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zzz [600]
3 years ago
14

Pleaseeee help me with thissss pleaseeee

Mathematics
1 answer:
Viefleur [7K]3 years ago
6 0

Answer:

Step-by-step explanation:

56. 6x = 132

      x = 22

57. 2/3 = -8x

     -24x = 2

        x = -2/24= -1/12

58. 5/11x = 55

      5x = 605

       x = 121

59. 4/5 = 10/16x

       4/5 = 5/8x

        23 = 25x

        23/25

60. 3 2/3x = 2/9

      11/9x = 2/9

      11x = 2

      x = 2/11

61. 4 4/5x = 1 1/5

      24/5x = 6/5

      24x = 6

     x = 6/24 = 1/4

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a car travels along the highway to yuma at steady speed. when it begins, it is 400 miles from yuma. After 7 hours, it is 59 mile
klemol [59]

Answer:

y= -50x + 400

Step-by-step explanation:

-50 miles from the place which means they are going further and further away so it is negative. And its 400 because y=mx+b the b is 400 because thats where they started.  Also it said it was correct on A-p-e-x.

4 0
3 years ago
3. The product of 7/10 and another factor
kicyunya [14]
The correct answer is d 7/7
4 0
4 years ago
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
What is the equation of the line that passes through (–2, 3) and is parallel to 2x + 3y = 6?
sergeinik [125]
If they are parallel then the coefficients of x and y will remain the same.  Only the constant will change.

2x + 3y = ?
2(-2) +3(3) = -4 + 9 = 5 so ? is 5

ANSWER: 2x + 3y = 5
6 0
3 years ago
Tabo earns k4200 per month. His budget is k2500 for food, k300 for rent, k150 for clothes and k100 for other things. (a) how muc
Mademuasel [1]

Answer:

a) K3050 and b) K1150

Step-by-step explanation:

a) Add all his expenditures

i.e: K300+K150+K100

=K3050

b) Subtract the remaining of his spendings from the total

i.e: K4209 - K3050

=K1150

3 0
2 years ago
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