Help me determine if relations 1, 2, 3, and 4 are functions or not please!

What is the volume if this rectangular prism: 5/2 cm by 3/4 cm by 1/3 cm

aleksley [76]

Answer:

5/8 cm

Step-by-step explanation:

5/2*3/4=15/8

15/8*1/3=15/24=5/8

7 0

3 years ago

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There are 24 people at a party and each person eats 1/8 of a cake. There was not any cake leftover. How many cakes were at the p

sasho [114]

Answer: 3 cakes were at the party

Step-by-step explanation:

Multiply 1/8*24 to find the exact number of cakes that were at the party

Hope this helps :)

5 0

2 years ago

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Find the squares of (y+3)

emmasim [6.3K]

Answer:

Since we know that

(a+b)

2

=a

2

+2ab+b

2

(a−b)

2

=a

2

−2ab+b

2

Now,

(i)

(x+3y)

2

=x

2

+2×x×3y+(3y)

2

=x

2

+6xy+9y

2

(ii)

(2x−5y)

2

=(2x)

2

−2×2x×5y+(5y)

2

=4x

2

−20xy+25y

2

(iii)

(a+

5a

1

)

2

=a

2

+2×a×

2a

1

+(

2a

1

)

2

=a

2

+1+

4a

2

1

(iv)

(2a−1a)=(1a)

2

(1a)

2

=1a

2

Answer

8 0

2 years ago

X is less than or equal to -5

erma4kov [3.2K]

Answer:

x<=-5

Step-by-step explanation:

6 0

3 years ago

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Suppose that a spherical droplet of liquid evaporates at a rate that is proportional to its surface area: where V = volume (mm3

Alex

Answer:

V = 20.2969 mm^3 @ t = 10

r = 1.692 mm @ t = 10

Step-by-step explanation:

The solution to the first order ordinary differential equation:

$\frac{dV}{dt} = -kA$

Using Euler's method

$\frac{dVi}{dt} = -k *4pi*r^2_{i} = -k *4pi*(\frac {3 V_{i} }{4pi})^(2/3)\\ V_{i+1} = V'_{i} *h + V_{i} \\$

Where initial droplet volume is:

$V(0) = \frac{4pi}{3} * r(0)^3 = \frac{4pi}{3} * 2.5^3 = 65.45 mm^3$

Hence, the iterative solution will be as next:

i = 1, ti = 0, Vi = 65.45

$V'_{i} = -k *4pi*(\frac{3*65.45}{4pi})^(2/3) = -6.283\\V_{i+1} = 65.45-6.283*0.25 = 63.88$

i = 2, ti = 0.5, Vi = 63.88

$V'_{i} = -k *4pi*(\frac{3*63.88}{4pi})^(2/3) = -6.182\\V_{i+1} = 63.88-6.182*0.25 = 62.33$

i = 3, ti = 1, Vi = 62.33

$V'_{i} = -k *4pi*(\frac{3*62.33}{4pi})^(2/3) = -6.082\\V_{i+1} = 62.33-6.082*0.25 = 60.813$

We compute the next iterations in MATLAB (see attachment)

Volume @ t = 10 is = 20.2969

The droplet radius at t=10 mins

$r(10) = (\frac{3*20.2969}{4pi})^(2/3) = 1.692 mm\\$

The average change of droplet radius with time is:

Δr/Δt = $\frac{r(10) - r(0)}{10-0} = \frac{1.692 - 2.5}{10} = -0.0808 mm/min$

The value of the evaporation rate is close the value of k = 0.08 mm/min

Hence, the results are accurate and consistent!

5 0

3 years ago