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3 years ago

5

Graph the equation y = -3x on a graph

1 answer:

Dafna1 [17]3 years ago

3 0

Put a dot on the -3 from the y-axis and draw a straight line from left to right.

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Philip tiene 8 bolas rojas, 3 bolas verdes, 6 bolas amarillas, 13 bolas negras y 15 bolas azules en su bolsa. ¿cuál es la media,

Zielflug [23.3K]

Answer:

Media: 9
Mediana: 8
Moda: 8, 3, 6, 13, 15
Rango: 12

Step-by-step explanation:
Ojala que te ayude ^^

3 0

2 years ago

In Exercises 45–48, let f(x) = (x - 2)2 + 1. Match the<br> function with its graph

MA_775_DIABLO [31]

Answer:

45) The function corresponds to graph A

46) The function corresponds to graph C

47) The function corresponds to graph B

48) The function corresponds to graph D

Step-by-step explanation:

We know that the function f(x) is:

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

45)

The function g(x) is given by:

$g(x)=f(x-1)$

using f(x) we can find f(x-1)

$g(x)=((x-1)-2)^{2}+1=(x-3)^{2}+1$

If we take the derivative and equal to zero we will find the minimum value of the parabolla (x,y) and then find the correct graph.

$g(x)'=2(x-3)$

$2(x-3)=0$

$x=3$

Puting it on g(x) we will get y value.

$y=g(3)=(3-3)^{2}+1$

$y=g(3)=1$

<u>Then, the minimum point of this function is (3,1) and it corresponds to (A)</u>

46)

Let's use the same method here.

$g(x)=f(x+2)$

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

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

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2x$

$0=2x$

$x=0$

Evaluatinf g(x) at this value of x we have:

$g(0)=(x)^{2}+1$

$g(0)=1$

<u>Then, the minimum point of this function is (0,1) and it corresponds to (C)</u>

47)

Let's use the same method here.

$g(x)=f(x)+2$

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

$g(x)=(x-2)^{2}+3$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2(x-2)$

$0=2(x-2)$

$x=2$

Evaluatinf g(x) at this value of x we have:

$g(2)=(2-2)^{2}+3$

$g(2)=3$

<u>Then, the minimum point of this function is (2,3) and it corresponds to (B)</u>

48)

Let's use the same method here.

$g(x)=f(x)-3$

$g(x)=(x-2)^{2}+1-3$

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

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2(x-2)$

$0=2(x-2)$

$x=2$

Evaluatinf g(x) at this value of x we have:

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

$g(2)=-2$

<u>Then, the minimum point of this function is (2,-2) and it corresponds to (D)</u>

<u />

I hope it helps you!

<u />

8 0

3 years ago

The Floyds have considered three options to ensure they have at least $1,375 saved for their trip.

ANTONII [103]

I am guessing you are talking about Option 3. They start with $125. So they times it by 1.25 as we are using the multiplier method. So in the First month, it will be 156.25. Second it will be 195.3125. So in August it will be $476.837157. Of couse round it and you will get $476.84.

I believe there is a more efficent way of doing this but I have forgotten.

4 0

3 years ago

Write the slope intercept equation of the function f whose graph satisifies the given conditions. The graph of f passes through

MrMuchimi

Let's find slope first.

(1,-2)
(-8,6)

$\\ \sf\longmapsto m=\dfrac{6+2}{-8-1}$

$\\ \sf\longmapsto m=\dfrac{8}{-9}$

Slope intercept form

$\\ \sf\longmapsto y=mx+b$

$\\ \sf\longmapsto y=-\dfrac{8}{9}x-2$

3 0

3 years ago

What is the volume of the extra large tube

Fantom [35]

If by base of 10, you mean that the diameter of the base is 10, then the radius of the base is 5.

volume = pi * r^2 * h

volume = pi * 5^2 * 10

volume = 250pi, or approximately 785

6 0

3 years ago