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In-s [12.5K]
3 years ago
8

Which method(s) can be used to solve a system of equations? Select all that apply. distributing graphing substitution formulas a

ddition modeling
Mathematics
1 answer:
katovenus [111]3 years ago
6 0
Distributing, formulas, and addition
You might be interested in
how many different four letter permutations can be formed using four letters out of the first 12 in the alphabet?
Verizon [17]

Answer:

11,800 different four letter permutations can be formed using four letters out of the first 12 in the alphabet

Step-by-step explanation:

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

P_{(n,x)} = \frac{n!}{(n-x)!}

In this question:

Permutations of four letters from a set of 12 letters. So

P_{(12,4)} = \frac{12!}{(12-4)!} = 11800

11,800 different four letter permutations can be formed using four letters out of the first 12 in the alphabet

3 0
2 years ago
Can somebody please help i need the help to help me pass my math class thx so much if you can
natulia [17]

Answer:

Step-by-step explanation:

1. x = 11.2

2. 13

3. 9

4. -7/15

5. 15b

6. -12x + 16

7. 4x + 4

8. 11x-10

9. 10a + 5

10. -x, 15, and 2b

Hope that helps, and good luck!

4 0
3 years ago
Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview
svetlana [45]

Answer:  The required solution is

y=(-2+t)e^{-5t}.

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

y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.

So, the general solution of the given equation is

y(t)=(A+Bt)e^{-5t}.

Differentiating with respect to t, we get

y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.

According to the given conditions, we have

y(0)=-2\\\\\Rightarrow A=-2

and

y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.

Thus, the required solution is

y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.

6 0
3 years ago
A card is selected at random from a deck of 52 playing cards, the probability that a face card selected is ______________
ratelena [41]

Answer:

A

Step-by-step explanation:

There are 12 face cards in a deck out of 52, so 12/52 would be the probability

hope this helps :)

7 0
3 years ago
I WILL MARK BRAINLIEST FOR THIS!!!!!!!!!!!
igomit [66]

Answer:

She gets a job as a lawyer after graduation.

Step-by-step explanation:

6 0
2 years ago
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