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3 years ago

7

Helllllpppppp!!!!!!!pleaseeee

1 answer:

dalvyx [7]3 years ago

3 0

In your point, y is 3 and x is 2. You have to fill in these values for x and y into each equation in A-D to see which one is "true". Let's start with A. y = -3x. Filling in 3 for y and 2 for x gives us 3 = -3(2). Does 3 equal -6? Of course not. So let's try B. y = x - 1. 3 = 2 - 1. Does 3 = 1? Of course not! Let's try C. y = 3x...3 = 3(2). 3 doesn't equal 6 either, so...last one. y = x + 1 is 3=2+1. Does 3 = 3? Yes! Finally! So the point on the coordinate plane is one the line represented by D.

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How do you solve this?

posledela

Notice that the given info consists of boundary lines for an area in the xy plane. We are interested ONLY in values of x and y that are 0 or greater (positive). Graph 5x + 3y $\leq$ 37 and 3x + 5y less than or equal to 35.

Find the points of intersection of all four straight lines (including the x- and y-axes). There will be 4 such points (incl. the origin).

Next, evaluate the objective function 2x + 14y at each of these 4 points. Which of the four results is the largest? the smallest? Label them as 'maximum' and 'minimum.'

Questions? Just comment on this discussion.

4 0

4 years ago

Let f (x)=1/x and g (x)=x^2+5x. A. Find (f×g)(x). B. Find the domain and range of (f×g)(x).

Ira Lisetskai [31]

we are given

$f(x)=\frac{1}{x}$

$g(x)=x^2+5x$

(A)

(f×g)(x)=f(x)*g(x)

now, we can plug it

$(fXg)(x)=\frac{1}{x} (x^2+5x)$

we can simplify it

$(fXg)(x)=\frac{1}{x} (x(x+5))$

$(fXg)(x)=x+5$

(B)

Domain:

Firstly, we will find domain of f(x) , g(x) and (fxg)(x)

and then we can find common domain

Domain of f(x):

$f(x)=\frac{1}{x}$

we know that f(x) is undefined at x=0

so, domain will be

$(-\infty,0)$ ∪ $(0,\infty)$

Domain of g(x):

$g(x)=x^2+5x$

Since, it is polynomial

so, it is defined for all real values of x

now, we can find common domain

so, domain will be

$(-\infty,0)$ ∪ $(0,\infty)$ ..............Answer

Range:

Firstly, we will find range of f(x) , g(x) and (fxg)(x)

and then we can find common range

Range of f(x):

$f(x)=\frac{1}{x}$

we know that range is all possible values of y for which x is defined

since, horizontal asymptote will be at y=0

so, range is

$(-\infty,0)$ ∪ $(0,\infty)$

Range of g(x):

$g(x)=x^2+5x$

Since, it is quadratic equation

so, its range will be

$[-6.25,\infty)$

now, we can find common range

so, range will be

$(-6.25,0)$ ∪ $(0,\infty)$ .............Answer

6 0

3 years ago

Please answer 7 and 8, thank you guys for helping! <3

elena-14-01-66 [18.8K]

7: since they went down further you go further from zero: -28-40=-68 or D.
8: they lost 6 and then 8 so 6+8=14 but they lost them so -14 or A :)

6 0

4 years ago

Read 2 more answers

Find the measure of KL

SVETLANKA909090 [29]

Answer:

4.2

Step-by-step explanation:

Because RM and ML are the same, that makes KL and KR congruent. Hope his helps. :)

8 0

3 years ago

Read 2 more answers

3 PART QUESTION!

Illusion [34]

Answer:

See below

Step-by-step explanation:

<u>Given relationship:</u>

d = rt

a) r = 3, d = 4

t = d/r = 4/3 h = 1 h 20 min

b) d = 4 miles, rate = r

t = 4/r, domain is r and must be greater than zero

c) d = 4 miles, time = t

r = d/t = 4/t, domain is t and it must be greater than zero

d) t = 3 h, rate = r

d = rt = 3r, domain is r and can be anything greater than zero

5 0

3 years ago