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iragen [17]
2 years ago
5

Anne needs 2 kinds of fruits to prepare a fruit salad. She has 6 kinds of fruits to choose from. In how many ways can she choose

the fruits? Identify if the situation involves combinations or permutations.
Mathematics
2 answers:
Y_Kistochka [10]2 years ago
8 0
There are 15 ways she can choose the fruits.  Since order is not important, this is a combination.

The combination is given by
_6C_2=\frac{6!}{2!4!}=\frac{720}{48}=15
baherus [9]2 years ago
8 0

The order doesn't matter. It's the combination then.

k objects can be chosen out of n objects, when the order doesn't matter, in C(n,k)=\dfrac{n!}{k!(n-k)!} ways.

So, the answer is C(6,2)=\dfrac{6!}{2!4!}=\dfrac{5\cdot6}{2}=15 ways.

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Consider the initial value problem: 2ty′=8y, y(−1)=1. Find the value of the constant C and the exponent r so that y=Ctr is the s
VikaD [51]

The correct question is:

Consider the initial value problem

2ty' = 8y, y(-1) = 1

(a) Find the value of the constant C and the exponent r such that y = Ct^r is the solution of this initial value problem.

b) Determine the largest interval of the form a < t < b on which the existence and uniqueness theorem for first order linear differential equations guarantees the existence of a unique solution.

c) What is the actual interval of existence for the solution obtained in part (a) ?

Step-by-step explanation:

Given the differential equation

2ty' = 8y

a) We need to find the value of the constant C and r, such that y = Ct^r is a solution to the differential equation together with the initial condition y(-1) = 1.

Since Ct^r is a solution to the initial value problem, it means that y = Ct^r satisfies the said problem. That is

2tdy/dt - 8y = 0

Implies

2td(Ct^r)/dt - 8(Ct^r) = 0

2tCrt^(r - 1) - 8Ct^r = 0

2Crt^r - 8Ct^r = 0

(2r - 8)Ct^r = 0

But Ct^r ≠ 0

=> 2r - 8 = 0 or r = 8/2 = 4

Now, we have r = 4, which implies that

y = Ct^4

Applying the initial condition y(-1) = 1, we put y = 1 when t = -1

1 = C(-1)^4

C = 1

So, y = t^4

b) Let y = F(x,y)................(1)

Suppose F(x, y) is continuous on some region, R = {(x, y) : x_0 − δ < x < x_0 + δ, y_0 −ę < y < y_0 + ę} containing the point (x_0, y_0). Then there exists a number δ1 (possibly smaller than δ) so that a solution y = f(x) to (1) is defined for x_0 − δ1 < x < x_0 + δ1.

Now, suppose that both F(x, y)

and ∂F/∂y are continuous functions defined on a region R. Then there exists a number δ2

(possibly smaller than δ1) so that the solution y = f(x) to (1) is

the unique solution to (1) for x_0 − δ2 < x < x_0 + δ2.

c) Firstly, we write the differential equation 2ty' = 8y in standard form as

y' - (4/t)y = 0

0 is always continuous, but -4/t has discontinuity at t = 0

So, the solution to differential equation exists everywhere, apart from t = 0.

The interval is (-infinity, 0) n (0, infinity)

n - means intersection.

7 0
1 year ago
Which table describes the behavior of the graph of f(x) = 2x3 – 26x – 24?
In-s [12.5K]

Answer:

For x∈ {-∞,-3} y<0, below x-axis

x∈ {-3,-1} y>0, above x-axis

x∈ {-1,4} y<0, below x-axis

x∈ {4,∞} y>0, above x-axis

Step-by-step explanation:

f(x)=2x^{3}-26x-24

2x^{3}-26x-24=0

2x^{3}-26x-24=0\\2x^{3}-2x-24x-24=0\\2x(x^{2} -1)-24(x+1)=0\\2x(x+1)(x-1)-24(x+1)=0\\(x+1)(2x^{2} -2x-24)=0\\=> x+1=0 => x_{1}=-1\\

=> 2x^{2} -2x-24=0\\=>2(x^{2} -x-12)=0\\=> x^{2} -x-12=0\\=> x^{2} -4x+3x-12=0\\=> x(x-4)+3(x-4)=0\\=> (x-4)(x+3)=0\\

=> x-4=0\\=> x_{2}=4\\=> x+3=0\\=>x_{3}=-3\\

For x∈ {-∞,-3} y<0, below x-axis

x∈ {-3,-1} y>0, above x-axis

x∈ {-1,4} y<0, below x-axis

x∈ {4,∞} y>0, above x-axis



7 0
2 years ago
Read 2 more answers
Jamal works in a city and sometimes takes a taxi to work. The taxicabs charge $1.50 for the first 1/5 mile and $0.25 for each ad
Natalija [7]
1.50+.25x=3.75
.25x=2.25
x=9
9/5+1/5=10/5
2 miles
4 0
2 years ago
Read 2 more answers
Sue played four games of golf for these games her modal score was 98 and her mean score was 100 her range of score was 10 what w
earnstyle [38]

Answer : Remaining two observation becomes 97 and 107.

Explanation :

Since we have given that

Mean = 100

Modal value = 98

Range = 10

As we know that ,

Range = Highest-Lowest

Let highest observation be x

Let lowest observation be y

So equation becomes x-y=10 ----equation 1

So, observation becomes

x,98,98,y

Now, we use the formula of mean i.e.

Mean = \frac{\text{Sum of observation}}{\text{N.of observaton}}

So, mean =\frac{x=98=98+y}{4}=400\\\frac{196+x+y}{4}=100\\x+y=400-196\\x-y=204

So our 2nd equation becomes

x+y=204

On using elimination method of system of linear equation on these two equation we get,

x=97

and

x+y=204\\y=204-x\\y=204-97\\y=107

Hence , remaining two observation becomes 97 and 107.

8 0
2 years ago
To find the area of parallelogram RSTU, Juan starts by drawing a rectangle around it. Each vertex of parallelogram RSTU is on a
shusha [124]

Answer:

C. (18+4)

Step-by-step explanation:

The answer on ed

6 0
2 years ago
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