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Semenov [28]
3 years ago
6

What is 3÷34 A. 2.25 B. 3 C. 4 D. 12

Mathematics
2 answers:
harina [27]3 years ago
6 0

Answer:

I think it is 12

(D)

blsea [12.9K]3 years ago
4 0

ANSWER:

its D= 12

its easy

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A rectangle has a length of the fifth root of 16 inches and a width of 2 to the 1 over 5 power inches. Find the area of the rect
gayaneshka [121]
Hello,

It 's difficult to translate that in french.

l=16 ^(1/5)
w=2^(1/5)
Area=32^(1/5)=(2^5)^(1/5)=2 (in²)


7 0
3 years ago
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Robert is a salesperson. He receives 7% commission on each piece of electronic equipment he sells. How much commission will he r
Shalnov [3]

Answer:

He receives $476 in comission.

Step-by-step explanation:

This question can be solved using a rule of three.

$6,800 is 100% = 1. x, which is his commision, is 7% = 0.07. Then

6800 - 1

x - 0.07

x = 6800*0.07

x = 476

He receives $476 in comission.

7 0
3 years ago
Which variable is most important to the following problem?
Ksenya-84 [330]

Answer:

B.) the number of tents the army orders

Step-by-step explanation:

The word problem does not speak of any prices or tent sizes so it cant be A or C

4 0
3 years ago
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Use lagrange multipliers to find the shortest distance, d, from the point (4, 0, −5 to the plane x y z = 1
Varvara68 [4.7K]
I assume there are some plus signs that aren't rendering for some reason, so that the plane should be x+y+z=1.

You're minimizing d(x,y,z)=\sqrt{(x-4)^2+y^2+(z+5)^2} subject to the constraint f(x,y,z)=x+y+z=1. Note that d(x,y,z) and d(x,y,z)^2 attain their extrema at the same values of x,y,z, so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.

The Lagrangian is

L(x,y,z,\lambda)=(x-4)^2+y^2+(z+5)^2+\lambda(x+y+z-1)

Take your partial derivatives and set them equal to 0:

\begin{cases}\dfrac{\partial L}{\partial x}=2(x-4)+\lambda=0\\\\\dfrac{\partial L}{\partial y}=2y+\lambda=0\\\\\dfrac{\partial L}{\partial z}=2(z+5)+\lambda=0\\\\\dfrac{\partial L}{\partial\lambda}=x+y+z-1=0\end{cases}\implies\begin{cases}2x+\lambda=8\\2y+\lambda=0\\2z+\lambda=-10\\x+y+z=1\end{cases}

Adding the first three equations together yields

2x+2y+2z+3\lambda=2(x+y+z)+3\lambda=2+3\lambda=-2\implies \lambda=-\dfrac43

and plugging this into the first three equations, you find a critical point at (x,y,z)=\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right).

The squared distance is then d\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)^2=\dfrac43, which means the shortest distance must be \sqrt{\dfrac43}=\dfrac2{\sqrt3}.
7 0
3 years ago
5x - 4y = 19 <br> x + 2y = 8
klemol [59]

Answer:

(5x-4y)=19

5×19-4×19

95-76

19

Step-by-step explanation:

x+2y=8

x=8-2

x=6

7 0
3 years ago
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