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2 years ago

Write an equation in slope-intercept form that represents the graphed function.

Mathematics

1 answer:

damaskus [11]2 years ago

8 0

Answer: Y = 1/4x - 2
Slope intercept form is y = mx + b
M is slope, and b is y intercept

Solving slope:
change in y over change in x is slope
The change in y from (0,-2) and (4,-1) is up one over four.
This can be represented as 1/4

Solving y-intercept:
Y-intercept is when x = 0 or when the line touches the y axis
As you can see on the graph, the y intercept is (0,-2) or -2

Final:
Y = 1/4x - 2

You might be interested in

1st question: Reuben received 80 dollars. Write a signed number to represent this change.

lesya692 [45]

Answer:

1) +80

2) -1520

Step-by-step explanation:

Positive Signed numbers are used in real world situations to indicate the increase. Examples include; increase in weight, deposit of money into an account, increase in income, profit, height above sea level, etc. They are all represented by positive signed numbers.

Whereas, negative signed numbers show decrease. Examples include; loss of weight, withdrawal of money from an account, loans, liabilities, losses, temperature below zero degrees, height below sea level.

1) We are told that Reuben received 80 dollars. Thus, it means he received an increase in the amount of money he already had. Thus, the signed number will be +80.

2) We are told that A car corporation produced 1520 fewer cars this month than last.

Thus, the signed number will be negative since there is a decrease in number of cars.

Thus, signed number = -1520

7 0

3 years ago

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

3 years ago

Question 2 of 5

AlekseyPX

Given:

The different recursive formulae.

To find:

The explicit formulae for the given recursive formulae.

Solution:

The recursive formula of an arithmetic sequence is $f(n)=f(n-1)+d, f(1)=a,n\geq 2$ and the explicit formula is $f(n)=a+(n-1)d$ , where a is the first term and d is the common difference.

The recursive formula of a geometric sequence is $f(n)=rf(n-1), f(1)=a,n\geq 2$ and the explicit formula is $f(n)=ar^{n-1}$ , where a is the first term and r is the common ratio.

The first recursive formula is:

$f(1)=5$

$f(n)=f(n-1)+5$ for $n\geq 2$ .

It is the recursive formula of an arithmetic sequence with first term 5 and common difference 5. So, the explicit formula for this recursive formula is:

$f(n)=5+(n-1)(5)$

$f(n)=5+5(n-1)$

Therefore, the correct option is A, i.e., $f(n)=5+5(n-1)$ .

The second recursive formula is:

$f(1)=5$

$f(n)=3f(n-1)$ for $n\geq 2$ .

It is the recursive formula of a geometric sequence with first term 5 and common ratio 3. So, the explicit formula for this recursive formula is:

$f(n)=5(3)^{n-1}$

Therefore, the correct option is F, i.e., $f(n)=5(3)^{n-1}$ .

The third recursive formula is:

$f(1)=5$

$f(n)=f(n-1)+3$ for $n\geq 2$ .

It is the recursive formula of an arithmetic sequence with first term 5 and common difference 3. So, the explicit formula for this recursive formula is:

$f(n)=5+(n-1)(3)$

$f(n)=5+3(n-1)$

Therefore, the correct option is D, i.e., $f(n)=5+3(n-1)$ .

6 0

3 years ago

Read 2 more answers

A card is drawn at random from a standard pack of playing cards. Then a fair coin is flipped. What is the probability of selecti

serg [7]

Answer:

The order doesn’t matter, as the two events are probabilistically independent of one another. There are 13 spades in a standard 52 card deck, so the probability of drawing the number 5 is 13/52 or 1/4. The probability of heads is 1/2. The probability of both occurring is the product of the two probabilities, or 1/4*1/2 = 1/8 or 12.5%.

7 0

3 years ago

Read 2 more answers

HELPPP PRETYY PLEASEE

seropon [69]

Hello!

1) PEMDAS. Almost any expression ever

2) A coefficient is a numerical or constant quantity placed before and multiplying the variable in an algebraic expression.

3) 8.38889 yards

4) idk sorry :'(

5) X= 11

6) X = -5

7) 3(x + 5).

I hope it helps!

8 0

3 years ago