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earnstyle [38]
3 years ago
12

Bobby biked 12/3 hours on monday 21/3 hours on tuesday and 22/3 hours on wednesday. what is the total number of hours bobby spen

t biking?
Mathematics
1 answer:
serg [7]3 years ago
8 0
The fractions you're given can be simplified down, so you get (12 ÷ 3) 4 hours on Monday, (21 ÷ 3) 7 hours on Tuesday, and (22 ÷ 3) 7 1/3 hours on Wednesday. You can now add these up and get your answer as 18 1/3 hours, or 18.33.

I hope this helps!
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Simplify.<br> (8 + 14m) + m<br> O<br> 8 + 15 m<br> 23 m<br> 9 + 14 m<br> 22 m
olganol [36]

Answer:

Hi there!

Your answer is:

A) 8+15m

Step-by-step explanation:

Since we can't do anything within the parentheses, they go away.

8+14m+m

Combine like terms

8+15m

Hope this helps

5 0
3 years ago
Can someone please help me find this answer.
Stels [109]

Answer:

-2x+4

Step-by-step explanation:

multiply all the components of the bracket by - 1.

then you arrive at the answer above

8 0
3 years ago
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
Two binders costs$4.36. how many bindees can buy with $55?​
Reil [10]

Answer:

Step-by-step explanation:

First you need to find the price for one binder. If 2 binders cost 4.36 you have to divide by 2 to find the price for 1. You should get 2.18. Then you have to see how many times you can multiply 2.18 without having the product go over 2.18. 2.18 • 22 = 47.96, the number closest to 50 without going over. Therefore, she can buy 22 binders.

6 0
3 years ago
Read 2 more answers
A plant nursery in Wildgrove keeps records of how many of its plants are annuals, biennials,
yanalaym [24]

Step-by-step explanation:

Complete answer: Annual plants are the ones that flower only once in their lifetime and then they die. Biennial plants are the ones that flower twice in their lifetime and perennial plants are the ones that flower many times in their life cycle. ... Perennial - grasses, alfalfa, etc.

8 0
2 years ago
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