All categories

2 years ago

The sum of two consecutive integers is one less than three times the smaller integer. Find the two integers

Mathematics

2 answers:

In-s [12.5K]2 years ago

6 0

Hello!

Let's represent these integers with L and s, for the large and small integers.

L=s+1

L+s=3s-1

Let's plug our L-value into the second equation and solve for s.

s+1+s=3s-1

We combine like terms.

2s+1=3s-1

We subtract 3s from both sides.

-s+1=-1

We subtract 1 from both sides.

-s=-2

We divide both sides by -1.

s=2

As they are consecutive numbers, the larger number would be 3.

Therefore, our integers are 2 and 3.

I hope this helps!

Ulleksa [173]2 years ago

4 0

$n+n+1=3n-1\\\\
2n+1=3n-1\\\\
3n-2n=1+1\\\\
n=2\\
n+1=3$

You might be interested in

Sitting at the edge of a cliff, you throw a stone vertically upward with a velocity of 30 m/s. The height of the stone above the

Oksanka [162]

Answer:

Step-by-step explanation:

Given that,

The function f(t) = 3t − 0.5t² is the height of the motion of the stone in 10m units

And it's initial velocity is 30m/s

u=30m/s

Using calculation

Let find it turning point

f(t) = 3t − 0.5t²

Then, f'(t) =3-t

f(t)=0

So, 0=3-t

Then t=3secs

So let know the inflection point to show if t=3secs is the maximum point or minimum point

Then, we need second integral of f(t)

f'(t) = 3 − t

f''(t)=-1

Since f''(t) is negative this shows that the point is the maximum point

So at t=3secs is the maximum point,

So now to know the maximum point

f(t) = 3t − 0.5t²

t=3secs

f(t)=3(3)-0.5(3)²

f(t)=9-4.5

f(t)=4.5m

Since f(t) is in 10m units

Then,

The height is f(t) =10×4.5

f(t) =45m

Then the maximum height of the stone is 45m and the time to reach the maximum height is t=3secs

Now, Using equation of motion

Time to reach max height.

v=u-gt. Upward motion g=9.81m/s2

Final velocity is v=0m/s

0=30-9.81t

-30=-9.81t

t=-30/-9.81

t=3.06secs

t=3.1secs. To 1d.p

If we have used g=10m/s², it will have the same value as the graph and the maximum and minimum point calculation

Maximum height, using equation of motion

v²=u²-2gH

0²=30²-2×9.81H

-30²=-19.62H

Then,

H=-30²/-19.62

H=45.87

a. From the graph it shows that the maximum value of the is 4.5m at time t=3secs

b. Now, the maximum height of stone as shown on the graph is 4.5m

And since is 10 meter unit scale

Then the height becomes 4.5×10

Maximum height is 45m.

c. The stone will spend two times the time it uses to reach t maximum height to return to the height of the cliff

Time to return back to the cliff is

T=2t

T=2×3

T=6secs

d. If the stone hits the ground after t=7secs

So after the ball reached a maximum height of 45m, he has spent 3secs, so we can calculate the distance the stone travelled from the maximum to the ground

The time is of travel will be 4secs

Then,

Initial velocity is 0, from the point of return at rhe maximum height

S=ut+½gt²

S=0+½×9.81×4²

S=78.48m

So the total distance from the maximum height down to the bottom of the cliff is 78.48m

Then, the height of the cliff is the height from the maximum height to the bottom of the cliff minus the maximum height reach by the stone

Then,

H(cliff)=78.48-45.87

H(cliff)=33.48m

From the graph

At t=7secs

The stone is at a distance of -3.5m

Showing that the height of the cliff

Since it is of 10m units

Then the height of the cliff is

3.5×10=35m

Also

f(t) = 3t − 0.5t² at t=7secs

F(7)=3(7)-0.5(7)²

F(7)=21-0.5×49

F(7)=21-24.5

Then,

f(7)=-3.5m

This shows the downward motion

Since it is 10m unit

The the height of cliff downward is

3.5×10=35m

Height of cliff =35m

3 0

2 years ago

The cost of taking your pet aboard the air flight with you in the continental US varies according to the airlines. The five numb

melisa1 [442]

Answer: The lowest value: 100 and highest value: - 150 .

If we were to build the box plot for this data, the box would stretch between $Q_1= 100$ and $Q_3= 125$ .

Step-by-step explanation:

We know that the box-plot is the graphical way to represent the five -number summary (Minimum value , First quartile $(Q_1)$ , Median , Third Quartile $(Q_3)$ , Maximum value).