A boy's mother was 32 years old when he was born. Let a represent the boy's age and m represent the mother's age.

Write 5 and 2/5 as a decimal

Alenkasestr [34]

Answer:

5.4.

Step-by-step explanation:

If we divide 2 by 5 we get 0.4

So 5 and 2/5 = 5.4

7 0

4 years ago

Read 2 more answers

Two large and 1 small pumps can fill a swimming pool in 4 hours. One large and 3 small pumps can also fill the same swimming poo

kolezko [41]

Let x and y be the unit rates at which one large pump and one small pump works, respectively.

Two large/one small operate at a unit rate of

(1 pool)/(4 hours) = 0.25 pool/hour

so that

2x + y = 0.25

One large/three small operate at the same rate,

(1 pool)/(4 hours) = 0.25 pool/hour

x + 3y = 0.25

Solve for x and y. We have

y = 0.25 - 2x ==> x + 3 (0.25 - 2x) = 0.25

==> x + 0.75 - 6x = 0.25

==> 5x = 0.5

==> x = 0.1

==> y = 0.25 - 2 (0.1) = 0.25 - 0.2 = 0.05

In other words, one large pump alone can fill a 1/10 of a pool in one hour, while one small pump alone can fill 1/20 of a pool in one hour.

Now, if you have four each of the large and small pumps, they will work at a rate of

4x + 4y = 4 (0.1) + 4 (0.05) = 0.6

meaning they can fill 3/5 of a pool in one hour. If it takes time t to fill one pool, we have

(3/5 pool/hour) (t hours) = 1 pool

==> t = (1 pool) / (3/5 pool/hour) = 5/3 hours

So it would take 5/3 hours, or 100 minutes, for this arrangement of pumps to fill one pool.

6 0

3 years ago

The total stopping distance T for a vehicle is T = 2.5x + 0.5x 2 , where T is in feet and x is the speed in miles per hour. Appr

Kobotan [32]

Answer:

% change in stopping distance = 7.34 %

Step-by-step explanation:

The stooping distance is given by

$T = 2.5 x + 0.5 x^{2}$

We will approximate this distance using the relation

$f (x + dx) = f (x)+ f' (x)dx$

dx = 26 - 25 = 1

T' = 2.5 + x

Therefore

$f(x)+f'(x)dx = 2.5x+ 0.5x^{2} + 2.5 +x$

This is the stopping distance at x = 25

Put x = 25 in above equation

2.5 × (25) + 0.5× $25^{2}$ + 2.5 + 25 = 402.5 ft

Stopping distance at x = 25

T(25) = 2.5 × (25) + 0.5 × $25^{2}$

T(25) = 375 ft

Therefore approximate change in stopping distance = 402.5 - 375 = 27.5 ft

% change in stopping distance = $\frac{27.5}{375}$ × 100

% change in stopping distance = 7.34 %

6 0

3 years ago

Idk how to do this 4x+y=1

ivolga24 [154]

4x+y=1
4x+y-1=1-1
4x+y-1=0

6 0

3 years ago

6ab + a(a – b – 6b) + 4a(2b +a) – ab , para a= -1 e b=2 ME AJUDEM! É URGENTE

Marianna [84]

Answer:

-81

Step-by-step explanation:

replace the numbers you were given in a or b.

so

6(-12) -1(-1 – 2 – 6(2)) + 4(-1)(2(2) -1) -12

6 0

3 years ago