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2 years ago

Put the following equation of a line into slope-intercept form, simplifying all fractions.

Mathematics

2 answers:

natita [175]2 years ago

5 0

Answer: y=3x+3

Step-by-step explanation:

12x -4y= -12

subtract 12x from both sides

-4y= -12x-12

Divide everything by -4 to make y positive and alone

y=3x+3

kicyunya [14]2 years ago

4 0

Answer:

3x - y = -3

Step-by-step explanation:

divide everything by 4, the greatest common factor. That's all you need to do.

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Mary earns $25 per hour at her job. Next month, she will receive a 10% raise. Complete the two expressions that represent how mu

Natali5045456 [20]

25 + _0.10__*25

25 + __2.5___

In the 1st blank, the answer is 0.1 or 0.10 because you have to change percents to decimals to find the raise. 10% as a decimal = 0.1 or 0.10

In the 2nd blank, the answer is 2.5 because on the top, it's 0.1*25, which equals 2.5, and 2.5 is the raise. All you have to do is multiply 10%, or 0.1, by 25, and you get 2.5.

Hope this helped☺☺

6 0

3 years ago

Stephanie surveyed the students in her class to find out their favorite color. her results are in the table. what percent of Ste

wariber [46]

Hi there!

To solve this problem, we need to find the amount of classmates out of the entire class who did not choose blue as their favorite color and then convert it into a percentage.

First, let's find the fraction:

Since there are 25 total people in her class, since

6 + 9 + 10 = 25

25 is the denominator of the fraction;

Since there are 16 people who did not choose blue,

6 + 10 = 16

16 is the numerator of the fraction.

So, our fraction is 16/25.

Now, we need to convert it into a percentage. To do this, we need to make the denominator 100 by multiplying it by a certain number, and then multiplying the numerator by that same number so the fraction stays equal:

16/25
= 16*4/25*4
= 64/100
= 64%

So, the answer is 64% of Stephanie's classmates.

Hope this helps!

3 0

2 years ago

Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a

8_murik_8 [283]

Answer:

a) 3/5·((-2)^n + 4·3^n)
b) 3·2^n - 5^n
c) 3·2^n + 4^n
d) 4 - 3 n
e) 2 + 3·(-1)^n
f) (-3)^n·(3 - 2n)
g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form $a[n]=r^n$ , then it will satisfy ...

$r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}$

Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

$r^2-c_1r-c_2=0$

The quadratic formula tells us values of r that satisfy this are ...

$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

$a[n]=pr_1^n+qr_2^n$

We can find p and q by solving the initial condition equations:

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

$p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}$

_____

Using these formulas on the first recurrence relation, we get ...

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

$a[n]=(p+qn)r^n$