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3 years ago

Solve each equation by factoring, and state the solutions.

Mathematics

1 answer:

Elenna [48]3 years ago

7 0

Answer:

(a) $x^2+25=(x-5)^2+2\times x\times 5=(x-5)^2+10x$

(b) x = -5, -5

Step-by-step explanation:

We have given that the equation

(a) $x^2+25=0$

We have to factorize the equation

We know that $(a-b)^2=a^2+b^2-2ab$

So $(a-b)^2+2ab=a^2+b^2$

So $x^2+25=(x-5)^2+2\times x\times 5=(x-5)^2+10x$

(B) We have given equation $x^2+10x+25$

It can be factorize as $x^2+10x+25=x^2+5x+5x+25=x(x+5)+5(x+5)$

Now taking x+5 as common

$(x+5)(x+5)$ = 0

x= -5 , -5

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4 years ago

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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:

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

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

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

$\implies \sf I=3528-3200$

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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

$13,957 is invested, part at 7% and the rest at 6%. If the interest earned from the amount invested at 7% exceeds the interest e

ch4aika [34]

Answer:

The Amount invested at 7% interest is $12,855

The Amount invested at 6% interest = $1,102

Step-by-step explanation:

Given as :

The Total money invested = $13,957

Let The money invested at 7% = $p_1$ = $A

And The money invested at 6% = $p_2$ = $13957 - $A

Let The interest earn at 7% = $I_1$

And The interest earn at 6% = $I_2$

$I_1$ - $I_2$ = $833.73

Let The time period = 1 year

Now, From Simple Interest method

Simple Interest = $\dfrac{\textrm principal\times \textrm rate\times \textrm time}{100}$

Or, $I_1$ = $\dfrac{\textrm p_1\times \textrm 7\times \textrm 1}{100}$

Or, $I_1$ = $\dfrac{\textrm A\times \textrm 7\times \textrm 1}{100}$

And

$I_2$ = $\dfrac{\textrm p_2\times \textrm 6\times \textrm 1}{100}$

Or, $I_2$ = $\dfrac{\textrm (13,957 - A)\times \textrm 6\times \textrm 1}{100}$

∵ $I_1$ - $I_2$ = $833.73

So, $\dfrac{\textrm A\times \textrm 7\times \textrm 1}{100}$ - $\dfrac{\textrm (13,957 - A)\times \textrm 6\times \textrm 1}{100}$ = $833.73

Or, 7 A - 6 (13,957 - A) = $833.73 × 100

Or, 7 A - $83,742 + 6 A = $83373

Or, 13 A = $83373 + $83742

Or, 13 A = $167,115

∴ A = $\dfrac{167115}{13}$

i.e A = $12,855

So, The Amount invested at 7% interest = A = $12,855

And The Amount invested at 6% interest = ($13,957 - A) = $13,957 - $12,855

I.e The Amount invested at 6% interest = $1,102

Hence,The Amount invested at 7% interest is $12,855

And The Amount invested at 6% interest = $1,102 . Answer

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3 years ago

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