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zhuklara [117]
3 years ago
12

Find the slope of the graph below. Help !!

Mathematics
1 answer:
In-s [12.5K]3 years ago
5 0

Answer:

the slope is -1/4

Step-by-step explanation:

the change in y over the change in x from point (-2,-2) to (2,-3) is how to find the slope

y1-y2/x1-x2 = m

-2--3/-2-2 = -1/4

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Solve for b in the proportion.<br><br> 48<br> b<br> = <br> 39<br> 26<br><br><br> b =
vitfil [10]

Answer:

b=32

Step-by-step explanation:

4 0
3 years ago
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How many ML are in 7L
sergeinik [125]
Hello,
7l=7×1000=7000 ml

Bye :-)
8 0
4 years ago
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Find the point(s) on the surface z^2 = xy 1 which are closest to the point (7, 11, 0)
leonid [27]
Let P=(x,y,z) be an arbitrary point on the surface. The distance between P and the given point (7,11,0) is given by the function

d(x,y,z)=\sqrt{(x-7)^2+(y-11)^2+z^2}

Note that f(x) and f(x)^2 attain their extrema, if they have any, at the same values of x. This allows us to consider the modified distance function,

d^*(x,y,z)=(x-7)^2+(y-11)^2+z^2

So now you're minimizing d^*(x,y,z) subject to the constraint z^2=xy. This is a perfect candidate for applying the method of Lagrange multipliers.

The Lagrangian in this case would be

\mathcal L(x,y,z,\lambda)=d^*(x,y,z)+\lambda(z^2-xy)

which has partial derivatives

\begin{cases}\dfrac{\mathrm d\mathcal L}{\mathrm dx}=2(x-7)-\lambda y\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dy}=2(y-11)-\lambda x\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dz}=2z+2\lambda z\\\\\dfrac{\mathrm d\mathcal L}{\mathrm d\lambda}=z^2-xy\end{cases}

Setting all four equation equal to 0, you find from the third equation that either z=0 or \lambda=-1. In the first case, you arrive at a possible critical point of (0,0,0). In the second, plugging \lambda=-1 into the first two equations gives

\begin{cases}2(x-7)+y=0\\2(y-11)+x=0\end{cases}\implies\begin{cases}2x+y=14\\x+2y=22\end{cases}\implies x=2,y=10

and plugging these into the last equation gives

z^2=20\implies z=\pm\sqrt{20}=\pm2\sqrt5

So you have three potential points to check: (0,0,0), (2,10,2\sqrt5), and (2,10,-2\sqrt5). Evaluating either distance function (I use d^*), you find that

d^*(0,0,0)=170
d^*(2,10,2\sqrt5)=46
d^*(2,10,-2\sqrt5)=46

So the two points on the surface z^2=xy closest to the point (7,11,0) are (2,10,\pm2\sqrt5).
5 0
4 years ago
A cell phone company $500 for a new phone and $60 for a monthly plan. If C(t) is a rational function that represents the average
lisov135 [29]

The range of a function is the set of possible values that can be obtained from the dependent variable. The range of the function is: R: (500, \infty)

Given that:

Phone = \$500

Monthly\ Plan = \$60

Let the number of months be t. So, the function C(t) is calculated as follows:

C(t) = Phone + Monthly\ Plan \times t

C(t) = 500 + 60 \times t

C(t) = 500 + 60t

The range is calculated as follows:

The smallest possible value of t is 0 i.e. when no monthly subscription is done.

So, we have:

C(0) = 500 + 60\times 0= 500 + 0 = 500

And the highest is \infty i.e. for a large value of t

So, we have:

C(\infty) = 500 + 60\times \infty= 500 + \infty = \infty

Hence, the range of the function is:

R: (500, \infty)

Read more about range of functions at:

brainly.com/question/13824428

5 0
3 years ago
A total of 708 tickets were sold for the school play they were either adult tickets or student tickets there were 58 more studen
zhenek [66]
X = adult tickets and x+58 = student tickets

Create the equation from the problem.
x + x + 58 = 708

Combine like terms.
2x +58 = 708

Subtract 58.
2x = 650

Divide by 2.
x = 325

Therefore, your answer is:
there were 325 adult tickets sold.

(Adding 58 to that number will get you the student tickets sold which is 383 if they ask.)
3 0
3 years ago
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