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2 years ago

Which function has lowest rate of change?

Mathematics

1 answer:

Delvig [45]2 years ago

8 0

Answer:

A: Algebraic Function

Step-by-step explanation:

Algebraic function is given as;

5x – 2y = 15

Since we are dealing with rate of change, let's make y the subject first and find dy/dx.

Thus;

2y = 5x - 15

y = (5x/2) - 15/2

dy/dx = 5/2

dy/dx = 2.5

The verbal function is given as;

Cost of 5 pencils is 15 dollars

Thus, rate of 1 pencil = 15/5 = 3 pencils per dollar

We can see that the lowest rate of change is 2.5 and it is the algebraic function.

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Simplify: (2x – 3)2 4x2 + 9 4x2 – 12x + 9 4x2 – 12x + 6 4x2 – 6

Alexxandr [17]

Answer:

$4x^2-12x+9$

Step-by-step explanation:

we have

$(2x-3)^2$

we know that

$(a-b)^2=a^2-2ab+b^2$

In this problem

$a=2x\\b=3$

substitute in the formula

$(2x-3)^2=(2x)^2-2(2x)(3)+(3)^2$

$(2x-3)^2=4x^2-12x+9$

7 0

2 years ago

Read 2 more answers

Which of the following sets are subspaces of R3 ?

Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

$\to A={(x,y,z)|3x+8y-5z=2} \\\\\to for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)$

The set A is not part of the subspace $R^3$

for point B:

$\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0$

$\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon B$

The set B is part of the subspace $R^3$

for point C: $\to C={(x,y,z)|x$

In this, the scalar multiplication can't behold

$\to for (-2,-1,2) \varepsilon C$

$\to -1(-2,-1,2)= (2,1,-1)$ ∉ C

this inequality is not hold

The set C is not a part of the subspace $R^3$

for point D:

$\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)$

The scalar multiplication s is not to hold

$\to for (-4, 1,2)\varepsilon D\\\\\to -1(-4,1,2) = (4,-1,-2)$ ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace $R^3$

For point E:

$\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\$

The $x_1, x_2$ is the arbitrary, in which $ax_1+bx_2$ is arbitrary

$\to a(x_1,0,0)+b(x_2,0,0) \varepsilon E$

The set E is the part of the subspace $R^3$

For point F:

$\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon F \\\\\to a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))$

The $x_1, x_2$ arbitrary so, they have $ax_1+bx_2$ as the arbitrary $\to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F$

The set F is the subspace of $R^3$

5 0

2 years ago

The graph of y= f(X) is shown below, what point could be used to find f(3)?

balandron [24]

Answer:

Point A

Step-by-step explanation:

The long that would be used to find f(3) is the point that shows the value of y when x = 3.

Looking at the graph given, the only point that shows the value of y when x = 3 is point A.

When x = 3, y = 0.

Therefore, f(3) = 0.

8 0

2 years ago

For breakfast, Randy had two Egg McMuffins and a hash brown, to taling 750 calories . Jack only had on e Egg McMuffin and a hash

Delvig [45]

Let us assign variables.
Let x = egg mcmuffins
Let y = hash brown

Randy had:
2x + y = 750

Jack had
x + y = 450

Using systems of linear equation, we can get the value for x and y by elimination. Subtract the equations.
2x + y = 750
x + y = 450
--------------
x = 300

Get y
x + y = 450
300 + y = 450
y = 150.

So there are 300 calories for McMuffins and 150 calories for hash brown.

4 0

2 years ago

[10 Points, Algebra 2]

tensa zangetsu [6.8K]

5/ 1+1= 5/2
1+4/1^2-3(1)-4= 5/-6

5/2 - 5/-6 does not equal zero

4 0

2 years ago