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2 years ago

Please help me Im lost on this :(

Mathematics

2 answers:

kkurt [141]2 years ago

7 0

Answer:

Step-by-step explanation:

the quadratic formula is

$\frac{ - b + \sqrt{b ^{2} - 4ac} }{2a}$

Make the quadratic equal to 0, so it will be 3x^2 + 8x -2=0. Then substitute the values into the equation.

xeze [42]2 years ago

6 0

Answer: C) 3x^2+8x-10=-8

Step-by-step explanation:

3x^2+8x-10x=-8

3x^2+8x-2=0

a=3 b=8 c=-2

Plug it into quadratic formula.

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Can someone explain this please???

xeze [42]

A function is a rule that assigns exactly one output to a given input. The input is taken from a set called the domain, and the corresponding output belongs to a set called the range.

1. In this exercise, we're calling the pool of patients 1-8 the domain, and the pool of nurses A-D the range. The given table describes a function because any patient is assigned to only one nurse.

2. This wouldn't be a function if at least one patient was assigned to more than one nurse. If this were to happen in practice, the patient could be, say, given the same dose of some medicine twice if the nurses aren't careful.

3. Making the nurse pool the domain and the patient pool the range would give a relation that is not a function, since more than one patient is assigned to one nurse.

4 0

3 years ago

What is the perimeter of the shaded triangle

astraxan [27]

What units are you using

8 0

3 years ago

Create your own scenario for a system of equations and solve it. Please do not just write two equations, but apply what you have

BARSIC [14]

Answer:

Dr. Potter provides vaccinations against polio and measles.

Each polio vaccination consists of 444 doses, and each measles vaccination consists of 222 doses. Last year, Dr. Potter gave a total of 606060 vaccinations that consisted of a total of 184184184 doses.

How many polio vaccinations and how many measles vaccinations did Dr. Potter give last year?

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Mr White can travel for 6 hours while taking 3 breaks of 10 minutes each. Once, he had to travel for 36 hours. How many minutes

alexgriva [62]

180 min

Explanation:

3 × 10 min = 30 min

In 6 hours, Mr. White takes a total of 30 min in breaks.

36 h = 6 h × 6

In 36 hours , there are 6 sets of 6 h .

Hence we can multiply the number of minutes in breaks taken in

6 h ours by 6 to obtain the number of minutes in breaks taken in 36 h ours.

30 min × 6 = 180 min

5 0

3 years ago

A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.

Usimov [2.4K]

Answer:

a) $\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

b) $v(t=3) = 3(3)^2 -18(3) +15=-12$

c) $t =1s, t=5s$

d) $[0,1) \cup (5,\infty)$

e) $D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

f) $a(t) = \frac{dv}{dt}= 6t -18$

g) The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

Step-by-step explanation:

For this case we have the following function given:

$f(t) = s = t^3 -9t^2 +15 t$

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

$\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

$v(t=3) = 3(3)^2 -18(3) +15=-12$

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

$3t^2 -18 t +15 =0$

We can divide both sides of the equation by 3 and we got:

$t^2 -6t +5=0$

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

$(t-5)*(t-1) =0$

And for this case we got $t =1s, t=5s$

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

$t^2 -6t +5 >0$

$(t-5) *(t-1)>0$

So then the intervals positive are $[0,1) \cup (5,\infty)$

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

$D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6$

And if we replace we got:

$D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

$a(t) = \frac{dv}{dt}= 6t -18$

Part g and h

The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

5 0

3 years ago

Please help me Im lost on this :(​

Please help me Im lost on this :(