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3 years ago

What does -2 + -3 =?

Mathematics

2 answers:

Mice21 [21]3 years ago

6 0

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

Tema [17]3 years ago

5 0

It's equals -5 because you add both negatives.

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Name a radius of circle P.

Lemur [1.5K]

A:PN

C is diameter

MN =2*PN

5 0

3 years ago

Can you guys help out I these questions for a test

DaniilM [7]

the first x < 3 so x could be from - 1 till 3 in this graph

the other x > - 7 so x could be from from - 7 till - 3 in this graph

5 0

3 years ago

Determine whether the given signal is a solution to the difference equation. Then find the general solution to the difference eq

Ulleksa [173]

Answer:

The answer to this question can be defined as follows:

Step-by-step explanation:

The given equation is:

$y_{k + 2} + 8y_{k +1} - 9y_{k} = 20k + 12..(1)$

put,

$y_k = k^2\\\\y_{k+2}=(k+2)^2\\\\y_{k+1}=(k+1)^2\\\\$

$(k+2)^2+8(K+1)^2-9k^2 = 20k+12\\\\=20k+12= 20K+12\\\\$

hence y_k=k^2 is its solution.

Now,

$\to y_{k+2}+ 8y_k + 1 - 9y_k = 20k + 12$

the symbol form is:

$(E^2+8E-9)_{yk}=20k+12$

$\to m^2+8m-9=0\\\\\to m^2+(9-1)m-9=0\\\\\to m^2+9m-m-9=0\\\\\to m(m+9)-(m+9)=0\\\\\to (m+9)(m-1)=0\\\\\to m=-9 \ \ \ \ \ \ \ m=1\\$

The general solution is:

$y_k = c_1(-9)^k + c_2(`1)^k\\\\y_k =c_1(-9)^k+c_2$

The complete solution is:

$y_k=(y_k)_c+(y_k)_y\\\\y_k= c_1(-9)^k+c_2+k^2$

The answer is option b: $y_k = k^2 + c_1(-9)^k + c_2$

After solve the complete solution is:

$\bold{y_1=c_1(-9)^k+c_2+k^2.....}$

5 0

3 years ago

A+1=1 find the value of b^a

Ray Of Light [21]

Answer:

Step-by-step explanation:

a+1=1

a=1-1

a=0

b^a which means:

b^0

whenever power is 0 answer is always one

so value of b^a is 1

6 0

3 years ago

Read 2 more answers

Let C(n, k) = the number of k-membered subsets of an n-membered set. Find (a) C(6, k) for k = 0,1,2,...,6 (b) C(7, k) for k = 0,

vladimir1956 [14]

Answer:

(a) $C(6,0) = 1$ , $C(6,1) = 6$ , $C(6,2) = 15$ , $C(6,3) = 20$ , $C(6,4) = 15$ , $C(6,5) = 6$ , $C(6,6) = 1$ .

(b) $C(7,0) = 1$ , $C(7,1) = 7$ , $C(7,2) = 21$ , $C(7,3) = 35$ , $C(7,4) = 35$ , $C(7,5) = 21$ , $C(7,6) = 7$ , $C(7,7)=1$ .

Step-by-step explanation:

In this exercise we only need to recall the formula for C(n,k):

$C(n,k) = \frac{n!}{k!(n-k)!}$

where the symbol $n!$ is the factorial and means

$n! = 1\cdot 2\cdot 3\cdot 4\cdtos (n-1)\cdot n$ .

By convention 0!=1. The most important property of the factorial is $n!=(n-1)!\cdot n$ , for example 3!=1*2*3=6.

(a) The explanations to the solutions is just the calculations.

$C(6,0) = \frac{6!}{0!(6-0)!} = \frac{6!}{6!} = 1$

$C(6,1) = \frac{6!}{1!(6-1)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6$

$C(6,2) = \frac{6!}{2!(6-2)!} = \frac{6!}{2\cdot 4!} = \frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15$

$C(6,3) = \frac{6!}{3!(6-3)!} = \frac{6!}{3!\cdot 3!} = \frac{5!\cdot 6}{6\cdot 6} = \frac{5!}{6} = \frac{120}{6} = 20$

$C(6,4) = \frac{6!}{4!(6-4)!} = \frac{6!}{4!\cdot 2!} = frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15$

$C(6,5) = \frac{6!}{5!(6-5)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6$

$C(6,6) = \frac{6!}{6!(6-6)!} = \frac{6!}{6!} = 1$ .

(b) The explanations to the solutions is just the calculations.

$C(7,0) = \frac{7!}{0!(7-0)!} = \frac{7!}{7!} = 1$

$C(7,1) = \frac{7!}{1!(7-1)!} = \frac{7!}{6!} = \frac{6!\cdot 7}{6!} = 7$

$C(7,2) = \frac{7!}{2!(7-2)!} = \frac{7!}{2\cdot 5!} = \frac{6!\cdot 7}{2\cdot 5!} = \frac{5!\cdot 6\cdot 7}{2\cdot 5!} = \frac{6\cdot 7}{2} = 21$

$C(7,3) = \frac{7!}{3!(7-3)!} = \frac{7!}{3!\cdot 4!} = \frac{6!\cdot 7}{6\cdot 4!} = \frac{5!\cdot 6\cdot 7}{6\cdot 4!} = \frac{120\cdot 7}{24} = 35$

$C(7,4) = \frac{7!}{4!(7-4)!} = \frac{6!\cdot 7}{4!\cdot 3!} = frac{5!\cdot 6\cdot 7}{4!\cdot 6} = \frac{120\cdot 7}{24} = 35$

$C(7,5) = \frac{7!}{5!(7-2)!} = \frac{7!}{5!\cdot 2!} = 21$

$C(7,6) = \frac{7!}{6!(7-6)!} = \frac{7!}{6!} = \frac{6!\cdot 7}{6!} = 7$

$C(7,7) = \frac{7!}{7!(7-7)!} = \frac{7!}{7!} = 1$

For all the calculations just recall that 4! =24 and 5!=120.

6 0

3 years ago