Help what do I do do i just write y-3x+5 or solve it

Mathematics

2 answers:

bixtya [17]2 years ago

4 0

Answer:

y = -3x +5

Step-by-step explanation:

In general, do what the question asks you to do: write the equation of the line.

The slope and y-intercept of the line are given, so the easiest, most direct way to write the equation is to use "slope-intercept form." That form looks like ...

y = mx + b . . . . . . where m is the slope, b is the y-inercept

You are given m = -3 and b = 5, so the equation you are asked to write can be ...

y = -3x +5 . . . . . . this equation is the answer to the question

Additional comment

There are many forms of the equation for a line. Another one in common use for problems like this is the "point-slope form." You would use that one for a simple, direct answer to writing an equation for a line with a given slope through a given point.

y -k = m(x -h) . . . . . . . line with slope m through point (h, k)

If you are given the x- and y-intercepts, you can use "intercept form" to write the equation:

x/a +y/b = 1 . . . . . for x-intercept 'a' and y-intercept 'b'

Another useful form is "standard form", which has ...

ax +by = c . . . . where a > 0, and a, b, c are mutually prime

For example, the equation for your line could be written in standard form as ...

3x +y = 5

Rina8888 [55]2 years ago

3 0

Answer:

y=-3x+5

Step-by-step explanation: Yes you have it correct. The question is asking for you to put it into an equation that fits the slope and y-intercept which is slope-intercept form. Additional note, when you have a problem like this and it doesn't say what type of equation form to put it in, put it in slope-intercept form for simplicity's sake.

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2. Harry just deposited $1500 into a savings account giving 6% interest compounded quarterly.a) How much will be in the account

Studentka2010 [4]

The formula for compound interest is:

$\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ \text{where,} \\ A=\text{ Final amount} \\ r=\text{ Interest rate} \\ n=\text{ Number of times interest applied per period} \\ t=\text{ Number of time period elapsed} \\ P=\text{ Intial principal balance} \end{gathered}$

Given data:

$\begin{gathered} P=\text{ \$1500} \\ r=6\text{ \%}=0.06 \\ n=4\text{ times (compounded quarterly)} \end{gathered}$

a. After ten years, that is t = 10 years, the amount in the account will be

$\begin{gathered} A=1500(1+\frac{0.06}{4})^{4\times10} \\ A=\text{ }1500(1+0.015)^{40} \\ A=\text{ }1500(1.015)^{40} \\ A=\text{ \$2721.03} \end{gathered}$

b. After twenty years, that is t = 20 years, the amount in the account will be:

$\begin{gathered} A=1500(1+\frac{0.06}{4})^{4\times20} \\ A=1500(1.015)^{4\times20} \\ A=1500(1.015)^{80} \\ A=\text{ \$}4935.99 \end{gathered}$

c. The time it takes for Harry's initial account value to double will be:

$\begin{gathered} A=2\text{ x initial value = 2 }\times\text{ \$1500 = \$3000} \\ 3000=1500(1.015)^{4t} \\ (1.015)^{4t}=\frac{3000}{1500} \\ (1.015)^{4t}=2 \\ \text{ Find the logarithm of both sides} \\ \ln (1.015)^{4t}=\ln 2 \\ 4t=\frac{\ln 2}{\ln 1.015} \\ 4t=46.56 \\ t=\frac{46.56}{4}=11.64 \end{gathered}$

Therefore, the time it takes Harry's initial account to double is approximately 11 years

8 0

1 year ago

5y + 3 = 8y − 5 + 2y

DIA [1.3K]

Answer:

$\fbox{y = 8/5 }$