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3 years ago

-3(1-2x)=3x+3(x-3)+6

Mathematics

1 answer:

deff fn [24]3 years ago

6 0

Let's solve this problem step-by-step.

−3(1−2x)=3x+3(x−3)+6

Step 1: Simplify both sides of the equation.

−3(1−2x)=3x+3(x−3)+6

(−3)(1)+(−3)(−2x)=3x+(3)(x)+(3)(−3)+6(Distribute)

−3+6x=3x+3x+−9+6

6x−3=(3x+3x)+(−9+6)(Combine Like Terms)

6x−3=6x+−3

6x−3=6x−3

Step 2: Subtract 6x from both sides.

6x−3−6x=6x−3−6x

−3=−3

Step 3: Add 3 to both sides.

−3+3=−3+3

0=0

So, 0=0 or all real numbers.

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3 years ago

The simple interest on a certain sum of money for 2 years at 5% per annum is Rs 320. What will be the compound interest on the s

Brilliant_brown [7]

Answer:

Rs 328

Step-by-step explanation:

Find the principal amount invested.

Simple Interest Formula

I = Prt

where:

I = interest earned
P = principal
r = interest rate (in decimal form)
t = time (in years)

Given:

I = Rs 320
r = 5% = 0.05
t = 2 years

Substitute the given values into the formula and solve for P:

⇒ 320 = P(0.05)(2)

⇒ 320 = P(0.1)

⇒ P = 3200

Compound Interest Formula

$\large \text{$ \sf I=P\left(1+\frac{r}{n}\right)^{nt} -P$}$

where:

I = interest earned
P = principal amount
r = interest rate (in decimal form)
n = number of times interest applied per time period
t = number of time periods elapsed

Given:

P = 3200
r = 5% = 0.05
n = 1 (annually)
t = 2 years

Substitute the given values into the formula and solve for I:

$\implies \sf I=3200\left(1+\frac{0.05}{1}\right)^{2} -3200$

$\implies \sf I=3200\left(1.05\right)^{2} -3200$

$\implies \sf I=3200\left(1.1025\right) -3200$

$\implies \sf I=3528-3200$

$\implies \sf I=328$

Therefore, the compound interest on the same sum for the same time at the same rate is Rs 328.

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2 years ago