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erma4kov [3.2K]
3 years ago
10

(5+5x)-(1-3x) simplify please and thank you

Mathematics
1 answer:
algol [13]3 years ago
5 0

Answer:  8x + 4

Step-by-step explanation:  

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What is the simple interest earned on an investment of $12,625 in mutual funds for 12 years at a rate of 4%
vfiekz [6]

Answer:

Interest earned= $6,060

Step-by-step explanation:

Giving the following formula:

Principal (P)= $12,625

Interest (r)= 4%= 0.04

Period of time (t)= 12 years

<u>To calculate the interest earned after twelve years, we need to use the following formula:</u>

I= P*r*t

I= (12,625*0.04*12)

I= $6,060

Interest earned= $6,060

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2 years ago
Solve for y.<br><br> 7y - 14 = 35
Radda [10]

Answer:

it's 7

Step-by-step explanation:

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8 0
3 years ago
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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

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3 years ago
Consider the expression. What is the result of applying the quotient of powers rule to the expression?
fenix001 [56]

Answer:

Let us consider the expression:

\frac{a^5\cdot b^6}{a^2\cdot b^9}     (1)

Now, the quotient of power rules says that the numbers that have same base can be find by subtracting if powers are with same sign

And adding if powers are with opposite sign.

We will solve equation (1) by this quotient of power rule.

So, it can be rewritten as:

a^{5-2}\cdot b^{6-9}

\Rightarrow a^{3}\cdot b^{-3}.

 

8 0
3 years ago
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The complement of an angle is 25°. What is the measure of the angle?
Varvara68 [4.7K]
65 degrees.

90 - 25 = 65



3 0
3 years ago
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