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scoundrel [369]

3 years ago

5

Find three consecutive odd positive integers such that 5 times the sum of all three is 72more than the product of the first and

second integers.

1 answer:

Alenkasestr [34]3 years ago

8 0

Let the positive odd numbers be x,x+2,x+4.
According to the question,
5(3x+30)=72+x^2+4x
x^2-13x+42=0
(x-6) and (x-7)=0
x=6 or x=7

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Pls help me i need help

Llana [10]

Answer:

3x was subtracted from the left side, but 3x was subtracted from the right side. The Subtract Property of Equality states that you can subtract the same number from each side and the equation will remain true. But 3x and 3 are not the same number (unless x is 1).

x = -8/5

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

$(7x+3x)+(21)=5$
$10x + 21 = 5$

Step 2: Subtract 12 from both sides.

$10x + 21 - 21 = 5 - 21$
$10x = -16$

Step 3: Divide both sides by 10.

$\frac{10x}{10} = \frac{-16}{10}$
$x = -\frac{8}{5}$

8 0

3 years ago

Read 2 more answers

The library is 1.75 miles directly north of the school. The park is 0.6 miles directly south of the school. How far is the libra

kumpel [21]

If something is placed the way these places are(directly NORTH and the park is SOUTH) all you have to do is add your two values, lining up your decimal points. 1.75
+.6
2.35
Let me know if this helps by making me your brainliest answer

4 0

3 years ago

Read 2 more answers

Prove that the statement (ab)^n=a^n * b^n is true using mathematical induction.

mamaluj [8]

Answer:

see below

Step-by-step explanation:

(ab)^n=a^n * b^n

We need to show that it is true for n=1

assuming that it is true for n = k;

(ab)^n=a^n * b^n

( ab) ^1 = a^1 * b^1

ab = a * b

ab = ab

Then we need to show that it is true for n = ( k+1)

or (ab)^(k+1)=a^( k+1) * b^( k+1)

Starting with

(ab)^k=a^k * b^k given

Multiply each side by ab

ab * (ab)^k= ab *a^k * b^k

( ab) ^ ( k+1) = a^ ( k+1) b^ (k+1)

Therefore, the rule is true for every natural number n

5 0

3 years ago

Read 2 more answers

(5) Find the Laplace transform of the following time functions: (a) f(t) = 20.5 + 10t + t 2 + δ(t), where δ(t) is the unit impul

Aloiza [94]

Answer

(a) $F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}$

(b) $F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}$

Step-by-step explanation:

(a) $f(t) = 20.5 + 10t + t^2 +$ δ(t)

where δ(t) = unit impulse function

The Laplace transform of function f(t) is given as:

$F(s) = \int\limits^a_0 f(s)e^{-st} \, dt$

where a = ∞

=> $F(s) = \int\limits^a_0 {(20.5 + 10t + t^2 + d(t))e^{-st} \, dt$

where d(t) = δ(t)

=> $F(s) = \int\limits^a_0 {(20.5e^{-st} + 10te^{-st} + t^2e^{-st} + d(t)e^{-st}) \, dt$

Integrating, we have:

=> $F(s) = (20.5\frac{e^{-st}}{s} - 10\frac{(t + 1)e^{-st}}{s^2} - \frac{(st(st + 2) + 2)e^{-st}}{s^3} )\left \{ {{a} \atop {0}} \right.$

Inputting the boundary conditions t = a = ∞, t = 0:

$F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}$

(b) $f(t) = e^{-t} + 4e^{-4t} + te^{-3t}$

The Laplace transform of function f(t) is given as:

$F(s) = \int\limits^a_0 (e^{-t} + 4e^{-4t} + te^{-3t} )e^{-st} \, dt$

$F(s) = \int\limits^a_0 (e^{-t}e^{-st} + 4e^{-4t}e^{-st} + te^{-3t}e^{-st} ) \, dt$

$F(s) = \int\limits^a_0 (e^{-t(1 + s)} + 4e^{-t(4 + s)} + te^{-t(3 + s)} ) \, dt$

Integrating, we have:

$F(s) = [\frac{-e^{-(s + 1)t}} {s + 1} - \frac{4e^{-(s + 4)}}{s + 4} - \frac{(3(s + 1)t + 1)e^{-3(s + 1)t})}{9(s + 1)^2}] \left \{ {{a} \atop {0}} \right.$

Inputting the boundary condition, t = a = ∞, t = 0:

$F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}$

3 0

3 years ago

One warehouse had 185 tons of coal, another one had 237 tons. The first warehouse delivered 15 tons of coal to its clients daily

sdas [7]

Given

One warehouse had 185 tons of coal, another one had 237 tons

The first warehouse delivered 15 tons of coal to its clients daily, the second one delivered 18 tons of coal to its clients daily.

Find out in how many days will the second warehouse have 1.5 times more coal than the first one.

To proof

Let us assume that the number of days of delivered the coal = x

Take two cases

First case

As given in the question

One warehouse had coal = 185 tons

The first warehouse delivered 15 tons of coal to its clients daily

(As first warehouse delivered 15 tons of coal to its clients daily than the total amount of coal decrease daily as days increase)

Than the equation becomes for first warehouse

= ( 185 - 15x)

( this equation represent the amount of coal left after x days)

Second case

second warehouse had coal = 237 tons

the second one delivered 18 tons of coal to its clients daily.

(As first warehouse delivered 18 tons of coal to its clients daily than the total amount of coal decrease daily as days increase)

Than the equation becomes for the second warehouse

= ( 237 - 18x)

( this equation represent the amount of coal left after x days)

As given in the question

the second warehouse have 1.5 times more coal than the first one

Thus

( 237 -18x ) = 1.5× ( 185 - 15x)

237 - 18x = 277.5 - 22.5x

22.5x - 18x = 277.5 - 237

4.5x = 40.5

x = 9

Therefore the second warehouse have 1.5 times more coal than the first one in 9 days.

Hence proved

7 0

4 years ago