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FromTheMoon [43]
3 years ago
8

The diagram shows the solid glass case for a clock. The case is a cuboid with a cylinder

Mathematics
1 answer:
Karolina [17]3 years ago
5 0

Answer:

241.5cm³

Step-by-step explanation:

Volume of the glass = Volume of the cuboid - Volume of the cylinder

Volume of the cuboid = Length * width * height

Volume of the cuboid  =  8cm * 4cm * 10cm

Volume of the cuboid  = 320cm³

Volume of the cylinder = πr²h

Radius r = 5/2 = 2.5cm

height = 4cm

Volume of the cylinder = 3.14(2.5)²*4

Volume of the cylinder = 78.5cm^3

Volume of the glass = 320 - 78.5

Volume of the glass = 241.5cm³

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5x-2=3x+40 answer plz :')
noname [10]

X=21

5x(-3x) -2= 3x(-3x) +40

2x-2(+2)=40(+2)

2x=42

X=21

7 0
3 years ago
What is the pattern? 128; 640; 3,200
nikklg [1K]

Answer:

128×5=640

640×5=3200

Multiply by 5

6 0
4 years ago
Solve. 21 + 3x = 3(-x + 3) - 12<br>x= ​
mars1129 [50]
X = -4.
explanation:
21+3x=3(-x+3)-12
21+3x=-3x+9-12
21+6x=9-12
21+6x=-3
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4 0
3 years ago
Circle A has a radius of 9 centimeters and Circle B has a diameter of 19 centimeters. What is the relationship between the circu
IRISSAK [1]

Answer:

18:19

Step-by-step explanation:

Given that,

The radius of circle A = 9 cm

The diameter of circle B = 19 cm

We know that,

Radius = diameter/2

The radius of circle B = 19/2 cm

The circumference of circle A is :

C_A=2\pi r_A ....(1)

And that of circle B is :

C_B=2\pi r_B ....(2)

Dividing equation (1) and (2) :

\dfrac{C_A}{C_B}=\dfrac{2\pi r_A}{2\pi r_B}\\\\=\dfrac{r_A}{r_B}\\\\=\dfrac{9}{\dfrac{19}{2}}\\\\\dfrac{C_A}{C_B}=\dfrac{18}{19}

Hence, the ratio of circumferences of Circle A and Circle B is 18:19.

7 0
3 years ago
In this problem, x = c1 cos t + c2 sin t is a two-parameter family of solutions of the second-order de x'' + x = 0. find a solut
nordsb [41]

If you are already given the general solution, finding the particular solution means to impose some conditions in order to fix the coefficients of the general solution.

The first thing we need to impose is that the function must output -1 when the input is zero. So, if we plug zero as input we have

x(t) = c_1 \cos(t) + c_2 \sin(t) \implies x(0) = c_1 \cos(0) + c_2 \sin(0) = c_1

But we want x(0)=-1, which implies x(0)=c_1=-1

So, we fixed the first coefficient. To fix the second one, we can use the second piece of information (and of course the already-found value for c_1).

We want the first derivative to output 3 when the input is zero. So, first of all, let's compute the first derivative, and evaluate it in zero:

x(t) = -\cos(t) + c_2 \sin(t) \implies x'(t) = \sin(t) + c_2\cos(t)

x'(0) = \sin(0) + c_2\cos(0) = c_2

And as before, since x'(0)=c_2 and we want x'(0)=3, we deduce c_2=3.

So, once we fix both coefficients, the general solution becomes

x(t) = c_1 \cos(t) + c_2 \sin(t) \mapsto -\cos(t)+3\sin(t)

3 0
3 years ago
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