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4 years ago

Help me out some one

Mathematics

1 answer:

Olegator [25]4 years ago

7 0

Y = mx + b is the slope-intercept form of the equation of a line,
where m = slope, and b = y-intercept.

In problems 1 and 3, your equations are written in the y= mx + b form, so you can read the slope and y-intercept directly.

1.
m = -5/2
b = -5

3.
m = -1
b = 3

5.
For problem 5, you need to solve for y to put the equation
in y = mx + b form. Then you can read m and b just like we did
for problems 1 and 3.

4x + 16y = 8

16y = -4x + 8

y = -4/16 x - 8/16

y = -1/4 x - 1/2

m = -1/4

b = -1/2

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Help with these two math questions thank u :)

Tju [1.3M]

Answer:

1.) - (4, 4) 2.) - (3,5) Hope this helps!!

4 0

3 years ago

A kite descends to the ground from a height of 32 feet. Use an integer to describe the descent.

nydimaria [60]

32-x=0; with x equaling the amount descended or -32.

8 0

3 years ago

Y = x^2+ 7x - 5 can be written in the form y = (x + a)^2+b

BARSIC [14]

Answer:

see explanation

Step-by-step explanation:

The equation of a parabola in vertex form is

y = (x - h)² + k (h, k) are the coordinates of the vertex

Given y = x² + 7x - 5

To express in vertex form use the method of completing the square

add/subtract ( half the coefficient of the x- term )²

y = x² + 2( $\frac{7}{2}$ )x + $\frac{49}{4}$ - $\frac{49}{4}$ - 5

y = (x + $\frac{7}{2}$ )² - $\frac{49}{4}$ - $\frac{20}{4}$

y = (x + $\frac{7}{2}$ )² - $\frac{69}{4}$

Hence

a = $\frac{7}{2}$ and b = - $\frac{69}{4}$

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:

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