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Sunny_sXe [5.5K]
2 years ago
7

30 points and mark as brainly

Mathematics
1 answer:
Alex17521 [72]2 years ago
3 0

Answer:

16. B

17. B

18. B

19. B

20. B

I am not sure if they are all correct but I hope I helped you a tiny bit

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Help with mathh pleaseeee
myrzilka [38]

Answer:

4.8h - 2.9d - 16

General Formulas and Concepts:

<u>Algebra I</u>

  • Combining Like Terms

Step-by-step explanation:

<u>Step 1: Define</u>

8h - 7.9d - 16 + 5d - 3.2h

<u>Step 2: Simplify</u>

  1. Combine like terms (h):                    4.8h - 7.9d - 16 + 5d
  2. Combine like terms (d):                    4.8h - 2.9d - 16
7 0
2 years ago
Alysa is solving the question below
Korvikt [17]

Answer:

Step-by-step explanation:

Since y=y for a solution to the system of equations she should just solve

3x-12=-2x+8

4 0
2 years ago
Find the value of the power.
marin [14]
The value of of the power is 5, because it is 3^5.
7 0
2 years ago
Read 2 more answers
Any 10th grader solve it <br>for 50 points​
kkurt [141]

Answer:

\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0  is proved for the sum of pth, qth and rth terms of an arithmetic progression are a, b,and c respectively.

Step-by-step explanation:

Given that the sum of pth, qth and rth terms of an arithmetic progression are a, b and c respectively.

First term of given arithmetic progression is A

and common difference is D

ie., a_{1}=A and common difference=D

The nth term can be written as

a_{n}=A+(n-1)D

pth term of given arithmetic progression is a

a_{p}=A+(p-1)D=a

qth term of given arithmetic progression is b

a_{q}=A+(q-1)D=b and

rth term of given arithmetic progression is c

a_{r}=A+(r-1)D=c

We have to prove that

\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)=0

Now to prove LHS=RHS

Now take LHS

\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)

=\frac{A+(p-1)D}{p}\times (q-r)+\frac{A+(q-1)D}{q}\times (r-p)+\frac{A+(r-1)D}{r}\times (p-q)

=\frac{A+pD-D}{p}\times (q-r)+\frac{A+qD-D}{q}\times (r-p)+\frac{A+rD-D}{r}\times (p-q)

=\frac{Aq+pqD-Dq-Ar-prD+rD}{p}+\frac{Ar+rqD-Dr-Ap-pqD+pD}{q}+\frac{Ap+prD-Dp-Aq-qrD+qD}{r}

=\frac{[Aq+pqD-Dq-Ar-prD+rD]\times qr+[Ar+rqD-Dr-Ap-pqD+pD]\times pr+[Ap+prD-Dp-Aq-qrD+qD]\times pq}{pqr}

=\frac{Arq^{2}+pq^{2} rD-Dq^{2} r-Aqr^{2}-pqr^{2} D+qr^{2} D+Apr^{2}+pr^{2} qD-pDr^{2} -Ap^{2}r-p^{2} rqD+p^{2} rD+Ap^{2} q+p^{2} qrD-Dp^{2} q-Aq^{2} p-q^{2} prD+q^{2}pD}{pqr}

=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2}-pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}

=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2} -pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}

\neq 0

ie., RHS\neq 0

Therefore LHS\neq RHS

ie.,\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0  

Hence proved

5 0
2 years ago
Hi can anyone help me with these 3 questions would really appreciate it thanks
Virty [35]

Answer:

1) 8

Step-by-step explanation:

3+2 is 5

4^2 is 16

16×5=80

2×5=10

80÷10=8

7 0
3 years ago
Read 2 more answers
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