All categories

2 years ago

Someone help plz plz plz plz

Mathematics

2 answers:

Troyanec [42]2 years ago

6 0

Answer:

slope is -55 balance decreases by $55 each month

Step-by-step explanation:

solve for slope using slope formula:

(620-895)/(6-1) = -275/5 = -55

decreases by $55 since that's the slope

julia-pushkina [17]2 years ago

4 0

Hi! i know someone has already answered but i solved this on paper, i got the same answer! the slope should be -55 and the balance goes down 55 per month since the slope of the line is negative!

hope this helps and sorry if i was too late!

You might be interested in

What is p(x)???????????????

lys-0071 [83]

Bxbx apls ops sna e and

7 0

3 years ago

A pyramid has a regular hexagonal base with side lengths of 4 and a slant height of 6. Find the lateral area of the pyramid.

aleksandr82 [10.1K]

It should be 72 in^2.

7 0

4 years ago

Read 2 more answers

The diagonals of quadrilateral ABCD intersect at point E, AE=5x-10,EC=2x+5, and BE= y/3.. If ABCD is a square, find the value of

enot [183]

Check the picture below.

5 0

3 years ago

Read 2 more answers

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:

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