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Vera_Pavlovna [14]

3 years ago

6

ginger the dog eats 1/2 cup of food each meal. if Elsie has 17 cups of food ,how many meals can she feed ginger before she runs

out?

1 answer:

V125BC [204]3 years ago

5 0

Answer:

34

Step-by-step explanation:

you just mutiply 17 by 0.5

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Is 6 a solution to the equation? Show your work.

meriva

Answer:

yes I think 6 is a solution to the equation

7 0

3 years ago

Read 2 more answers

Given the system of constraints name all vertices. Then find the maximum value of the given objective

larisa86 [58]

Answer:

$35$

Step-by-step explanation:

The constraints are

The red line represents the function

$y\leq \dfrac{1}{3}x+1$

At $y=0$

$0=\dfrac{1}{3}x+1\\\Rightarrow -1=\dfrac{1}{3}x\\\Rightarrow x=-3$

At $x=0$

$y=0+1\\\Rightarrow y=1$

Two points are $(-3,0),(0,1)$

The blue line represents the function

$5\geq y+x$

at $y=0$

$5=x$

at $x=0$

$y=5$

Two points are $(5,0),(0,5)$

The other two constraints are $x\geq 0$ , $y\geq 0$ . So, the point has to be in the first quadrant

From the graph it can be seen there are two points where the function will be maximum let us check them.

$(3,2)$

$7x-2y=7\times 3-2\times 2=17$

$(5,0)$

$7x-2y=7\times 5-2\times 0=35$

So, the maximum value of the function is $35$ .

3 0

3 years ago

How many different ways can the line be named? What are those names?

boyakko [2]

Answer: A. 3 ways: k, DE, ED (both DE and ED have line markers over top)

To name a line, we just need two points on the line. We list them in any order because the line extends forever in both directions. Contrast this with a ray where order does matter. The little k is another way to name a line, potentially simplifying things.

Choice B is close, but it mentions ray DE instead of line DE. Choice C is missing line ED. Choice D is a similar story as choice B. These facts allow us to rule out B through D.

5 0

3 years ago

Read 2 more answers

4x = -36<br> plz helpppp

san4es73 [151]

Hi there! :)

4x=-36

Divide both sides by 4 in order to isolate x:

x=-9

Hope this helps you and have a nice day!

~Just a joyful teen

$SilentNature$

8 0

3 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