrel [369]

Answer:

Step-by-step explanation:

The average rate of change of a function f between a and b (a< b) is :

R =[f(b)-f(a)] ÷ (a-b)

●●●●●●●●●●●●●●●●●●●●●●●●

Let R be the average rate of change of this function

f(6) = 2×6^2 - 7×6 = 72-42 = 30

f(2) = 2×2^2 - 7×2 = 8-14 = -6

R = [f(6) - f(2)]÷ (6-2)

R = [30-(-6)] ÷ 4

R = -36/4

R = -9

■■■■■■■■■■■■■■■■■■■■■■■■■

The average rate of change of this function is -9

8 0
3 years ago
Read 2 more answers
Of 100 students,65 are members of a mathematics club and 40 are members of a physics club. if 10 are members of neither club the
alexandr1967 [171]

a. 15 students are members of both clubs

b. 50 students only are members of the mathematics club

c. 25 students only are members of the physics club

Step-by-step explanation:

The given is:

  • There are 100 students
  • 65 are members of a mathematics club
  • 40 are members of a physics club
  • 10 are members of neither club

We need to find how many students are members of

a. both clubs?

b. only mathematics club

c. only physics club

∵ The total number of students = 100

∵ 10 are members of neither club

- Subtract 10 from 100 to find the members of the mathematics

  club or the physics club

∵ 100 - 10 = 90

∴ There are 90 members of the mathematics club or physics club

∴ n(mathematics or physics) = 90

∵ 65 students are members of a mathematics club

∵ n(mathematics) = 65

∵ 40 students are members of a physics club

∴ n(physics) = 40

∵ n(mathematics or physics) = n(mathematics) + n(physics) - n(both)

∴ 90 = 65 + 40 - n(both)

∴ 90 = 105 - n(both)

- Add n(both) to each side

∴ n(both) + 90 = 105

- Subtract 90 from each side

∴ n(both) = 15

∴ 15 students are members of both clubs

a. 15 students are members of both clubs

∵ 65 students are members of the mathematics club

∵ 15 of them are members of the physics club

∴ n(mathematics only) = 65 - 15 = 50

∴ 50 students only are members of the mathematics club

b. 50 students only are members of the mathematics club

∵ 40 students are members of the physics club

∵ 15 of them are members of the mathematics club

∴ n(physics only) = 40 - 15 = 25

∴ 25 students only are members of the physics club

c. 25 students only are members of the physics club

Learn more:

You can learn more about word problems in brainly.com/question/8907574

#LearnwithBrainly

5 0
3 years ago
3. Solve the system of equations using linear combination. <br><br> a + c = 9<br> 8a + 4.5c = 58
lana66690 [7]
Hey I have posted a picture the method I use to solve simultaneous equations is change the sign and follow the new sign.
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5 0
3 years ago
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Solve the Differential equation (x^2 + y^2) dx + (x^2 - xy) dy = 0
natita [175]

Answer:

\frac{y}{x}-2ln(\frac{y}{x}+1)=lnx+C

Step-by-step explanation:

Given differential equation,

(x^2 + y^2) dx + (x^2 - xy) dy = 0

\implies \frac{dy}{dx}=-\frac{x^2 + y^2}{x^2 - xy}----(1)

Let y = vx

Differentiating with respect to x,

\frac{dy}{dx}=v+x\frac{dv}{dx}

From equation (1),

v+x\frac{dv}{dx}=-\frac{x^2 + (vx)^2}{x^2 - x(vx)}

v+x\frac{dv}{dx}=-\frac{x^2 + v^2x^2}{x^2 - vx^2}

v+x\frac{dv}{dx}=-\frac{1 + v^2}{1 - v}

v+x\frac{dv}{dx}=\frac{1 + v^2}{v-1}

x\frac{dv}{dx}=\frac{1 + v^2}{v-1}-v

x\frac{dv}{dx}=\frac{1 + v^2-v^2+v}{v-1}

x\frac{dv}{dx}=\frac{v+1}{v-1}

\frac{v-1}{v+1}dv=\frac{1}{x}dx

Integrating both sides,

\int{\frac{v-1}{v+1}}dv=\int{\frac{1}{x}}dx

\int{\frac{v-1+1-1}{v+1}}dv=lnx + C

\int{1-\frac{2}{v+1}}dv=lnx + C

v-2ln(v+1)=lnx+C

Now, y = vx ⇒ v = y/x

\implies \frac{y}{x}-2ln(\frac{y}{x}+1)=lnx+C

5 0
3 years ago
What is an equation of the line that passes through the point (-6, – 7) and is perpendicular to the line 6x + 5y = 30?
fomenos

Answer :6x + 5y = 30

5y = 30 - 6x

y = 6 - (6/5) x

So slope = -6/5

Perpendicular line has slope 5/6   (product of the slopes is -1)

So y = 5/6 x            +   b

Passed through (-6, -7) so           -7  =   -5  + b

Sp b = -2

So y = (5/6)x - 2

Step-by-step explanation:

3 0
3 years ago
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