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3 years ago

6

Please help me. i don't get it.

1 answer:

vova2212 [387]3 years ago

8 0

Do you know how to find cubic metres??

You might be interested in

The equation of line fis - 3y + 4x = 9 write the equation of a line that is parallel to line and passes through the point (- 12,

ollegr [7]

Answer:

$y = - 4x - 42$

Step-by-step explanation:

The equation of the line that is parallel to the line we are trying to find is

$- 3y + 4x = 9 \\ or \\ - 3y = - 4x + 9 \:(standard form) \\ \\therefore \: the \: slope \: of \: the \: line \\ \: is \: - 4$

We can recall that when two lines are parallel it means that they have the same slope. Therefore the slope of the line we are trying to find is also -4. We now know that:

$y = 6 \: \: \: x = - 12 \: \: \: slope = - 4$

Therefore the equation of the line is:

$y = mx + c \\ 6= ( - 4)( - 12) + c \\ 6 = 48 + c \\ 6 - 48 = c \\ - 42 = c$

$y = mx + c \\ y = - 4x - 42$

3 0

2 years ago

F(x)=-2x-1 and g(x)=x^2-1. Find (fog)(-5).

serg [7]

Answer: - 49

Step-by-step explanation:

f(x) = -2x - 1

g(x) = $x^{2}$ - 1

fg(x) = f ( $x^{2}$ - 1) , we just put in the value of g(x) , the next thing is to substitute $x^{2}$ - 1 for the value of x in f(x) , that is

fg(x) = -2( $x^{2}$ - 1) - 1

fg(x) = -2 $x^{2}$ + 2 - 1

fg(x) = -2 $x^{2}$ + 1

Therefore : fg(-5) means we will substitute -5 for x in fg(x) , that is

fg(-5) = -2 ( $(-5)^{2}$ + 1

fg(-5) = -2(25) + 1

fg(-5) = -50 + 1

fg(-5) = -49

7 0

4 years ago

2, -2, 2, -2, common ratio

adelina 88 [10]

To get -2 from 2, you would multiple by -1.

To get 2 from -2, you also would multiple by -1.

Ad infinitum.

So the common ratio is -1.

Hope this helps.

8 0

4 years ago

Explain how to write a quadratic equation given the following three points on the graph (5,31) (3,11) (0,11)

masha68 [24]

Given:

The graph of a quadratic function passes through the points (5,31) (3,11) (0,11).

To find:

The equation of the quadratic function.

Solution:

A quadratic function is defined as:

$y=ax^2+bx+c$ ...(i)

It is passes through the point (0,11). So, substitute $x=0,y=11$ in (i).

$11=a(0)^2+b(0)+c$

$11=c$

Putting $c=11$ in (i), we get

$y=ax^2+bx+11$ ...(ii)

The quadratic function passes through the point (5,31). So, substitute $x=5,y=31$ in (ii).

$31=a(5)^2+b(5)+11$

$31-11=a(25)+5b$

$20=25a+5b$

Divide both sides by 5.

$4=5a+b$ ...(iii)

The quadratic function passes through the point (3,11). So, substitute $x=3,y=11$ in (ii).

$11=a(3)^2+b(3)+11$

$11-11=a(9)+3b$

$0=9a+3b$

Divide both sides by 3.

$0=3a+b$ ...(iv)

Subtracting (iv) from (iii), we get

$4-0=5a+b-3a-b$

$4=2a$

$\dfrac{4}{2}=a$

$2=a$

Putting $a=2$ in (iv), we get

$0=3(2)+b$

$0=6+b$

$-6=b$

Putting $a=2,b=-6$ in (ii), we get

$y=(2)x^2+(-6)x+11$

$y=2x^2-6x+11$

Therefore, the required quadratic equation is $y=2x^2-6x+11$ .

7 0

3 years ago

Someone plz help asap

zheka24 [161]

Answer:

As the slope is -2, so the rate of change of the function is:

rate of change of function = -2

Step-by-step explanation:

The slope is basically the rate of change of a function.

Let us take the two points from the graph

(0, 5)

(4, -3)

Taking the slope between two points to determine the rate of change of a function

$\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\left(x_1,\:y_1\right)=\left(0,\:5\right),\:\left(x_2,\:y_2\right)=\left(4,\:-3\right)$

$m=-2$

As the slope is -2, so the rate of change of the function is:

rate of change of function = -2

5 0

3 years ago