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3 years ago

Factor x^2 + 4x - 21

Mathematics

2 answers:

babymother [125]3 years ago

5 0

Answer:

(x - 3)(x + 7)

Step-by-step explanation:

x^2 + 4x - 21

=> x^2 + 7x - 3x - 21

=> x(x + 7) -3(x + 7)

=> (x + 7)(x - 3)

ser-zykov [4K]3 years ago

3 0

Answer:

$(x+7)(x-3)$

Step-by-step explanation:

So we want to factor the expression:

$x^2+4x-21$

When factoring, we want to find two numbers that equals (a)(c) and add up to b. We can then substitute it for b.

So, we want two numbers that when multiplied equals (1)(-21)=-21 and adds up to 4.

We can use 7 and -3.

So:

$x^2+4x-21\\=x^2-3x+7x-21$

Factor out an x from the first two terms, and a 7 from the third and fourth terms:

$=x(x+3)+7(x-3)$

Grouping:

$=(x+7)(x-3)$

And we are done :)

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A town has a population of 4000 and grows at 3.5% every year. To the nearest year, how long will it be until the population will

Romashka [77]

18.273 years will take to reach the population of 7500, if a town has a population of 4000 and grows at 3.5% every year.

Step-by-step explanation:

The given is,

Population of 4000

Rate of grow 3.5% every year

Population after few years 7500

Step:1

Formula to calculate the population with the grow rate every year,

$F=P(1+r)^{t}$ ..........................(1)

Where,

F - Population after t years

P - Initial population

r - Rate of grow

t - No.of years

From given,

F = 7500

P = 4000

r = 3.5%

Equation becomes (1),

$7500=4000(1+0.035)^{t}$

$\frac{7500}{4000} = (1+0.035)^{t}$

1.875 = $(1+0.035)^{t}$

$1.875 = (1.035)^{t}$

Take log on both sides,

$log 1.875 = (t)log 1.035$

Substitute log values,

$0.2730= (t) 0.014940349$

$t = \frac{0.273}{0.014940349}$

= 18.273

t = 18.273 year

t = 18 years 3 months 2 weeks

Result:

18.273 years will take to reach the population of 7500, if a town has a population of 4000 and grows at 3.5% every year.

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3 years ago

A man invests a total of $9,493in two savings accounts. One account yields 9% simple interest and the other 10% simple interest.

Tju [1.3M]

$ 1850 was invested in 9% account

<u>Solution:</u>

Given that

Total amount invested by man in two saving accounts = $9493

Simple interest on one account =9%

Simple interest on second account = 10%

Total interest earned = $930.80

Need to determine amount invested in 9 % account.

Let assume amount invested in account where Simple Interest is 9% = x

And assume amount invested in account where Simple Interest is 10% = y

As total amount invested in two accounts is $9493

=> x + y = 9493

=> y = 9493 - x ------(1)

$\text { Simple Interest }=\frac{\text { Amount Invested } \times \text {rate of interest } \times \text {time}}{100}$

$\begin{array}{l}{\text { Simple interest when rate of interest is } 9 \%=\frac{x \times 9 \times 1}{100}=\frac{9 x}{100}} \\\\ {\text { Simple interest when rate of interest is } 10 \%=\frac{y \times 10 \times 1}{100}=\frac{10 y}{100}}\end{array}$

As total interest earned = $930.80

$\begin{array}{l}{\Rightarrow \frac{9 x}{100}+\frac{10 y}{100}=930.80} \\\\ {\Rightarrow 9 x+10 y=930.80 \times 100} \\\\ {\Rightarrow 9 x+10 y=93080}\end{array}$

On substituting value of y from equation(1) in above equation , we get

9x + 10 (9493 – x) = 93080

=> 9x + 94930 – 10x = 93080

=> -x = 93080 – 94930

=> -x = -1850

=> x = 1850

Amount invested in account where Simple Interest is 9% = x = $1850

Hence $1850 was invested in 9% account.

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Answer:

Step-by-step explanation:

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