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PilotLPTM [1.2K]
3 years ago
15

Which one of them is right ?

Mathematics
1 answer:
Alla [95]3 years ago
5 0
They are both correct, therefore the answer to this problem is answer choice C. Both Ivy and Andrey.
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Macs farm has a total of 18 horses and 27 cows. what is the ratio of horses to cows in simplest form?
Evgesh-ka [11]
The ratio is basically finding the quotient of two numbers. If you want to find the ratio of horses to cows, divide the number of horses to that of cows.

Ratio = 18/27

When expressing it to its simplest form, make sure it is reduced. 18 and 27 are both factors of 9. So,

Ratio = 9*2/9*3
Cancelling out 9, the ratio at its simplest form is: <em>2/3</em>
7 0
3 years ago
Franklin rolls a pair of six-sided fair dice with sides numbered 1 through 6.
Zarrin [17]

Answer:

The answer is 7/36.

Step-by-step explanation:

First, you find out how many possible outcomes there are from rolling a pair of dice. On one cube, you can roll a 1,2,3,4,5, or 6; so there are 6 outcomes. Since there are two cubes, you multiply 6 by itself to get a total of 36 possible outcomes. Next, you find the probability of the sum of the numbers rolled being an even number; the possibilities are 2,4,6,8,10, or 12, which is 6/36. The probability of rolling a multiple of 5; the one possibility is just 5, since we already accounted for rolling a 10 as an even number. So that is 1/36. The word <u>or</u> says that we add the two probabilities, so the final answer is 6/36+1/36=7/36.

6 0
3 years ago
Use spherical coordinates. Find the volume of the solid that lies within the sphere x^2 + y^2 + z^2 = 81, above the xy-plane, an
Natasha_Volkova [10]

Answer:

The volume of the solid is 243\sqrt{2} \ \pi

Step-by-step explanation:

From the information given:

BY applying sphere coordinates:

0 ≤ x² + y² + z² ≤ 81

0  ≤ ρ²   ≤   81

0  ≤ ρ   ≤  9

The intersection that takes place in the sphere and the cone is:

x^2 +y^2 ( \sqrt{x^2 +y^2 })^2  = 81

2(x^2 + y^2) =81

x^2 +y^2 = \dfrac{81}{2}

Thus; the region bounded is: 0 ≤ θ ≤ 2π

This implies that:

z = \sqrt{x^2+y^2}

ρcosФ = ρsinФ

tanФ = 1

Ф = π/4

Similarly; in the X-Y plane;

z = 0

ρcosФ = 0

cosФ = 0

Ф = π/2

So here; \dfrac{\pi}{4} \leq \phi \le \dfrac{\pi}{2}

Thus, volume: V  = \iiint_E \ d V = \int \limits^{\pi/2}_{\pi/4}  \int \limits ^{2\pi}_{0} \int \limits^9_0 \rho   ^2 \ sin \phi \ d\rho \   d \theta \  d \phi

V  = \int \limits^{\pi/2}_{\pi/4} \ sin \phi  \ d \phi  \int \limits ^{2\pi}_{0} d \theta \int \limits^9_0 \rho   ^2 d\rho

V = \bigg [-cos \phi  \bigg]^{\pi/2}_{\pi/4}  \bigg [\theta  \bigg]^{2 \pi}_{0} \bigg [\dfrac{\rho^3}{3}  \bigg ]^{9}_{0}

V = [ -0+ \dfrac{1}{\sqrt{2}}][2 \pi -0] [\dfrac{9^3}{3}- 0 ]

V = 243\sqrt{2} \ \pi

4 0
3 years ago
2 – (–8) + (–3) = ?
vodka [1.7K]
A minus before a parenthesis means we need to change the sign.

So 2 + 8 - 3 = 7
5 0
3 years ago
Use the method of reduction of order to find a second independent solution of the given differential equation. t2y'' + 3ty' + y
Charra [1.4K]

I assume you mean y_1=t^{-1}, and not y_1=t-1, since this doesn't satisfy the ODE.

Assume a second solution of the form y_2=vy_1, where v is a function of t. Then

{y_2}'=v'y_1+v{y_1}'

{y_2}''=v''y_1+2v'{y_1}'+v{y_1}''

Substituting into the ODE gives

t^2\left(\dfrac{v''}t-\dfrac{2v'}{t^2}+\dfrac{2v}{t^3}\right)+3t\left(\dfrac{v'}t-\dfrac v{t^2}\right)+\dfrac vt=0

\implies tv''+v'=0

\implies(tv')'=0

\implies tv'=C

\implies v'=\dfrac Ct

\implies v=C_1\ln|t|+C_2

\implies y_2=\dfrac{\ln t}t

where we omit the second term because it's already accounted for by y_1.

3 0
3 years ago
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