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just olya [345]
2 years ago
14

A plumber has a fixed call out rate of $80 and has an hourly rate of $60

Mathematics
1 answer:
ANTONII [103]2 years ago
3 0

Answer:

do FYI kkkkkkjjj ghost oh go j

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Can somebody help me with the rest of the question
kari74 [83]

Shop #1 has the best price for 21 pizzas because it is $105, while Shop #2 costs $147 for 21 pizzas which is a lot more money. they would save $42, which is 147-105=42. hope this helps:)

5 0
3 years ago
Super easy to get pointsss!!! HELPPPP
jeyben [28]

Answer:

3x3x12= volume of large prism= 108

1x1x12= volume of hole= 12

(Volume of prism)-(volume of hole)= 96

Step-by-step explanation:

8 0
3 years ago
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How do you round 1.038 to the nearest hundredth
Talja [164]
You have to go past the decimal point in since the first number in the tenths place is a zero you can't really round here so you go to the hundreds place and not so three you round down because 3 is closer to 0 then it is 10 so it would be 1. 4



3 0
3 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
3 years ago
PLZZZZ HELP WITH THIS QUESTION
Sauron [17]

A. 1/5 fish

We know that 2/5 of Mike's fish are clownfish. Therefore, 3/5 are not clownish as 5/5 – 2/5 = 3/5. Also, if you look at the model, there are five pieces. If we assume that model represents the whole of Mike's fish and you take away two pieces, you are left with 3/5. So we know that the remaining fish is 3/5

Next, we know that of these 3/5 fish, 1/3 is damselfish, so we need to find 1/3 of 3/5. To do this, we must multiply 1/3 by 3/5 as "of" means multiply in Math.

So: 1/3 • 3/5 = 1 • 3/3 • 5 = 3/15       3 ÷ 3 = 1 and 15 ÷ 3 = 5    3/15 = 1/5

1/5 of Mike's fish are damsel fish


B. 2/5 fish

Now we know that 1/5 of Mike's fish is damselfish, and 2/5 is clownfish. To find the fraction of his fish that are neither, therefore, we must subtract their sum from the whole.

First, we add 1/5 and 2/5 together. Adding the numerators, 1 and 2, we get 1/5 + 2/5 = 3/5

Next, we subtract: 5/5 – 3/5 = 2/5, so 2/5 of his fish are neither clownfish or damselfish

And if you look at the model again, you can see that if you cross out 1 piece for the damselfish, and 2 pieces for the clownfish, you are left with 2/5

8 0
3 years ago
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