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BARSIC [14]
3 years ago
13

Work out the percentage change when a price of £80 is increased to £100.

Mathematics
1 answer:
Mars2501 [29]3 years ago
3 0

Answer:

125%

Step-by-step explanation:

80/100=8

80*x=100

100/80=1.25

1.25*100=125

x=125%

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What os the volume of the closet??
Mars2501 [29]

Answer:

82 7/8

Step-by-step explanation:

Use the volume formula

17/4 x 13/4 x 6

1326/16

Simplify

82 7/8

8 0
3 years ago
Select the correct answer.<br> What is the approximate measure of exterior angle B in the polygon?
puteri [66]

Answer:

C

Step-by-step explanation:

the sum of the exterior angles of a polygon = 360°

sum the 4 exterior angles and equate to 360

8x + 9x + 5 + 5x + 4 + 9x + 3 = 360 , that is

31x + 12 = 360 ( subtract 12 from both sides )

31x = 148 ( divide both sides by 31 )

x ≈ 11.23 ( to 2 dec. places )

then exterior angle at B is

9x + 3 = 9(11.23) + 3 ≈ 104° → C

7 0
2 years ago
The cost of a Black Forest cake at two bakeries is as follows: Julie's Bakery: cost $15, delivery charge $0.20 per mile; Bakers
Otrada [13]
15 + 0.2x = 20 + 0.15x
0.05x = 5
x = 100 miles
7 0
3 years ago
The 22 students in student council want to raise $491.60 for a fundraiser. They have already raised $189.32. If each student rai
Novay_Z [31]

Answer: $13.74

Step-by-step explanation:

The students have already raised $189.32 so 491.60-189.32= 302.28 is the remainder the 22 students must raise in order to final an equal amount each individual student must make divide your remainder (302.28) by the number of students (22)

4 0
3 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
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