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3 years ago

Find the perimeter of the figure.

Mathematics

1 answer:

iren2701 [21]3 years ago

5 0

Answer:

below

Step-by-step explanation:

p = 2( a + b)

p = 2(24 +16)

p =80 in

p semicircle

=πr

= 3.142 *8

= 25.136

p of figure

p =80 +25.136

p=105.136 in

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Divide using long division. -15x3 + 37x2 - 8x - 6/3x-5

Mazyrski [523]

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3 years ago

PLSSSS HELPPP find the surface area and volume of the prism ILL GIVE BRAINLIESTT

Lady bird [3.3K]

Answer: 94 and 60, 96 and 42

Step-by-step explanation:

1. For surface area, you need to find the area the sides and base.

Base: 3*4= 12

Side 1: 4*5= 20

Side 2: 3*5= 15

Multiply each by 2.

12*2= 24

20*2= 40

15*2= 30

Add them up.

30+40+24= 94 is the surface area.

For the volume, you multiply each given number.

3*4*5

length * width * height = volume

60 is the volume.

2. For surface area, you need to find the area the sides and base. This is a triangular prisim so the formula is differrent.

Base: 3*7= 21

Side 1: (area of a triangle is 1/2bh) 3*4= 12 / 2= 6

Side 2: (area of a triangle is 1/2bh) 3*4= 12/2= 6

Side 3 (the back): 4*7= 28

Side 4 (the shaded): 5*7= 35

Add them up. This isn't a rectangular prisim so we aren't multiplying the found areas again, in the triangula rpriism we already found ALL areas.

21+6+6+28+35

= 96 is the surface area.

For the volume, you multiply each given number. You follow for triangles,

1/2 BAH

7*4*3=84

84/2

= 42 is the volume.

8 0

2 years ago

Read 2 more answers

Solve differential equation: y'''+4y''-16y'-64y=0 y(0)=0, y'(0)=26, y''(0)=-16

Ipatiy [6.2K]

Answer: The required solution of the given differential equation is

$y(x)=3e^{4x}-3e^{-4x}+2xe^{-4x}.$

Step-by-step explanation: We are given to solve the following differential equation :

$y^{\prime\prime\prime}+4y^{\prime\prime}-16y^\prime-64y=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\\y(0)=0,~y^\prime(0)=26,~y^{\prim\prime}(0)=-16.$

Let, $y=e^{mx}$ be an auxiliary solution of equation (i).

Then, $y^\prime=me^{mx},~~y^{\prime\prime}=m^2e^{mx},~~y^{\prime\prime\prime}=m^3e^{mx}.$

Substituting these values in equation (i), we get

$m^3e^{mx}+4m^2e^{mx}-16me^{mx}-64e^{mx}=0\\\\\Rightarrow (m^3+4m^2-16m-64)e^{mx}=0\\\\\Rightarrow m^3+4m^2-16m-64=0,~~~~~~~~~[\textup{since }e^{mx}\neq 0]\\\\\Rightarrow m^2(m-4)+8m(m-4)+16(m-4)=0\\\\\Rightarrow (m-4)(m^2+8m+16)=0\\\\\Rightarrow (m-4)(m+4)^2=0\\\\\Rightarrow m-4=0,~~(m+4)^2=0\\\\\Rightarrow m=4,~m=-4,~-4.$

So, the general solution is given by

$y(x)=Ae^{4x}+Be^{-4x}+Cxe^{-4x}.$

Then, we have

$y^\prime=4Ae^{4x}-4Be^{-4x}-4Cxe^{-4x}+Ce^{-4x},\\\\y^{\prime\prime}=16Ae^{4x}+16Be^{4x}+16Cxe^{-4x}-4Ce^{-4x}-4Ce^{-4x}\\\\\Rightarrow y^{\prime\prime}=16Ae^{4x}+16Be^{4x}+16Cxe^{-4x}-8Ce^{-4x}.$

With the conditions given, we get

$y(0)=A+B+C\times 0\\\\\Rightarrow A+B=0\\\\\Rightarrow A=-B~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

$y^\prime(0)=4A-4B+C\\\\\Rightarrow 4A-4B+C=26\\\\\Rightarrow 4(A+A)+C=26~~~~~~~~~~~~~~~~[\textup{using equation (i)}]\\\\\Rightarrow C=26-8A~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)$

and

$y^{\prime\prime}(0)=16A+16B-8C\\\\\Rightarrow 16A-16A-8C=-16~~~~~~~~~~~~[\textup{using equation (ii)}]\\\\\Rightarrow -8C=-16\\\\\Rightarrow C=2.$

From equation (iii), we get

$C=26-8A\\\\\Rightarrow 2=26-8A\\\\\Rightarrow 8A=24\\\\\Rightarrow A=3.$

From equation (ii), we get

$B=-3.$

Therefore, the required solution of the given differential equation is

$y(x)=3e^{4x}-3e^{-4x}+2xe^{-4x}.$

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3 years ago

Mr Leonard wants to top-up his central heating oil tank. The oil tank can take up to 1200 litres of oil. There are already 450 l

garik1379 [7]

Mr. Leonard gets £ 45.84375 as discount

Solution:

The oil tank can take up to 1200 liters of oil

There are already 450 liters of oil in the tank

The remaining oil which is to be added is given by :

Remaining oil = 1200 - 450 = 750 liters

The price of oil is 81.5 p per liter

Then calculate the total price of 750 L of oil:

$Total\ Price = 750 \times 81.5 = 61125$

Thus total price is 61125 p

Mr Leonard gets a 7.5% discount on the price of the oil

Therefore,

Discount amount = 7.5 % of 61125

$Discount\ amount = 7.5 \% \times 61125\\\\Discount\ amount = \frac{7.5}{100} \times 61125\\\\Discount\ amount = 0.075 \times 61125\\\\Discount\ amount = 4584.375$