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2 years ago

Pls help (this is for a frend)

Mathematics

1 answer:

son4ous [18]2 years ago

4 0

$~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill & \$5\\ P=\textit{original amount deposited}\\ r=rate\to 1.25\%\to \frac{1.25}{100}\dotfill &0.0125\\ t=years\dotfill &\frac{1}{2} \end{cases} \\\\\\ 5 = (P)(0.0125)(\frac{1}{2})\implies \cfrac{5}{(0.0125)(\frac{1}{2})}=P\implies 800=P$

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Y=-4.3x-1.3<br> Y=1.7x+4.7

Contact [7]

Answer:

0.566666666

Step-by-step explanation:

5 0

3 years ago

Stephen can deliver 14 newspapers in 20 minutes. How many HOURS will it take him to deliver 131 newspaper

LiRa [457]

Answer:

approx. 3 hours

Step-by-step explanation:

14 newspapers in 20 minutes means he can deliver 42 newspapers an hour.

60/42 = 1.4285 newspapers per minute

131 x 1.4285 = 187.1335 minutes = 3.1188 hours

7 0

2 years ago

Determine whether the improper integral converges or diverges, and find the value of each that converges.

Ksju [112]

Answer:converge at $I=\frac{1}{3}$

Step-by-step explanation:

Given

Improper Integral I is given as

$I=\int^{\infty}_{3}\frac{1}{x^2}dx$

integration of $\frac{1}{x^2}$ is - $\frac{1}{x}$

$I=\left [ -\frac{1}{x}\right ]^{\infty}_3$

substituting value

$I=-\left [ \frac{1}{\infty }-\frac{1}{3}\right ]$

$I=-\left [ 0-\frac{1}{3}\right ]$

$I=\frac{1}{3}$

so the value of integral converges at $\frac{1}{3}$

8 0

3 years ago

Write the polynomial f(x)=x^4-10x^3+25x^2-40x+84. In factored form

Verizon [17]

<h2>Steps:</h2>

So firstly, to factor this we need to first find the potential roots of this polynomial. To find it, the equation is $\pm \frac{p}{q}$ , with p = the factors of the constant and q = the factors of the leading coefficient. In this case:

$\textsf{leading coefficient = 1, constant = 84}\\\\p=1,2,3,4,6,7,12,14,21,28,42,84\\q=1\\\\\pm \frac{1,2,3,4,6,7,12,14,21,28,42,84}{1}\\\\\textsf{Potential roots =}\pm 1, \pm 2,\pm 3,\pm 4,\pm 6, \pm 7,\pm 12,\pm 14,\pm 21,\pm 28,\pm 42,\pm 84$

Next, plug in the potential roots into x of the equation until one of them ends with a result of 0:

$f(1)=(1)^4-10(1)^3+25(1)^2-40(1)+84\\f(1)=1-10+25-40+84\\f(1)=60\ \textsf{Not a root}\\\\f(2)=2^4-10(2)^3+25(2)^2-40(2)+84\\f(2)=16-10*8+25*4-80+84\\f(2)=16-80+100-80+84\\f(2)=80\ \textsf{Not a root}\\\\f(3)=3^4-10(3)^3+25(3)^2-40(3)+84\\f(3)=81-10*27+25*9-120+84\\f(3)=81-270+225-120+84\\f(3)=0\ \textsf{Is a root}$

Since we know that 3 is a root, this means that one of the factors is (x - 3). Now that we know one of the roots, we are going to use synthetic division to divide the polynomial. To set it up, place the root of the divisor, in this case 3 from x - 3, on the left side and the coefficients of the original polynomial on the right side as such:

3 | 1 - 10 + 25 - 40 + 84
_________________

Firstly, drop the 1:

3 | 1 - 10 + 25 - 40 + 84
↓
_________________
1

Next, multiply 3 and 1, then add the product with -10:

3 | 1 - 10 + 25 - 40 + 84
↓ + 3
_________________
1 - 7

Next, multiply 3 and -7, then add the product with 25:

3 | 1 - 10 + 25 - 40 + 84
↓ + 3 - 21
_________________
1 - 7 + 4

Next, multiply 3 and 4, then add the product with -40:

3 | 1 - 10 + 25 - 40 + 84
↓ + 3 - 21 + 12
_________________
1 - 7 + 4 - 28

Lastly, multiply -28 and 3, then add the product with 84:

3 | 1 - 10 + 25 - 40 + 84
↓ + 3 - 21 + 12 - 84
_________________
1 - 7 + 4 - 28 + 0

Now our synthetic division is complete. Now since the degree of the original polynomial is 4, this means our quotient has a degree of 3 and follows the format $ax^3+bx^2+cx+d$ . In this case, our quotient is $x^3-7x^2+4x-28$ .

So right now, our equation looks like this:

$f(x)=(x-3)(x^3-7x^2+4x-28)$

However, our second factor can be further simplified. For the second factor, I will be factoring by grouping. So factor x³ - 7x² and 4x - 28 separately. Make sure that they have the same quantity inside the parentheses:

$f(x)=(x-3)(x^2(x-7)+4(x-7))$