What is 16 to the one billionth power

John’s father is 5 times older than John and John is twice as old as his sister Alice. In two years time, the sum of their ages

oksano4ka [1.4K]

Answer:

John’s age is 8 years

Step-by-step explanation:

Let

x -----> John’s father's age

y -----> John’s age

z -----> John’s sister's age

we know that

$x=5y$ -----> equation A

$y=2z$

$z=y/2$ -----> equation B

$(x+2)+(y+2)+(z+2)=58$

$x+y+z+6=58$

$x+y+z=52$ -----> equation C

Substitute equation A and equation B in equation C and solve for y

$5y+y+y/2=52$

Multiply by 2 both sides to remove the fraction

$10y+2y+y=104$

$13y=104$

$y=8$

Find the value of x

$x=5(8)=40$

Find the value of z

$z=8/2=4$

therefore

John’s father's age is 40 years

John’s age is 8 years

John’s sister's age is 4 years

7 0

3 years ago

. CDEF is a trapezium

guajiro [1.7K]

9514 1404 393

Answer:

24 cm²

Step-by-step explanation:

To find the area of the trapezium, we must know the length CD. That means we must know the length BC. Fortunately, the perimeter of ABCF is given, so we have ...

P = 2(AB +BC)

BC = (P/2) -AB = (20 cm)/2 - 6 cm = 4 cm

Then CD is ...

CD = BD -BC = 9 cm -4 cm = 5 cm

The area of the trapezium is given by the formula ...

A = (1/2)(b1 +b2)h

A = (1/2)(5 cm + 3 cm)(6 cm) = 24 cm²

The area of trapezium CDEF is 24 cm².

3 0

2 years ago

3y + x = 12 What is the value of x when y = 3?

Andre45 [30]

Step-by-step explanation:

3y + x = 12

When y = 3, we have 3(3) + x = 12.

=> 9 + x = 12, x = 3.

6 0

2 years ago

Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.

finlep [7]

Answer:

a) $v = \frac{[L]}{[T]} = LT^{-1}$

b) $a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

c) $\int v dt = s(t) = [L]=L$

d) $\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

e) $\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

$v = \frac{ds}{dt}$

$a= \frac{dv}{dt}$

Part a

If we do the dimensional analysis for v we got:

$v = \frac{[L]}{[T]} = LT^{-1}$

Part b

For the acceleration we can use the result obtained from part a and we got:

$a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

Part c

From definition if we do the integral of the velocity respect to t we got the position:

$\int v dt = s(t)$

And the dimensional analysis for the position is:

$\int v dt = s(t) = [L]=L$

Part d

The integral for the acceleration respect to the time is the velocity:

$\int a dt = v(t)$

And the dimensional analysis for the position is:

$\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

$\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

7 0

3 years ago

Expand the expression 4(x-5)

Hunter-Best [27]

Answer:

4x-20

Step-by-step explanation:

5 0

3 years ago

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