I need help ASAP ill mark u a brainiest i need an answer thats it

What is the equation, in point-slope form, of the line that is parallel to the given line and passes through the point (−3, 1)?

UkoKoshka [18]

The equation of the line parallel to given line and passing through (-3, 1) will be y – 1 = (3/2)(x + 3). Then the correct option is D.

<h3>What is the equation of line?</h3>

The equation of line is given as

y = mx + c

Where m is the slope and c is the y-intercept.

The slope of the parallel lines are equal.

m = (2 + 4) / (2 + 2)

m = 6/4

m = 3/2

Then the equation of the line will be

y = (3/2)x + c

The line is passing through (-3, 1). Then the value of c will be

1 = (3/2)(-3) + c

c = 9/2 + 1

Then the equation will be

y = (3/2)x + 9/2 + 1

y – 1 = (3/2)(x + 3)

Then the correct option is D.

More about the equation of line link is given below.

brainly.com/question/21511618

#SPJ1

5 0

2 years ago

What is the value of c in the equation? 15.3 = StartFraction c Over 2.1 EndFraction 13.2 17.4 32.13 321.3

harina [27]

Answer:

32.13

Step-by-step explanation:

15.3= c/2.1

c= 15.3* 2.1

c= 32.13

3 0

4 years ago

Read 2 more answers

Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (naut

frozen [14]

Answer:

a) $\sqrt{144-288t+208t^2}$ b.) -12knots, 8 knots c) No e) $4\sqrt{13}$

Step-by-step explanation:

We know that the initial distance between ships A and B was 12 nautical miles. Ship A moves at 12 knots(nautical miles per hour) south. Ship B moves at 8 knots east.

a)

We know that at time t , the ship A has moved $12\dot t (n.m)$ and ship B has moved $8\dot t (n.m)$ . We also know that the ship A moves closer to the line of the movement of B and that ship B moves further on its line.

Using Pythagorean theorem, we can write the distance s as:

$\sqrt{(12-12\dot t)^2 + (8\dot t)^2}\\s=\sqrt{144-288t+144t^2+64t^2}\\s=\sqrt{144-288t+208t^2}$

b)

We want to find $\frac{ds}{dt}$ for t=0 and t=1

$\sqrt{144-288t+208t^2}|\frac{d}{dt}\\\\\frac{ds}{dt}=\frac{1}{2\sqrt{144-288t+208t^2}}\dot (-288+416t)\\\\\frac{ds}{dt}=\frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\\frac{ds}{dt}(0)=\frac{208\dot 0-144}{\sqrt{144-288\dot 0 + 209\dot 0^2}}=-12knots\\\\\frac{ds}{dt}(1)=\frac{208\dot 1-144}{\sqrt{144-288\dot 1 + 209\dot 1^2}}=8knots$

c)

We know that the visibility was 5n.m. We want to see whether the distance s was under 5 miles at any point.

Ships have seen each other = $s\leq 5\\\\\sqrt{144-288t+208t^2}\leq 5\\\\144-288t+208t^2\leq 25\\\\199-288t+208t^2\leq 0$

Since function $f(x)=199-288x+208x^2$ is quadratic, concave up and has no real roots, we know that $199-288x+208x^2>0$ for every t. So, the ships haven't seen each other.

d)

Attachedis the graph of s(red) and ds/dt(blue). We can see that our results from parts b and c were correct.

e)

Function ds/dt has a horizontal asympote in the first quadrant if

$\lim_{t \to \infty} \frac{ds}{dt}$

So, lets check this limit:

$\lim_{t \to \infty} \frac{ds}{dt}=\lim_{t \to \infty} \frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\=\lim_{t \to \infty} \frac{208-\frac{144}{t}}{\sqrt{\frac{144}{t^2}-\frac{288}{t}+208}}\\\\=\frac{208-0}{\sqrt{0-0+208}}\\\\=\frac{208}{\sqrt{208}}\\\\=4\sqrt{13}$

Notice that:

$4\sqrt{13}=\sqrt{12^2+5^2}$ =√(speed of ship A² + speed of ship B²)

5 0

4 years ago

You are choosing between two health clubs. Club A offers membership for a fee of $ 28$28 plus a monthly fee of $ 20.$20. Club B

eimsori [14]

Answer:

That is to say that at 3 months, the values of the club are equal.

Club A = 268 dollars in total

Club B = 295 dollars in total

Step-by-step explanation:

The first thing is to propose an equation for each club. Let m be the number of months.

Club A = 28 + 20 * m

Club B = 19 + 23 * m

Now, to know when it is the same, we must equal the value of Club A with Club B, it would look like this

28 + 20 * m = 19 + 23 * m

By rearranging the equation we have:

23 * m - 20 * m = 28-19

3 * m = 9

m = 3

That is to say that at 3 months, the values of the club are equal.

Now to know the total value, we will do it with an annual membership, which would be a total of 12 months, therefore it is to replace in the equation of each one:

Club A = 28 + 20 * 12 = 268 dollars in total

Club B = 19 + 23 * 12 = 295 dollars in total

3 0

3 years ago

Noone is helping with my questions and my mom said if I don't finish all my work I can't go to the amusement park :(

n200080 [17]

Answer:

how can i help you

Step-by-step explanation:

i will help

6 0

3 years ago