All categories

4 years ago

Please help if been stuck on this problem for forever

Mathematics

1 answer:

Sedbober [7]4 years ago

6 0

Answer:

the answers are:x1=-4 and x2=5

You might be interested in

The current l, in a circuit varies inversely with the resistance, R. If the resistance is 12 when the current is 3, what is the

ivolga24 [154]

Answer:

Step-by-step explanation:

Given I varies inversely with R then the equation relating them is

I = $\frac{k}{R}$ ← k is the constant of variation

To find k use the condition R = 12 when I = 3

k = IR = 3 × 12 = 36

I = $\frac{36}{R}$ ← equation of variation

When I = 6, then

6 = $\frac{36}{R}$ , hence

R = $\frac{36}{6}$ = 6

4 0

3 years ago

Put 21 boxes into sets of 3 how many sets did you make. plzzz help me thank u

natta225 [31]

21/3 = 7
Division problem
You made 7 sets

7 0

3 years ago

Read 2 more answers

4 times the square of one number minus 5 times another number

iVinArrow [24]

Answer:

4x^2-5y

Step-by-step explanation:

4x^2-5y

6 0

3 years ago

The perimeter of a rectangle is 126 inches. The length is 7 less than 4 times the width. Find the dimensions of the rectangle. P

motikmotik

I hope someone solves your problem :)

8 0

3 years ago

A biologist observed that a certain bacterial colony obeys the population growth law and that the colony triples every 4 hours.

MrRa [10]

Answer:

a) $P(t) = 2e^{0.275t}$

b) 54.225 square centimeters.

c) 2.52 hours

Step-by-step explanation:

The population growth law is:

$P(t) = P_{0}e^{rt}$

In which P(t) is the population after t hours, $P_{0}$ is the initial population and r is the growth rate, as a decimal.

In this problem, we have that:

The colony occupied 2 square centimeters initially, so $P_{0} = 2$

The colony triples every 4 hours. So

$P(4) = 3P_{0} = 6$

(a) An expression for the size P(t) of the colony at any time t.

We have to find the value of r. We can do this by using the P(4) equation.

$P(t) = P_{0}e^{rt}$

$6 = 2e^{4r}$

$e^{4r} = 3$

Applying ln to both sides, we get:

$4r = 1.1$

$r = 0.275$

$P(t) = 2e^{0.275t}$

(b) The area occupied by the colony after 12 hours.

$P(t) = 2e^{0.275t}$

$P(12) = 2e^{0.275*12}$

$P(12) = 54.225$

(c) The doubling time for the colony?

t when $P(t) = 2P_{0} = 2*2 = 4$ .

$P(t) = 2e^{0.275t}$

$4 = 2e^{0.275t}$

$e^{0.275t} = 2$

Applying ln to both sides

$0.275t = 0.6931$

$t = 2.52$

4 0

3 years ago

Please help if been stuck on this problem for forever​

Please help if been stuck on this problem for forever